Martin Flashman's Courses
MATH 344 Linear Algebra Fall, 2011
Class Notes and Summaries
 WEEK Monday Wednesday Friday week1 8-22 8-24 8-26 week 2 8-29 8-31 9-2 week 3 9-5 NO Class Labor Day 9-7 9-9 week 4 9-12 9-14 9-16 week 5 9-19 9-21 9-23 week 6 9-26 9-28 9-30 week 7 10-3 10-5 10-7 week 8 10-10 10-12 10-14 week 9 10-17 10-19 10-21 week 10 10-24 10-26 10-28 week 11 10-31 11-2 11-4 week 12 11-7 11-9 11-11 week 13 11-14 11-16 11-18 week 14 NoClasses 11-21 11-23 11-27 week 15 11-28 11-30 12-2 week 16 12-5 12-7 11-9 week 17 Final Exam 12-12 Exam

[Old notes]
8-22 discussion of course organization. [See syllabus on line. Background Assessment on Moodle ]
• Other texts in Linear Algebra:

• Finite Dimensional Vector Spaces by Paul Halmos
Linear algebra by Kenneth Hoffman and Ray Kunze.
Linear Algebra by Friedberg, Insel, and Spence
Linear Algebra by  Lang
Applied linear algebra by Ben Noble.
Linear Algebra and Its Applications by G. Strang
Topics in Abstract Algebra by I. Herstein
HSU Library Holdings
• On Proofs:

• How to Read and Do Proofs by D.Solow
The Keys to Advanced Mathematics by D. Solow
How to Solve It by G. Polya

• Motivational Question I:
• What can we say about powers of a square 2 by 2 matrix A and AX , where X is `(x,y)^{tr}`.
• What is the geometric interpretation of AX when X is  `(1,0)^{tr}` or `(0,1)^{tr}`?
• Examples:

A= (
 0 1 1 0

)
`A^n = I `  if `n` is even
`A^n =A` if `n` is odd.

A= (
 1 1 0 1

)

`A^n `  =
(
 1 n 0 1

)
Prove? [Use induction]

A  =
(
 r 0 0 s

)

1. r = 1/2;  s= 1/2   Discussion:  Same if  0< |r|, |s| < 1

`A^n`  `->`
(
 0 0 0 0

)
2. r= 2, s = 1/2  Discussion: ....
3. r= 1, s = -1   Discussion: `A^n = I `  if `n` is even; `A^n =A` if  `n `is odd

•   Using technology - Find the eigenvalues for

A= (

 0 1 1 0

)

8-29Complex Numbers C
Visualize complex numbers in a plane.
`||z|| = sqrt(a^2 + b^2) `
z = ||z||(cos(t) +i sin(t))    [called the "polar form" of the complex number]

The arithmetic of C.

The geometry of complex arithmetic:

If z = a+bi = ||z||(cos(t) +i sin(t)) and w  = c+di = ||w||(cos(s) +i sin(s)) then

z+w = (a+c)+(b+d)i which corresponds geometrically to the "vector " sum  of z and w in the plane, and

zw = ||z||(cos(t) +i sin(t)) ||w||(cos(s) +i sin(s))=  ||z|| ||w|| (cos(t) +i sin(t))(cos(s) +i sin(s))
= ||z|| ||w|| (cos(t) cos(s) - sin(t)sin(s) + (sin(t) cos(s) + sin(s)cos(t)) i)
= ||z|| ||w|| (cos(t+s) + sin(t+s) i)

So the magnitude of the product is  the product of the magnitudes of z and w   and the angle of the product is the sum of the angles of z and w.

Powers of complex numbers.
Based on the previous result on products,
`z^n =  ||z||^n  (cos(nt) + sin(nt) i) `
If ||z||<1, then the powers of z will spiral into 0.
If ||z||>1, then the powers of z will spiral outward without bound.
If ||z||=1 and z is not 1, then the powers of Z will oscillate around the circle of radius 1 without having any limit.

Notation: cos(t) + i sin(t) is sometimes written as cis(t).
Note: If we consider the series for ex = 1 + x + x2/2! +x3/3! + ...
then eix = 1 + ix + (ix)2/2! +(ix)3/3! + ... = 1 + ix -  x2/2! - ix3/3! + ...
... = cos(x) + i sin(x)
Thus `e^{i*pi} = cos(pi) + i sin(pi)= -1`; and so.... `ln(-1) = i *pi` because the definition of `ln` says `ln(a)=b` if and only if `e^b = a`
Furthermore: `e^{a+bi} = e^a*e^{bi} = e^a ( cos (b) + sin(b) i) `

8-31
Matrices with complex number entries.

A  =
(
 r 0 0 s

)

If r and s are complex numbers in the matrix A, then consider what happens to `A^n` as n gets large:
If ||r|| < 1 and ||s|| < 1, the powers of A will get close to the zero matrix;
if r=s=1 the powers of A will always be A.
If ||r|| < 1 and s = 1 the powers of A will get close to the matrix

A  =
(
 0 0 0 1

)

and if ||s|| < 1 and r = 1 the powers of A will get close to the matrix

A  =
(
 1 0 0 0

)

and otherwise the powers of A will diverge .

Polynomials with complex coefficients.

Because multiplication and addition make sense for complex numbers, we can consider polynomials with coefficients that are complex numbers and use a complex number for the variable
This makes a complex polynomial a function from the complex numbers to the complex numbers.

This can be visualized using one plane for the domain of the polynomial and a second plane for the co-domain/ target range of the polynomial.

The Fundamental Theorem of Algebra: If f is a non constant  polynomial with complex number coefficients then there is at least on complex number z*  where f (z*) = 0.

For more on complex numbers see: Dave's Short Course on Complex Numbers,

Fields: The structure used for "solving linear equations," finding inverses for matrices, and doing other parts of matrix algebra. The set of "scalars" for vectors.

Definition- Axioms See Fields.

9-2
Examples:
• Q = { x = m/n where m,n are integers and n is not 0.}
• Consider various of the field axioms for Q using x= m/n and y=r/s, x+y = [ms+nr]/[ns] and x*y= [mr]/[ns].
• Equality of rational numbers is important.  m/n=m'/n'  if and only if mn'=m'n.
• 0 = 0/1; 1 = 1/1. -(m/n) = (-m)/n.  Multiplicative inverse of m/n : n/m with m not 0.
• Other properties are tedious to verify.
•  R: infinite decimals, sequences of rational numbers. This is a topic best discussed in a course n Real Analysis. We will accept that R is a field.
•  C= { z= x+ yi: x, y are real numbers} discussed previously... as afield.
• F2=Z2 = {0,1} Smallest possible field. Important for computers. 1+1 =0 is key feature.
• Fp =Zp, where p is a prime number = [ 0,1,2,...., p-1} has p elements. (p-1) + 1= 0. All other addition and multiplication facts developed from this and multiplication as repeated addition.
• F4 : A FIELD WITH 4 ELEMENTS - KEY- 1+1= 0 So other elements are not called 2  and 3 but `alpha` and `beta` and `alpha +1 = beta`.
9-7
More Examples of fields:
• {`f `  where `f `is a rational function defined on all but a finite number of real numbers}
• algebraic numbers = {z: z is a complex number which is the root of a polynomial with integer coefficients}.
Polynomials with coefficients in a field. [We will study these in some detail later].
Matrices with entries in a field.

• A few more properties of Fields:
• Suppose F is a field.
• If  a and b are in F and ab = 0 then either a =0 or b=0.

• Proof: Case 1. If a=0 then we are done.
Case 2. If a is not 0, then a has an inverse... c where ca=1. Then b=1b= (ca)b=  c(ab) = c0 = 0.
Thus either  a =0 or b=0. IRMC   [I rest my case.]
• If  a, b and c are in F a+b = a+c implies b=c.

• Proof: Let k be the element of the field where k + a = 0 Then
b = 0 + b = (k +a)+b =k +(a+b)= k +(a+c) =(k +a)+c = 0 + c = c.
• If  a, b and c are in F with  a not equal to 0 ab = ac implies b=c.

• Proof: Similar to the last result.
• Let  n·1 stand for  1 + 1 + 1 + ... + 1 with n summands. Either (i) for all n,  n·1 is not 0 in which case we say the field has "characteristic 0" , or (ii) for some n, n·1=0. In the case n·1=0, there is a smallest n for which n·1=0, in which case we say the field has "characteristic n". For example: R, Q, and C all have characteristic 0, while Z2, Zp, where p is a prime number, and any finite field such as  F4, all have a non zero characteristic.
• If the characteristic of F is not zero, then it is a prime number.

• Proof: If n is not a prime, n = rs with 1<r,s<n. Let a = 1+1+...+1 r times and b= 1+1+...+1 s times. Then ab=n·1=0, so either a = 0 or b = 0, contradicting the fact the n was supposed to be the SMALLEST natural number for which n·1=0. IRMC.

• Solving equations with fields:
Suppose F is a field and `alpha` and `beta` are elements of F and X is a variable. We say that the equation `alpha` X = `beta` has a solution in F if there is an element of F, `gamma` where `alpha gamma = beta`.
Proposition: If `alpha \ne 0` then  there is an element of F, `gamma`,  where `alpha gamma = beta`.
Preface: We did some work on the side to "solve" the equation.
This analysis did the work of finding the candidate for `gamma` and seeing it was in F.
Proof: Suppose `alpha  in F` and `alpha  \ne 0.`
Then by the field axioms, there is an element of F, `alpha^{-1}` where `alpha^{-1} alpha = 1`.
Let  `gamma = alpha^{-1} beta`. Then `alpha gamma = alpha (alpha^{-1} beta) = (alpha alpha^{-1}) beta = 1 beta = beta`. IRMC!
• Systems of linear equations:
One linear equation with n unknowns with coefficients in a field F:
`alpha_1,alpha_2,...,alpha_n , and beta in F`
`alpha_1X_1 + alpha_2X_2 + ... + alpha_nX_n = beta `
k linear equations with n unknowns with coefficients in a field F:
`alpha_{11}X_1 + alpha_{12}X_2 + ... + alpha_{1n}X_n = beta_1 `
`alpha_{21}X_1 + alpha_{22}X_2 + ... + alpha_{2n}X_n = beta _2`
.
.
.
`alpha_{k1}X_1 + alpha_{k2}X_2 + ... + alpha_{kn}X_n = beta _k`
The techniques learned in your previous linear algebra course can still be used for F since those techniques relied solely on the arithmetic that makes sense in any field.

9-9
• Motivation Question II

• Coke or Pepsi?
• Initial Sample: out of 1000 people.... 600 liked coke, 400 liked Pepsi. ... v0 = (600,400)

• A. First re sample: 500 of 600 stayed loyal to Coke, 100 switched to Pepsi.
300 of 400 stayed loyal to Pepsi , 100 switched to Coke.  so  v1 = (600,400)
• B. Alternative Re sample: 400 of 600 stayed loyal to Coke, 200 switched to Pepsi.
•                       300 of 400 stayed loyal to Pepsi , 100 switched to Coke.  so  v1 = (500,500)

• Example A.
`v_0TA = v_1` where

TA= (
 5/6 1/6 1/4 3/4

)

Example B.
`v_0TB = v_1` where

TB= (
 2/3 1/3 1/4 3/4

)

Questions: If the pattern in case B continues with the proportions of those switching remaining the same:
what can we say about vn  when n is large?
Does the distribution of vn when n is large depend on v_0?
Is there some initial distribution that would remain unchanged under the pattern of switching in B?
i.e. is there a distribution v*0 = (c*,p*)  where v*0 = v*1 =  ... = v*n.
Solve the equation:  (c*, 1000-c*) TB = (c*, 1000-c*).

How are these questions related to Motivation Question I?

v0Tn = vn.

Vectors and vector spaces.
• Traditional vectors: vectors in R2, R2, and Rn.

A list or an n-tuple. ["list" is language used in Axler and by computer science.]
Examples: (3,5) is a list of 2 real numbers,  (3,4,5,3,5) is a list of 5 real numbers. ((2,3,5,2), (2,3), ( )) is a list of 3 lists.
The length of a list is the number n of places for objects in the list.
If v and w are lists, we say v = w if v and w have the same length n, v = (v , v , ..., v ), w = (w ,w , ..., w ) and vi = w for each i, where i = 1,2,...,n.
Example. Visualize R2, R3, and R4.
• Definition: If S is a set, Sn = {lists of length n with entries from the set S}

• = { v = (v , v , ..., v ) where  vis a member of S for each i, where i = 1,2,...,n..}
• Definition: Suppose F is a field.  We define an internal operation on Fn+, and an external operation between  F and Fn, * , by the following:

• If v =  (v , v , ..., v ) and w =  (w ,w , ..., w ) are in Fn  and a is in F then v + w = (v + w1, v2 + w , ..., vn+w ) and
a*v =  (av , av , ..., av ) using the addition and multiplication of F for the operations on the individual elements of F.
• Proposition: With the operations just defined on F the following properties are true:
[This should be familiar for F = R from your previous linear algebra course.]
• If v, w, and z are in F and a and b are in F, then v+w and a*v are in F [Closure] ;
• (v+w)+z = v+(w+z)  [Associativity 1]
• If we let 0 denote (0,0,...0) then 0+v=v+0 = v [Additive identity]
• For each v, there is a list in Fn , v', so that v+v' = v' + v =0. [Additive inverse]
• v+w=w+v [commutativity]
• 1*v = v  [Scalar Identity]
• a*( v + w) = (a*v) + (a*w) [distributive 1]
• (a+b)*v = (a*v) + (b*v)    [distributive 2]
• (ab)*v = a*(b*v)   [associativity 2]

• 9-12
• Abstract Vector Spaces
• Definition: Suppose V is a non empty set, F is a field, and there is an internal operation on V, +, and an external operation between  F and V, * .

• We say that V is a vector space over F,  " V is a vs/F", and we call the elements of V, "vectors", if the following properties are true:
• If v,w,and z are in V  and a and b are in F, then v+w and a*v are in V  [closure] ;
• (v+w)+z = v+(w+z)  [Associativity 1]
• If there is an element of V, denoted 0 so that 0+v=v+0 = v [Additive identity]
• For each v, there is an element of V, v', so that v+v' = v' + v =0. [Additive inverse]
• v+w=w+v [commutativity]
• 1*v = v [scalar identity]
• a*( v + w) = (a*v) + (a*w) [distributive 1]
• (a+b)*v = (a*v) + (b*v)    [distributive 2]
• (ab)*v = a*(b*v)   [associativity 2]

• Vector Space Examples in a more general context:
• Functions: Suppose S is a set. We define V(S,F) = { f :S -> F, the functions with domain S and target F}.
Note that if f and g are in V(S,F) then f=g means that for all s in S, f(s) = g(s).

• Note: When S = {1,2,...,n} we can give a 1 to 1 correspondence between V(S,F) and Fn by letting  f (k) = vk when v =(v , v , ..., v ) is in Fn .
Note: If S is the empty set, then V(S,F) is the empty set.
• Definition: Suppose S is a non empty set and F is a field We define an internal operation on V(S,F), +, and an external operation between  F and V(S,F), * , by the following:

• If v and w are in V(S,F) and a is in F then v + w is the function defined by (v+w)(s) = v(s) + w(s)  for all s in S and
a*v is the function defined by (a*v)(s) =  a v(s)  for all s in S using the addition and multiplication of F for the operations on the elements of F as appropriate.
• Proposition: With the operations just defined on V(S,F) is a vs/F, i.e., the following properties are true showing that V(S,F) is a vector space over the field F:
• If v,w,and z are in V(S,F)  and a and b are in F, then v+w and a*v are in V(S,F)  [closure] ;
• (v+w)+z = v+(w+z)  [Associativity 1]
• If we let 0 denote the function 0(s) = 0 for all in S then 0+v=v+0 = v [Additive identity]
• For each v, there is a function v' in V(S,F), v', so that v+v' = v' + v =0. [Additive inverse]
• v+w=w+v [commutativity]
• 1*v = v  [scalar identity]
• a*( v + w) = (a*v) + (a*w) [distributive 1]
• (a+b)*v = (a*v) + (b*v)    [distributive 2]
• (ab)*v = a*(b*v)   [associativity 2]

• Examples:
• S = N, the natural numbers = {0,1,2,3,...}. V(S,F) = F can be identified with infinite sequences of elements from the field.
• S= { (i,j): i=1,2,...n; j= 1,2,...,m} then we can give a one to one correspondence between V(S,F) and the collection of n by m matrices with entries in the field F. A matrix M corresponds to the function where f ((i,j)) = M i,j the entry in row i, column j of M.
• S = F. Then V(S,F) = V(F,F).
• S=F=R. Then V(S,F) = V(R,R)= {f:R->R}
• P(F) = { f in V(F,F) where f (x) = a0 + a1x + a2x2 + ...+ anxn with ai is in F for each i} = the polynomial functions from F to F.
• F[X] = { f in F, where f(n) = 0 for all but a finite number of n.}   X = (0,1,0,0,0,...).

Note the distinction between P(F) and F[X] when F = `F_2`.
• C0(R) = {f in V(R,R) where f is continuous}.
• C1(R) = {f in V(R,R) where f is continuously differentiable}.
• C(R) = {f in V(R,R) where f is n times differentiable for any n}.

• Simple properties of Vector spaces  [Reminder: Solow Appendix A has many helpful methods to start work on proofs.]
• is unique.  [Check out Solow A.12 on Uniqueness proofs.]
Proof: Suppose z is in V and z+v= v+z = v for all v in V.  Then 0 = 0 + z = z ! Thus there is only one element of a vector space that satisfies the "zero" property.
• -v is unique. Proof: Exercise.
• 0*v= 0. [0 is the zero vector, 0 is the zero scalar.]
Proof:  0*v = (0+0) * v = 0*v + 0*v .
Thus 0 =
0*v + -(0*v) =  (0*v + 0*v) + -( 0*v)  =0*v + (0*v + -( 0*v)) = 0*v + 0 = 0*v
• a*0 = 0.  Proof: Exercise.
• (-1)v = -v. Proof: Exercise.
• If v+z = w + z then v=w. [cancellation] Proof: Exercise.
• If a*v = 0, then either a = 0 or v = 0.  {See Solow A.11 on "either or" proofs.]
Proof: If
a = 0, then the statement is correct.
Suppose
a is not 0. Then there is a  b  in F where ba = 1.
Then v = 1*v = (
ba)* v =  b*(a*v) = b *0 = 0.
Comment: This proof is similar to the comparable statement about products in fields.

• 9-16
• Subspaces.
• Definition. W is a subspace of V, W < V, if  W is a subset of V and using the operations of V, W is also a vector space.
• Simple testing. Is W a subspace of V?
• Hard way: check all the axioms again!  :(
• Easier: Since the operation is the same, you don't need to check scalar identity, associativity, commutativity, or distributivity.
• Fact: If W<V and 0v is the zero for V and 0w is the zero for W then, 0v=0w.
Proof:
0v+0w = 0w but 0w+0w = 0w, so by cancellation: 0w=0w.
• Test 1: (3 steps) If W is non empty and closed under addition and scalar multiplication, then W < V.
• Proof: (i) Since w is not empty, choose w in W. Then 0w= 0, so 0 is in W and so W satisfies the additive identity property.

• (ii) If w is in W, then (-1)w =-w is also in W, so W satisfies the additive inverse property. IRMC.
• Easiest: (2 step) If W is non empty and if for any w and z in W and a in F, w+az is W  [super closure] then W<V.
• Proof: (i) Use a=1, so W is closed under addition.

• (ii)Use a= -1, w= z, then z + (-1)z = 0 is in W.
(iii) 0 + az = az is in W, so W is closed under scalar multiplication.  Now apply the previous test.  IRMC.

9-19
Examples
• F[X] = { f in F, where f(n) = 0 for all but a finite number of n.} < F
• Note: When F = R or C, F[X] can be identified with P(F).
Proof: Assume f and g are in F, and a is in F.  We need to show f+ a g is also in F . So consider  [ f+ ag](n). Since f and g are not 0 except at a finite number of n, [ f+ a g](n) will also be zero at all but a finite number of n.
• Suppose A is an n by k matrix with entries in F.
• N(A) ="the null space of A" = { v in Fn where vA= (0,0,...0) in Fk } <  Fn
• N(A) can be considered the subspace of solutions to the system of k homogenous linear equations in n unknowns determined by vA=(0,0,...,0 ).
• Proof: Clearly... 0A = 0, so 0 is in N(A). Suppose v and w are in N(A) and a is in F, then consider (v+aw)A = vA+(aw)A=vA+a(wA) = 0 +0 = 0, so (v+aw) is in N(A).
• R(A) = "the range space of A" = { w in Fk where for some v in Fn, vA= w  } <  Fk
• Proof:  Clearly... 0A = 0, so 0 is in N(A).Suppose z and w are in R(A) and a is in F, then there are v and u in  in Fk where vA=w and uA=z.  So (v+au)A = vA+(au)A=vA+a(uA) =w+az, so w+az is also in R(A).
• R(A) can be considered the collection of k dimensional vectors, w, for which the system of k linear equations in n unknowns determined by vA=w has a solution.
• {f in C(R) where f ''(x) - f(x) = 0} < C(R).
• Example: Solutions `(x_1,x_2,...,x_n)` to a single  homogenous linear equation  form a subspace of  `R^n`.

• Theorem: "The Intersection of a family of subspaces is a subspace."
Suppose V is a v.s. over a field F.
If U is a nonempty family of subspaces of V ,
then
U={v∈W  for all W∈U} is a subspace of V
• Proof: (i) O is a member of every subspace of V, so O is a member of ⋂U.
(ii) Suppose v and w are vectors in ⋂U and α∈F. Then v and w are vectors in  every subspace W that is in the family U. But since W is a subspace, v + α w is also a vector in all the subspaces W that are members of U. Hence  v + α w  is member  of ⋂U. So ⋂U is a subspace of V.
• Application of Theorem: N(A) Fn .
• Example: {f in C(R) where f is a solution to a homogeneous linear differential equation of order n} < C(R). And this holds for a family of homogeneous linear diff'l equations as well.

• Motivation:  Reconsider R(A)- the range space of a matrix A.
Suppose A is an n by k matrix with entries in F. v in Fn, vA is a vector in Fk which is a linear combination of the row  vectors of the matrix A determined by the entries of v.
This leads to considering the set of linear combinations of a collection of vectors (like the row vectors or column vectors of a matrix)- often described as the span of the set of vectors or the subspace generated by a set of vectors.

• Unfortunately this characterization of "span" is not as clear when the set of vectors is empty!
We will develop a more general definition of span that includes the empty set.

• Definition: Suppose S is a set of vectors in a v.s. V over the field F.
SPAN (S) =
$⋂$ {all W < V  where S $⊂$ W}.
• Facts : 1. SPAN(S) < V.
2. S $⊂$ SPAN(S).
3. If  S  $⊂$  W and W < V then  SPAN(S) <W.

• Fact: If U<W and W <V, then U<V.

• (Internal) Sums , Intersections,  and Direct Sums of Subspaces

• Suppose U1, U2,  ... , Un are all subspaces of V.
• Definition:  U1+ U2+  ... + Un = {v in V where v = u1+ u2+  ... + un for  uk in Uk , k = 1,2,...,n} called the (internal) sum of the subspaces.

• Facts: (i) U1+ U2+  ... + Un < V.
(ii)  Uk < U1+ U2+  ... + Un for each k, k= 1,2,...,n.
(iii) If W<V and Uk < W for each k, k= 1,2,...,n, then U1+ U2+  ... + Un <W.

So ...
U1+ U2+  ... + Un is the smallest subspace of V that contains Uk for each k, k= 1,2,...,n.
• Examples:

• U1 = {(x,y,z): x+y+2z=0} U2 = {(x,y,z): 3x+y-z=0}. U1 + U2 = R3.

Let Uk = {f in P(F): f(x) = akxk  where ak is in F} . Then U0+ U1+ U2+  ... + Un = {f : f (x) = a0 + a1x + a2x2 + ...+ anxn where a0 ,a1 ,a2,...,an are in F}.

The HW Exercise was presented in class: If U and W are subspaces of V and U $∪$ W  is also a subspace of V, then either U < W or W < U.

• Revisited Definition:  U1 $∩$ U2$∩$  ... $∩$ Un = {v in V where v is in Uk , for all k = 1,2,...,n} called the intersection of the subspaces.

• Facts:(i) U1$∩$ U2$∩$  ... $∩$ Un < V.
(ii)   U1$∩$U2$∩$  ... $∩$ Un < Uk for each k, k= 1,2,...,n.
(iii) If W<V and W < Uk for each k, k= 1,2,...,n, then W<U1$∩$ U2$∩$  ... $∩$ Un .
So ...
U1$∩$ U2$∩$  ... $∩$ Un is the largest subspace of V that is contained  in Uk for each k, k= 1,2,...,n.
Examples: U1 = {(x,y,z): x+y+2z=0} U2 = {(x,y,z): 3x+y-z=0}. U1 $∩$ U2 = {(x,y,z): x+y+2z=0 and 3x+y-z=0}= ...
Example: Let Uk = {f in P(F): f(x) = akxk  where ak is in F and characteristic of F is 0} then Uj$∩$Uk = {0} for j not equal to k.
Proof: Suppose f (x) =
axk and bxj for j not equal to k for all in F. Then since F has an infinite number of elements, the polynomial axk - bxj
= 0 for all x. Using x= 1 this means a = b  and using x = 2 = 1+1, this means a=b=0.
Remark: Example is not be true F = {0,1}. For then
x5 x3 for all x in F.

Definition: Suppose V is a v.s over F and  $U1$ and $U2$ are subspaces of V.
We say that V is the direct sum of  U1 and U2
and we write
V = $U1$ $⊕$
$U2$ if (1) $U1$ + $U2$ and (2) U1$∩$ U2 = {0}.

Example: Suppose A is the 5 by 5 diagonal real valued matrix with `A_{1,1} = A_{2,2} = 3` and `A_{3,3} = A_{4,4} = A_{5,5}=2`. `U_2` = {v where vA=2v} and `U_3`={v where vA =3v}. These are subspaces of `R^5` and `R^5 = U_2` ⊕ `U_3`.

Proposition: Suppose V = $U1$ $⊕$ $U2$  and $v∈V$,   v = $u1+u2=w1+w2$ with $ui$ and $wi$ are in $Ui$ for i = 1 and 2.
Then  $ui=wi$ for i = 1,2.
Conversely,  if V = $U1$ + $U2$ and if v = with $ui$ and $wi$ are in $Ui$ for i = 1 and 2.
implies  $ui=wi$ for i = 1,2 then V = $U1$  $⊕$  $U2$.

Proof:  From the hypothesis `u_1 + (-w_1) = w_2 + (-u_2) in U_1 and U_2 ,`
so it is in $U1∩U2$ = {0}. Thus ... $ui=wi$ for i = 1, 2.
Conversely: Suppose the hypothesis... and   then v =  v + 0  = 0 + v, so v = 0. Thus V = $U1$ $⊕$ $U2$.

9-26
To generalize the direct sum  to U1, U2,  ... , Un, we would start by assuming V = U1 + U2 +  ... + Un.

Direct Sums:
Suppose U1, U2,  ... , Un are all subspaces of V and U1+ U2+  ... + Un = V, we say V is the direct sum of U1, U2,  ... , Un if for any v in V, the expression of v as v = u1+ u2+  ... + un for  uk in Uk is unique, i.e., if v = u1'+ u2'+  ... + un' for  uk' in Uk then u1 = u1', u2=u2', ... , un=un'. In these notes we will write
V = U1
$⊕$ U2 $⊕$...$⊕$ Un

Examples:Uk = {v in F
n: v = (0,... 0,a,0, ... 0) where a is in F is in the kth place on the list.} Then U1$⊕$ U2$⊕$  ... $⊕$ Un = V.

Theorem:  V =  U1$⊕$ U2$⊕$  ... $⊕$ Un if and only if (i)U1+ U2+  ... + Un = V AND 0=u1+ u2+  ... + un for  uk in Uk implies u1=u2=...=un=0.
Theorem: V = U$⊕$W if and only if V = U+W and U$∩$W={0}.

Exercise:
Prove: If V = U1 ⊕ U2 ⊕...⊕ Un then  Ui∩ Uj = {0} for  all i and j that are not equal.

Examples using subspaces and direct sums in applications:
Suppose A is a square matrix (n by n) with entries in the field F.
For c in F, let Wc = { v in Fn where vA = cv}.
Fact: For any A and any c,  Wc< Fn . [Comment: for most c, Wc= {0}. ]
Definition: If Wc is not the trivial subspace, then c is called an eigenvalue or characteristic value for the matrix A and nonzero elements of Wc  are called eigen vectors or characteristic vectors for A.

Application 1 : Consider the coke and pepsi matrices:

Example A.
vA = cv? where

A= (
 5/6 1/6 1/4 3/4

)

Example B.
vB = cv where

B= (
 2/3 1/3 1/4 3/4

)

Questions: For which c is Wc non-trivial?
To answer this question we need to find (x,y) [not (0,0)] so that

Example A

(x,y) (
 5/6 1/6 1/4 3/4

) = c(x,y)

Example B

(x,y) (
 2/3 1/3 1/4 3/4

) = c(x,y)

Is R2 = Wc1 + Wc2 for these subspaces? Is this sum direct?

Focusing on Example B we consider for which will the matrix equation have a nontrivial solution (x,y)?
We consider the  equations:  2/3 x +1/4 y = cx and 1/3 x+3/4 y = cy.
Multiplying by 12 to get rid of the fractions and bringing the cx and cy to the left side we find that
(8-12
c)x + 3 y = 0 and 4x + (9-12c)y = 0

Multiplying by 4 and (8-12c) then subtracting the first equation from the second we have
((8-12c)(9-12c)  - 12 )y = 0. For this system to have a nontrivial  solution, it must be that
((8-12c)(9-12)
c  - 12 ) = 0 or
`72-(108+96) c+144c^2 -12 = 0`  `60-204c+144c^2 = 0`$.$
Clearly one root of this equation is 1, so factoring we have (1-c)(60-144c) = 0 and c = 1 and c = 5/12 are the two solutions... so there are exactly two distinct eigenvalues for example B,
c= 1 and c = 5/12  and two non trivial eigenspaces
W1  and W5/12 .

General Claim:
Proposition: If c is different from k, then Wc $∩$ Wk = {0}

Proof:?
Generalize?
What does this mean for  vn  when n is large?
Express `v_0 = v
_e + v~` with `v_e in W_1` and `v~ in W_{5/12}`.
Applying A we find
`v_1 = v_0 A = v_e A + v~ A = v_e  + 5/12 v~`.
Repeating this yields
` v_n = v_e + (5/12)^n v~`.  As `n -> oo` we have `v_n -> v_e`.

Does the distribution of vn when n is large depend on v0?

9-28
Application 2: For c a real number let

Wc = {f in C(R) where f '(x)=c f(x)} < C(R).
What is this subspace explicitly?
Let V={f in C(R) where f ''(x) - f(x) = 0} < C(R).
Claim: V = W1 $⊕$ W-1
Begin?
We'll come back to this later in the course!

If c is different for k, then Wc $∩$ Wk = {0}
Proof:...

Back to looking at things from the point of view of individual vectors:
Linear combinations:

Def'n.
Suppose S is a set of vectors in a vector space V over the field F. We say that a vector v in V is a linear combination of vectors in S if there are vectors u1, u2,  ... , un in S  and scalars a1, a2,  ..., an in F where v = a1u1+ a2u2+  ... + anun .
Comment: For many introductory textbooks: S is a finite set.

Recall. Span (S) = {v in V where v is a linear combination of vectors in S}
If S is finite and Span (S) = V we say that S spans V and V is a "finite dimensional" v.s.

Linear Independence.
Def'n. A set of vectors S is linearly dependent
if there are vectors u1, u2,  ... , un in S  and scalars
`α_1,α_2, ...,α_n in F` NOT ALL 0 where `0=α_1u_1+α_2u_2+ ...+α_n u_n` .
A set of vectors S is linearly independent  if it is not linearly dependent.

Other ways to characterize linearly independent.
A set of vectors S is linearly independent  if  whenever there are vectors u1, u2,  ... , un in S  and scalars
`α_1,α_2, ...,α_n in F` where `0=α_1u_1+α_2u_2+ ...+α_n u_n` , the scalars are all 0, i.e. `α_1=α_2= ...=α_n = 0` .

9-30
Examples: Suppose A is an n by m matrix: the row space of A= span ( row vectors of A) , the column space of A = Span(column vectors of A).
Relate to R(A)

Recall R(A) = "the range space of A" = { w in Fk where for some v in Fn, vA= w  } <  Fk
.
w is in R(A) if and only if w is a linear combination of the row vectors, i.e., R(A) = the row space of A.
If you consider  Av instead of vA, then R*(A) = the column space of A.

"Infinite dimensional" v.s. examples: P(F), F[X], F, C (R)
F[X] was shown to be infinite dimensional. [ If  p is in SPAN(p1,....,pn) then the degree of p is no larger than the maximum of the degrees of {p1,...pn}. So F[X] cannot equal SPAN(p1,...,pn) for any finite set of polynomials- i.e, F[X] is NOT finite dimensional.

Some Standard examples.

Bases- def'n.

Definition: A set B is called a basis for the vector space V over F if (i) B is linearly independent and (ii) SPAN( B)  = V.

Bases and representation of vectors in a f.d.v.s.

10-3
Suppose B is a finite basis
for Vwith its elements in a list, (
u1, u2,  ... , un) .  [A list is an ordered set.]
If v is in V, then
there are unique vectors
scalars
`α_1,α_2, ...,α_n in F` where  `v = α_1u_1+α_2u_2+ ...+α_n u_n .`

The scalars are called the coordinates of v w.r.t. B, and we will write

`v = [α_1,α_2, ...,α_n]_B.`

Linear Independence Theorems
Theorem 1 : Suppose S is a
linearly independent set and
`v`  is not an element of Span(S), then `S ∪ {v}` is also linearly independent.
Proof Outline:
Suppose vectors u1, u2,  ... , un in S  and scalars `α_1,α_2, ...,α_n,α in F` where `0=α_1u_1+α_2u_2+ ...+α_n u_n+α v` . If $α$ is not 0 then
`v=-α^{-1}(α_1u_1+α_2u_2+ ...+α_n u_n) in Span(S)`, contradicting the hypothesis. So $α=0$. But then `0=α_1u_1+α_2u_2+ ...+α_n u_n` and since S is linearly independent,
`α_1=α_2= ...=α_n=0`. Thus  `S ∪ {v}` is linearly independent.   EOP.

Theorem 2: Suppose S is a finite set of vectors with V = Span (S) and T is a subset of vectors in V. If  n( T) > n(S) then T is linearly dependent.
Proof Outline: Suppose n(S) = N. Then by the assumption  ... [Proof  works by finding N homogeneous linear equations with N+1 unknowns.]

10-5
Theorem 3:
Every finite dimensional vector space has a basis.
Proof outline:
• How to construct a basis, `B`, for a non trivial finite dimensional v.s., `V`. Since `V` is finite dimensional it has a subset S that is finite with `Span (S) = V`.
• Start with the empty set. This is linearly independent. Call this `B_0`.  If `span(B_0) = V` then you are done. `B_0` is a basis.
• If `Span(B_0)` is not `V` then there is a vector `v_1` in `V` where `v_1` is not in `Span(B_0)`. Apply Theorem 1 to obtain `B_1=B_0∪{v_1}` which is linearly independent. If `Span(B_1)=V` then `B_1` is a basis for `V`. Otherwise continue using Theorem 1 repeatedly until the resulting set of vectors has more then the number of sets in the spanning set. But by Theorem 2, this is a contradiction. So at some stage of the process, `Span(B_k) = V`, and `B_k` is a basis for `V`.   IRMC

Comment:The proof of the Theorem  also shows that given T, a  linearly independent subset of V and V a finite dimensional vector space, one can step by step add elements to  T, so that eventually you have a new set S where S is linearly independent with Span(S) = V and T   contained in  S. In other words we can construct a set B that is a basis for V with T contained in B.  This proves

Corollary: Every Linearly independent subset of a finite dimensional vector space can be extended to a basis of the vector space.

Theorem 4. If V is finite dimensional vs  and B and B' are bases for V, then `n(B) = n(B')`.

Proof: fill in ... based on the Theorem 2. `n(B) <= n(B')`  and `n(B') <= n(B)`  so...

Definition: The dimension of a finite dimensional v.s. over F is the number of elements in a(ny) basis for V.

Discuss dim({0})= ???.
What is Span of the empty set? Recall we characterized SPAN(S) = the intersection of all subspaces that contain S. Then Span (empty set) = Intersection of all subspaces= {0}.

The empty set is linearly independent!... so The empty set is a basis for {0} and the dimension of {0} is 0!

More Dimension Results:

Prop: A Subspace of a finite dimensional vector space is finite dimensional.

Proposition: Suppose dim(V) = n, S  a set of vectors with N(S) = n. Then
(1) If S is Linearly independent, then S is a basis.
(2) If Span(S) = V, then S is a basis.

Proof: (1) S is contained is a basis, B. If B is larger than S, then B has more than n elements, which contradicts that fact that any basis for V has exactly n elements. So B = S and S is a basis.
(2) Outline:V has a basis of n elements, B.  Suppose S in linearly dependent and show that there is a set with less than n elements that spans V. Hence B cannot be a basis. This,
S is a basis.
IRMC

Theorem: Sums, intersections and dimension: If U, W <V  are finite dimensional, then so is U+W and
dim(U+W) = Dim(U) + Dim(W)  - Dim(U$∩$W).

Proof: (idea) Build up bases of U and W from U$∩$W.... then check  that  the union of these bases is a basis for U+W

10-10
Example [from problem in Axler.]: Suppose p0,...,pm are in
Pm(R) and pi(2) = 0 for all i.
Prove {p0,...,pm} is linearly dependent.

Proof: Suppose {p0,...,pm} is linearly independent.
Notice that by the assumption for any coefficients

(a0p0+..+ampm )(2) = a0p0(2)+..+ampm(2) = 0
and since u(x)= 1 has u(2) = 1, u (= 1) is not in the SPAN(p0,...,pm).
Thus
SPAN(p0,...,pm)
is not Pm(R).

But SPAN ( 1,x, ..., xm) = Pm(F) and {
1,x, ..., xm } is linearly independent (proof?)
So dim (
Pm(R)) = m+1 thus we have SPAN (p0,...,pm)=Pm(R), a contradiction. So {p0,...,pm} is not linearly independent.
End of proof.

Example (visualize): In R2, P4(R). Any 5 polynomials of degree less than 5 that pass through (2,0) are linearly dependent.

Connect to Coke and Pepsi example: find a basis of eigen vectors using the B example for R2[Use the on-line technology]

Example B

(x,y) (
 2/3 1/3 1/4 3/4

) = c(x,y)

We considered the  equations:  2/3 x +1/4 y = cx and 1/3 x+3/4 y = cy and show that
there are exactly two distinct eigenvalues for example B,
c= 1 and c = 5/12  and two non trivial eigenspaces
W1  and W5/12 .
Now we can use technology to find eigenvectors in each of these subspaces.
Matrix calculator
, gives as a result that the eigenvalue 1 had an eigenvector (1,4/3)  while 5/12 had an eigenvector (1,-1). These two vectors are a basis for R2.

Linear Transformations: V and W vector spaces over F.
Definition: A function T:V $→$  W is a linear transformation if for any x,y in V and in F, T(x+y) = T(x) + T(y) and T(ax) = a T(x).

Examples: T(x,y) = (3x+2y,x-3y) is a linear transformation T: R2  -> R2.
G(x,y) = (3x+2y, x^2 -2y) is not a linear transformation.
G(1,1) = (5, -1) , G(2,2) = (10, 0)... 2*(1,1) = (2,2) but 2* (5,-1) is not (10,0)!
Notice that T(x,y)can be thought of as the result of a matrix multiplication

(x,y) (
 3 1 2 -2

)
So the two key properties are the direct consequence of the properties of matrix multiplication.... (v+w)A= vA+wA and (cv)A = c(vA).
For A a k by n matrix :  TA  (left argument) and
AT (right) are linear transformations on Fk and Fn.
TA  (x) = x A for x in Fk and AT(y) = A[y]tr for y in Fn and [y]tr indicates the entries of the vector treated as a one column matrix.

10-12
The set of all linear transformations from V to W is denoted L(V,W).

Consequences of the definition: If T:V->W is a linear transformation, then for any x and y in V and a in F,

(i) T(0) = 0.

(ii) T(-x) = -T(x)

(iii) T(x+ay) = T(x) + aT(y).

Quick test: If T:V->W is a function and (iii) holds for any x and y in V and a in F, then the function is a linear transformation.

D... Differentiation is a linear transformation: on polynomials, on ...

Example: (D(f))(x) = f' (x) or D(f) = f'.
(D(f + $α$ g))(x) = (f+$α$g)' (x) = f'(x) + $α$g'(x) = (f'+$α$g') (x)  or
D(f+
$α$g) = f'+ $α$g'= D(f) +$α$ D(g).

Theorem: T : V->W  linear, B a basis, gives TB:B ->W.
Suppose S:B -> W, then there is a unique linear transformation T_S:V->W such that T_SB=S.
Proof:
Let T_S(v) be defined as follows: Suppose v is expressed (uniquely) as a linear combination of elements of B, ie.
v =
a1u1+ a2u2+  ... + anun
... then let T_S(v)  = a1S(u1)+ a2S(u2)+  ... + anS(un) .
This is well defined since the representation of v is unique.
Uniqueness of T_S: If U is any linear transformation, U:V->W with U
B=S then U(v) = T_S(v) ,
so U must be
T_S.

Show T_S is linear:   Choose v and v' in V,
α in F, with v = a1u1+ a2u2+  ... + anunand v' = a'1u1+ a'2u2+  ... + a'nun

where T_S(v)  = a1S(u1)+ a2S(u2)+  ... + anS(un) and T_S(v')  = a'1S(u1)+ a'2S(u2)+  ... + a'nS(un)
Then check that
T_S(v+av') = T_S(v) + aT_S(v'). [Details omitted here.]

Finally... for any u in B, u = 1u, so  T_S(u) = 1S(u) = S(u) and T_SB= S.                             EOP

Comment: "A linear transformation is completely determined by what it does to a basis."
or... "T_
(TB)= T: V->W  and  T_SB= S: B-> W".

Example: T: P(F) $→$ P(F)....Use for a basis { xn for n = 0, 1,2,... } and S(xn) = nx n-1.
Or another example:
S(xn) = 1/(n+1) x n+1.

10-17
Key Spaces related to T:V->W
Null Space of T= kernel of T = {v in V where T(v) = 0 [ in W] }= N(T) < V
Range of T = Image of T = T(V) = {w in W where w = T(v) for some v in V} <W.

Comment: Connect these to matrix subspaces R(A) and N(A).

• N(A) ="the null space of A" = { v in Fn where vA= (0,0,...0) in Fk } <  Fn
• R(A) = "the range space of A" = { w in Fk where for some v in Fn, vA= w  } <  Fk

Major result of the day:
Theorem: Suppose T:V->W and V is a finite dimensional v.s. over F. Then N(T) and R(T) are also finite dimensional and Dim(V) = Dim (N(T)) + Dim(R(T)).
Proof:

Outline: start with a basis C for N(T) and extend this to a basis B for V.
Show that  T(B-C) is a basis for R(T).

Algebraic structure on L(V,W)
Definition of the sum and scalar multiplication:
T, U in L(V,W), a in F, (T+U)(v) = T(v) + U(v).
Fact: T+U is also linear.
(aT)(v) = a T(v) .
Fact:aT is also Linear.
Check:  Proposition: L(V,W) is a vector space over F.

10-19
Composition
: T:V -> W and U : W -> Z both linear, then  UT:V->Z where UT(v) = U(T(v)) is linear.

Note: If T':V-> W and U':W->Z are also linear, then  U(T+T') = UT + UT' and (U+U') T = UT + UT'. If S:Z->Y is also linear then S(UT) = (SU)T.

Linear Transformations and Bases
Theorem: If V and W are finite dimensional then so is `L(V,W)` and `dim(L(V,W)) = dim(V) dim(W)`.
Outline: Use bases for V and W to find a basis for L(V,W). That basis for L(V,W) also establishes  a function from L(V,W
) to the matrices that is a linear transformation!
More details will be supplied.
Details: Let `B={v_1,v_2, ..., v_k}` is a basis for V and `C= {w_1,w_2, ... ,w_q}` be a basis for W.  We can define a linear transformation `T_{i,j}: V ->W` by `T_{i,j}(v_r)=w_j` when `r=i` and otherwise
`T_{i,j}(v_r)=0_W`.
Claim:
`D = {T_{i,j} : i = 1,...,k ; j = 1,...,q}` is a basis for `L(V,W)` and thus `dim(L(V,W)) = dim(V) dim(W)`.

`D` is linearly independent:
Suppose `a_{i,j} in F` with `U =sum_{(i,j)} a_{i,j}T_{i,j} = 0.` Then with `r` fixed, `U(v_r)=sum_{i,j} a_{i,j}T_{i,j}(v_r) = sum_{j} a_{r,j}T_{r,j}(v_r) = sum_{j} a_{i,j}w_j= 0`Since C is a basis this means that `a_{r,j}=0` for all j and thus `a_{i,j}=0` for all `i and j`. This shows that D is linearly independent.

`D` spans `L(V,W)`: Suppose `R in L(V,W)`. Then for each `r` there are some `b_{r,j} in F`,  `R(v_r)= sum_{j} b_{r,j}w_j`.
This shows that
`R = sum_{(i,j)} b_{i,j}T_{i,j}`
and thus `R in SPAN(D)`.  EOP

Matrices  and Linear transformations.

10-24
Footnote on notation for Matrices:
If the basis for V is B and for W is C as above and `R in L(V,W); R: V ->W`,
the matrix of `R`,  with respect to those bases can be denoted
`M_B^C(R)` .
Note - this establishes a convention on the representation of a transformation.
The matrix for a vector `v` in the basis B is denoted
`M_B(v)` and for `T(V)` is denoted `M_C(T(v))`. If we treat these as  row vectors we have
`M_C(T(v)) = M_B(v) M_B^C(R) `
.
This all can be transposed using column vectors for the matrices of the vectors and transposing the matrix
`M_B^C(R)` we have with this transposed view:
`M_C(T(v))^{tr} =  M_B^C(R)^{tr} M_B(v)^{tr}`
To conform with the usual convention for function notation, the transposed version using column vectors matrices is used most frequently. When it is not ambiguous we will denote the transposed matrix by switching the position of B and C to the left of M: ` M_B^C(R)^{tr}= _C^BM(R) `
.
`So   _B^CM(R) _BM(v)=_CM(T(v))  `

Example: Suppose `T(x,y,z)=(2x+y-z, 3x+4z); T:R^3 -> R^2.` Let `B` and `C` be the standard ordered bases, `B={(1,0,0),(0,1,0),(0,0,1)} ; C={(1,0),(0,1)}`. Then
`T(1,0,0) = 2(1,0)+3(0,1)`
`T(0,1,0) = 1(1,0)+0(0,1)` and
`T(0,0,1) = -1(1,0) +4(0,1)`.

`M_B^C(T) = [(2,3),(1,0),(-1,4)]` or transposed ` [(2,1,-1),(3,0,4)]= _B^CM(T) `
Example: Suppose `T(f)=f'; T:P_2(R) ->P_1(R).` Let `B` and `C` be the  ordered bases, `B={1, x, x^2} ; C={1,x}`. Then
`T(1) = 0*1+0*x`
`T(x) = 1*1+ 0*x` and
`T(x^2) =0*1 + 2x`.

`M_B^C(T) = [(0,0),(1,0),(0,2)]` or transposed ` [(0,1,0),(0,0,2)]= _B^CM(T) `

The function `M : L(V,W) -> Mat (k,q; F)` is a linear transformation.

Recall definition of "injective" or "1:1" function.
Recall definition of "surjective" or "onto" function.

Theorem: T is 1:1 (injective) if and only if N(T) = {0}
Proof: =>  Suppose T is 1:1.  We now that T(0)=0 , so if T(v) = 0, then v = 0. Thus 0 is the only element of N(T) or N(T) = {0}.
<=  Suppose N(T) = {0}. If T (v) = T(w) then T(v-w) =T(v)-T(w) = 0 so v-w is in N(T).... ok, than must mean that v-w = 0,  so v=w and T is 1:1.

Theorem: T is onto if and only of the Range of T = W.
Theorem: T is onto if and only if for any (some) basis, B, of V, Span(T(B)) = W.
Theorem: If V and W are finite dimensional v.s. / F, dimV = dim W,  T : V $→$ W is linear, then T is 1:1 if and only if T is onto.
Proof: We know that dim V = dim N(T) + dim R(T).
=>  If T is 1:1, then  dim N(T) = 0, so dim V = dim R(T)  . Thus dim R(T) = dim W and T is onto.
<=   If T is onto, then dimR(T) = dim W. So dim N(T) = 0 and thus N(T) = {0} and T is 1:1.

Application to `M: L(V,W) -> Mat(k,q;F)`:
Assuming `dim(V) = k` and `dim(W)=q` then `dim(L(V,W)) = kq = dim (Mat(k,q;F)).` Then to show M is 1:1 it suffices to show that M is onto.  If we suppose `A in Mat(k,q;F)` with entries `A_{i,j}` and let `T = sum A_{i,j} T_{i,j}` then it should be direct to see that `M(T) = A.`

Definition: If `T:V -> W` is a linear transformation that is 1:1 and onto, then T is called a linear (vector space) isomorphism and W is said to be (linearly) isomorphic to V.

Cor.: Mat(k,q;F) is linearly isomorphic to L(V,W).

Remark: There is more "good" stuff happening in this isomorphism.  If Z is another vector space over F with dim(Z) = r and basis D, and `U in L(W,Z)` then
`M_B^C(T)M_C^D(U) = M_B^D(UT)` or using the transposed version (which is generally preferred for the niceness of its appearance) `thus... _C^DM(U)_B^CM(T) =  _B^DM(UT)`.
The importance of the Null Space of T, N(T), is in understanding what T does in general.

Example 1. D:P(R) -> P(R)... D(f) = f'. Then N(D) = { f: f(x) = C for some constant C.} [from calculus 109!]
Notice: If f'(x) = g'(x)  then f(x) = g(x) + C for some C.
Proof: consider D(f(x) - g(x)) = Df(x) - Dg(x) = 0, so f(x) -g(x) is in N(T).

Example 2: Solving  a system of homogeneous  linear equations. This was connected to finding the null space of a linear transformation connected to a matrix. Then what about a non- homogeneous system with the same matrix. Result: If z is a solution of the non- homogeneous system of linear equations and z ' is another solution, then z' = z + n where n is a solution to the homogeneous system.

Consider `p in R[x]` and `T in L(V,V)`, where `p = a_0 + a_1x + a_2x^2 + ... a_nx^n`. We define `p(T) in L(V,V)` by `p(T)= a_0Id + a_1T + a_2T^2 + ... + a_nT^n`

Example 3: More differential equations: `D:C^{oo}(R) -> C^{oo}(R)` where `Df(x) = f'(x)`. Suppose `p= x^2 -5x+4`. Then `N(p(T)) = {f : p(T)(f) = z}` where `z` denotes the zero function.  Then `N(p(T)) = {f:f''(x)-5f'(x)+4f(x) = 0` for all `x in R}`. This is the set of solutions to the homogeneous linear differential equation `f''(x)-5f'(x)+4f(x) = 0 `.
Note that `p = (x-4)(x-1)` which has roots (zeroes) 1 and 4 and that the solutions of the differential equation are all of the form `f = a_1e^x + a_2e^{4x}`. More on this example later in the course.

General Proposition: `T:V->W`. If b is a vector in W and a is in V with T(a) = b, then T-1({b}) = {v in V: v = a +n where n is in  N(T)} = a + N(T)

Comment: a + N(T) is called the coset of a mod N(T)...these are analogous to lines in R2. More on this later in the course.

Note:Why this called a "linear" transformation:
The geometry of linear: A line in R2 is {(x,y): Ax +By = C where A and B are not both 0} = {(x,y): (x,y) = (a,b) + t(u,v)}= L, line through (a,b) in direction of (u,v).

Suppose T is a linear transformation :
Let T(L) = L' = {(x'y'): (x',y')= T(x,y)}
T(x,y) = T(a,b) + t T(u,v).
If T(u,v) = (0,0) then L' = T(L) = {T(a,b)}.
If not then L' is also a line though T(a,b) in the direction of T(u,v).
[View this in winplot?]

10-31
Coke/Pepsi example B: T(x,y) =
(2/3 x +1/4 y, 1/3 x+3/4 y)
T(v0) = v
1, T(v1) = v2.... T(vk)=T(vk+1).
T(v*)=v* means a nonzero v* is an eigenvector with eigenvalue 1. T(1, 4/3) = (1,4/3).
Also `T(3/7, 4/7) = T
[(3/7)(1,4/3)] = 3/7T(1,4/3) =3/7(1,4/3) =(3/7,4/7)`.
`T(1,-1) =(5/12,-5/12 )= (5/12)(1,-1)` means that `(1,-1)` is an eigenvector with eigenvalue 5/12.

Invertibility of Linear Transformations

Def'n: T:V -> W, `T in L(V,W)`, is invertible if and only if
there is a linear transformation S :W -> V where TS = IdW and ST = IdV .

Fact:
If T is invertible then the S :W->V used in the definition is also invertible!
S is unique:  If S' satsifies the same properties as s, then
S = S Id = S(TS')  =(ST)S' = Id S' = S'
S is called "the inverse of T".
Prop: T is invertible iff  T is 1:1 and onto  (injective and surjective).
Outline of Proof:
(i) =>  Assume S... (a)show T is 1:1. [This uses ST = Id].(b) show T is onto [This uses TS = Id].
(ii) <=  Assume T is 1:1 and onto. Define S. [This uses that T is onto and 1:1] Show S is linear [This uses T is linear.] and TS =I and ST = I

Def: If there is a T:V->W that is invertible, we say W is isomorphic with V. (V=T W)
(i)V=Id V (ii)If V=T W then W=S V  (iii)If V=T W and W=U Z then V=UT Z.

Theorem: Suppose V and W are finite dimensional v.s./F. Then
V=T W if and only if dim(V) = dim(W).
Proof: =>: Suppose
V=T W. Then since there is a T: V`->` W that is an isomorphism, dim (V) = dim N(T) + dim R(T). But R(T) = W and N(T) = {0} so dim N(T) = 0 and dim(V) = dim(W).
<= (outline) Assume dim(V) = dim ( W) = n. Choose a basis B for V, B = {v1,v2,...,vn} and a basis C for W, C  = {w1,w2,...,wn}. Then use T defined by T(vi) = wi and show this is invertible.

Theorem: Suppose B= (v1,...,vn) and C=(w1,...,wm)  are finite bases  (lists) for V and W respectively. The linear transformation M: L(V,W) -> Mat(m,n,F) is an isomorphism.
Proof: Show injective by Null(M)= {Z} -where Z is the zero transformation.
Show M is onto by giving TA where M(
TA ) = A based on knowing A.
[OR use dimensions of these vector spaces are equal proved previously.]

Cor. [if isomorphism is established directly]: Dim L(V,W) = Dim(V) Dim(W).

11-2
Note on HW: For SOS problem 6.69, assume
`V = U ⊕ W`.

Prop. V a f.d.v.s.  If T is in L(V) then the following are equivalent:
(i) T is invertible.
(ii) T is 1:1.
(iii) T is onto.
Proof: (i) =>(ii). Immediate.
(ii)=>(iii) . Dim V = Dim(N(T)) + Dim(R(T)). Since T is 1:1, N(T)={0}, so Dim(N(T))= 0 and thus Dim V = Dim (R(T)) so R(T) = V and T is onto.
(iii) =>(i)
Dim V = Dim(N(T)) + Dim(R(T)) Since T is onto, R(T) = V... so Dim(N(T)) = 0. ... so N(T) = {0} and T is 1:1, so T is invertible.

Connection to square matrices:
A is invertible  is equivalent to....Systems of equations statements.

[Motivation]
Look at Coke/Pepsi example B: T(x,y) =
(2/3 x +1/4 y, 1/3 x+3/4 y)= (x,y)A
T(v0) = v
1, T(v1) = v2.... T(vk)=T(vk+1).
v2=T(v1) = TT(v0);... T(vk)=Tk(v0) = (x0,v0)Ak.
We considered the  equations:  2/3 x +1/4 y = cx and 1/3 x+3/4 y = cy and showed that
there are exactly two distinct eigenvalues for example B,
c= 1 and c = 5/12  and two non trivial eigenspaces
W1  and W5/12 .
Now we can use technology to find eigenvectors in each of these subspaces.
Matrix calculator
, gave as a result that the eigenvalue 1 had an eigenvector `(1,4/3)=v_1`  while 5/12 had an eigenvector `(1,-1)= v_2`. These two vectors are a basis for R2.
B=(v1,v2)
What is the matrix of T using this basis.
`M_B^B(T) = _B^BM(T) = ((1,0),(0,5/12))`

Using this basis and matrix makes it easy to see what happens when the transformation is applied repeatedly:

`M_B^B(T^n) = [M_B^B(T)]^n =((1,0),(0,5/12))^n=((1,0),(0,{5/12}^n))`

11-4
Recall this footnote on notation for Matrices:
If the basis for V is B and for W is C as above and `R in L(V,W); R: V ->W`,
the matrix of `R`,  with respect to those bases can be denoted
`M_B^C(R)` .

Note - this establishes a convention on the representation of a transformation.
The matrix for a vector `v` in the basis B is denoted
`M_B(v)` and for `T(V)` is denoted `M_C(T(v))`. If we treat these as  row vectors we have
`M_C(T(v)) = M_B(v) M_B^C(R) `.

This all can be transposed using column vectors for the matrices of the vectors and transposing the matrix `M_B^C(R)` we have with this transposed view:
`M_C(T(v))^{tr} =  M_B^C(R)^{tr} M_B(v)^{tr}`

To conform with the usual convention for function notation, the transposed version using column vectors matrices is used most frequently. When it is not ambiguous we will denote the transposed matrix by switching the position of B and C to the left of M:
` M_B^C(R)^{tr}= _C^BM(R) `.
`So   _B^CM(R) _BM(v)=_CM(T(v))  `
AND

If Z is another vector space over F with dim(Z) = r and basis D, and `U in L(W,Z)` then `M_B^C(T)M_C^D(U) = M_B^D(UT)` or using the transposed version (which is generally preferred for the niceness of its appearance)
`thus... _C^DM(U)_B^CM(T) =  _B^DM(UT)`.

Let's prove this!
Outline: Choose your bases, then find the matrix coefficients for T and U. Now apply U to T(v) where v is in the basis of V  and find the coefficient for a basis element of Z. compare this with how matrix multiplication works!

11-7
Change of basis [Again]: [Caveat: Some of the matrices in the work that follows may be transposed.]

Consider a linear operator T on a vector space V, `T: V -> V,  T in L(V,V)`.
So ... What is the rel
ationship between the very nice matrix we had for the coke and pepsi model for T that results from using the basis B of eigenvectors and the matrix for T that uses the standard basis, `E = (e_1,e_2)`? `M_B^B(T) = ((1,0),(0, 5/12))`
`M_E^E(T) =((2/3,1/3),(1/4,3/4))`.

The key to understanding the relationship between these matrices is the identity map!
We consider the matrix for the identity operator using B for the source space and
E for the target space.
`M_B^E(Id) =((1,4/3),(1,-1))`
And for the identity operator using E for the source and B for the target, MEB(Id).
Notice that
MEB(Id) MBE(Id) =MBB(Id* Id)=MBB(Id)= In ,the n by n identity matrix, and similarly MBE(Id) MEB(Id) =MEE(Id) = In . Thus both these matrices are invertible and each is the inverse of the other!

Furthermore:
`M_B^B(T) =((1,0),(0,5/12))`.

Now we see how to express `M_B^B(T)` in terms of `M_E^E(T)` and vice versa:

`M_B^E(Id)M_E^E(T)M_E^B(Id) = M_B^B(IdTId) = M_B^B(T)`
and
`M_E^B(Id)M_B^B(T)M_B^E(Id) = M_E^E(IdTId) = M_E^E(T)`

Now let `P = M_E^B(Id)` and `Q = M_B^E(Id) = P^{-1}`, then we have

`M_E^B(Id)M_B^B(T)M_B^E(Id) =  P M_B^B(T) P^{-1}  = M_E^E(T)`
or

PMBB(T)= MEE(T)P
and
QMEE(T)P=MBB(T).

Change of  Basis, Invertibility and similar matrices.
The previous example works in general:
The Change of Basis Theorem:
Suppose V is a f.d.v.s over F, dim(V) = n, and B and E are two bases for V. Suppose `T:V -> V` is a linear operator, then
MBE(Id)MEE(T)MEB(Id)=MBB(T)
and
MEB(Id)MBB(T)MBE(Id)=MEE(T).

If we let P =MEB(Id) and Q = MBE(Id) = P-1
[
QP =MBE(Id)MEB(Id) = MBB(Id)= In ]
then we have
PMBB(T)Q =PMBB(T)P-1=MEE(T)
or
PMBB(T)= MEE(T)P
and likewise  QMEE(T)P=MBB(T).

Def'n: We say that two n by n matrices A and B are similar if there is an invertible n by n  matrix P so that B = P-1AP and write A~B.

Proposition: i) A~A; ii) if A~B then B~A;iii)
if A~B and B~C then A~C.
Proof Outline: i) P= In . ii) Use Q=
P-1 . iii) If C = Q-1BQ then C = Q-1P-1APQ ...
Cor. Suppose V is a f.d.v.s over F, dim(V) = n, and B and E are two bases for V. Suppose `T:V -> V ` is a linear operator, then MBB(T) and MEE(T) are similar matrices.

There is a "converse" to the theorem based on the following
Proposition: Suppose P is an invertible n by n matrix.
The linear transformation
`T_P:F^n -> F^n`  defined by the matrix P where E is the standard ordered basis for Fnand MEE(TP) = P maps every basis B of Fn to a basis, TP(B)= B' .

11-9
Eigenvectors, Eigenvalues, Eigenspaces, Matrices, Diagonalizability, and Polynomials!

Definition: Suppose T is a linear operator on V, then a is an eigenvalue for T if there is a non-zero vector v where T(v) = av. The vector v is called an eigenvector for T.
Proposition: a is an eigenvalue for T  if and only if Null(T-aId)  is non-trivial.

Def'n:T is diagonalizable if V has a basis of eigenvectors.
T is diagonalizable if and only if M(T) is similar to a diagonal matrix, i.e., a matrix A where Ai,j=0 for indices i, j where i is not equal to j.

Fact: If T is diagonalizable with distinct eigenvalues  a1,...,an , then S = (T-
a1Id)(T-a2Id).... (T-anId) = 0.
Proof: It suffices to show that for any v in a basis for V, T(v) = 0.  Choose a basis for V of eigenvectors, and suppose v is an element of this basis with T(v) =
aj v. Then S(v)= (T-a1Id)(T-a2Id).... (T-anId)(v) = (T-a1Id)(T-a2Id).... (T-ajId)... (T-anId)(T-ajId)(v) = 0.

What about the Converse? If  there are distinct scalars a1,...,an where S(v) = (T-a1Id)(T-a2Id).... (T-anId)(v) = 0 for any v in V, is T diagonalizable? we will return to this later....!

A Quick trip into High School/Precalculus Algebra and Formal Polynomials: Recall... F[X]
F[X] = { f in F, where f(n) = 0 for all but a finite number of n}
=
{ formal polynomials with coefficients in F using the "variable" X}
<
F.
X = (0,1,0,0,....). example: 2+X + 5X2 +7 X4 = (2,1,5,0,7,0,0,...)

Notice: F[X] is an algebra over F... that is it has a multiplication defined on its elements...
Definition: If `f,g in F[X]` with `f = (a_0,a_1,...,a_n,0,0,...) with a_n not 0`  and `g = (b_0,b_1,...,b_k,0,0,...) with b_n not 0` then `f *g= (c_0,c_1,...,c_n,0,0,...)` where
`c_j = sum_{i=0}^{i=j}  a_ib_{j-i}`

In fact it has a multiplicative unity, 1 =(1,0,0,0...), and furthermore, this algebra has a commutative multiplication: if f,g are in F[X] then f*g = g*f.

Notice:
If f is not 0, then deg(f) = ...., and

Theorem: If f and g are not 0 = (0,0,0...),  then f*g is also non-zero, with deg(f*g) = deg(f) + deg(g).

11-14
Polynomials and Algebras:
If A is any algebra over F with unity, and `f in F[X]`,
`f  = (a_0 , a_1, ... ,a_n, 0 , 0 , ...)` then we have a function, f :A`->`A defined by
f
`(t) = a_0 I + a_1t+ ... +a_nt^n`  where I  is the unity for A and `t in A`.
In particular (i)A can be the field F itself, so f  `in P(F)`.
Example: F = Z2. f  = X2 + X in F[X]. Then f =(0,1,1,0,0,...) is not (0,0,0....) but f(t) = 0 for all t in F.

(ii) A can be L(V) where V is a finite dimensional vector space over F.
Then f (T) is also in L(V).
(iii) A can be M(n;F), the n by n matrices with entries from F.

Then f (M) is also in M(n;F).

The Division Algorithm, [proof?]
If g is not zero, for any f there exist unique q , r in F[X] where f  = q*g +r and either (i) r = 0 or (ii) deg(r) < deg(g).
The Remainder and Factor Theorems [Based on the DA]
Suppose c is in F,
for any f there exist unique q , r in F[X]
where f  = q*(X-c) +r and  r = f(c).

Suppose c is in F ,then  f (c) = 0 if and only if f = (X-c)*q for some q in F[X].

11-16
Roots and degree.
If c is in F and f (c) = 0, then c is called a root of f.
If f is not 0, and deg(f) = n then there can be at n distinct roots of f in F.

Factoring polynomials. A polynomial in F[X] is called reducible if there exist polynomials p and q, with deg(p)>0 and deg(q)>0 where f=p*q.
If deg(f )>0 and f is not reducible it is called irreducible (over F).
Example:
X2 + 1 is irreducible over R but Reducible over C.

(I)The FTof Alg for C[X].
Theorem: If f is non-zero in C[X] with deg(f)>0, then there is a complex number r where f (r) = 0

(II) The FT of Alg for R[X].

Theorem:
If f is non-zero in R[X] andis irreducible, then deg(f)= 1 or 2.

Proof  of II assuming (I):
If f is in R[X] and deg(f)>2, then f is in C[X].
If r is a root of f and r is a real number then f is reducible by the factor theorem.
If r=a +ib is not a real number, then because the complex conjugate of a sum(product) of complex numbers is the sum (product) of the conjugates of the numbers, and the complex conjugate of a real number is the same real number, we can show that f(a+bi) =0 = f(a-bi).  Now by the factor theorem (applied twice)
f = (X-(a+bi))*(X-(a-bi))*q=((X-a)2 + b2 )*q
and deg(q) = deg(f ) -2 >0. Thus f is reducible.

Back to Linear Algebra, Eigenvalues  and "the Minimal Polynomial for a Linear Operator":
Theorem: Suppose V is nontrivial f.d.v.s over the complex numbers
, C and  `T in L(V)`. Then T has an eigenvalue.
Comment: First consider this with the
Coke/Pepsi example B: [Corrected 11/17]
T(x,y) =
(2/3 x +1/4 y, 1/3 x+3/4 y).
Consider ( e1= (1,0), T(
e1) = (2/3,1/3), T(T(e1 ))=  (4/9+1/12, 2/9+3/16) ). This must be linearly dependent because it has 3 vectors in R2. This gives some coefficients in R not all zero, where a0 Id(
e1) + a1T(e1) + a2T2(e1)=0. Thus we have f in R[X] , f =  a0  + a1X + a2X2
and  f(T)(e1) = 0. In fact, we can use f =(X-1)(X-5/12) . f(T)(e1)=(T-Id)(T-5/12Id)(e1)= (T-Id)((T-5/12Id)(e1))=(T-Id)(2/3-5/12,1/3) = 0 Thus we find that (T-Id) (1/4,1/3)= 0, so (1/4,1/3) is an eigenvector for T with eigenvalue 1.
Now here is a
Proof (outline): Suppose dim V = n >0.
Consider v, a nonzero element of V, and the set (v, T(v), T2(v),
T3(v)....Tn(v)).
Since this set has n+1 vectors it must be linearly dependent. ...
...
This means there is a non-zero polynomial, f,  in C[X] where f (T)(v) = 0.
Let m = deg(f ).
Using the FT of Alg for C we have that f = a (X-c1)... (X-cm).
Now apply this to T as a product and ....
for some i and w (not 0), (T-ciId) (w) = 0. Thus T has an eigenvalue.

Theorem:V a fdvs /F, T in L(V,V) .
Then there is some non-zero polynomial f in F[X] where f
(T) = 0,
i.e., for all v in V,
f  (T)(v)= 0.
Proof (outline).
Suppose dim(V)=n. Consider the set (Id, T, T2,T3....Tn*n).
Since this set has n*n+1 vectors in L(V) where dim(L(V))= n*n, so it must be linearly dependent. ...
...
This means there is a non-zero polynomial, f,  in F[X] where f (T) = 0,
i.e., f(T)(v) = 0 for all v  in V.

11-18
Definition: Ann(T)={f in F[X] : f  (T) = 0}
.  The previous Theorem shows Ann(T) has a non trival element.
Prop: f, g in Ann(T), h in F[X] then f+g and h*f are in Ann(T).  [Ann(T) is an "ideal".]

Theorem (The minimal polynomial): There is a non zero monic polynomial in Ann(T) of smallest degree. This polynomial is unique and any polynomial in Ann(T) has this polynomial as a factor.

Proof: The previous theorem has shown Ann(T) has a nonzero polynomial element. Considering the degrees of the non-zero polynomials in Ann(T) there is a smallest positive degree, call it m and a polynomial g in Ann(T) with deg(g) = m. If g = bXm +....terms of lower degree, where b is not 0,
then f = 1/b*g is also in Ann(T) and f  is a monic polynomial.

Now suppose h is also in Ann(T) , then by the division algorithm, h = q*f + r where either r = 0 or deg(r)< m. But since h and f are in Ann(T), h - q*f = r is also in Ann(T). Since deg(f )=m, which is the lowest degree for an element of Ann(T), it must be that r = 0, so h =q*f. Now if h is also monic and deg(h) = m, then deg(q) = 0, and since h and f are both monic, q = 1, and h = f. Thus the non zero monic polynomial in Ann(T) of smallest degree is unique!  EOP

Prop. (Min'l Poly meets eigenvalues) Suppose m in F[X] is the mininal polynomial for T.  Then T has eigenvalue c if and only if X-c is a factor of m.
Proof: =>  Suppose  c is an eigenvalue for T.
Then W
=Null(T-c) is a nontrivial subspace of V and for w in Wc
T(w) = cw is also in Wc .  Let S(w)=T(w) for w in Wc . Notice that S is in L(W).
As a linear operator on
Wc ,   (X-c)(S) = 0, so X-c is the minimal polynomial for S.
But for any w in
Wc (and thus in V), m(S)(w) = m(T)(w) = 0, so m is in Ann(S), and X-c is a factor of m.

<=  Suppose
X-c is a factor of m.  m= (X-c)*q. Since m is the minimal polynomial for T and deg(q)= deg(m)-1
q
(T) is not the 0 operator. Thus there is some v in V where w =q(T)(v) is not 0.
But
m(T)(v)=...=(T-cId)(q(T)(v))=(T-cId)(w) = 0. So c is an eigenvalue for T.   EOP

11-28
Discussion of determinants and characteristic polynomial for a matrix. Cayly Hamilton theorem discussed with its implication that the degree of the minimal polynomial for an n by n matrix is no greater than n. [Details to be added. ]

Cor. T is invertible if and only if the constant for = m(0) is not 0.
Cor. If T is invertible then T-1 = -1/
m(0) (( m-m(0))/X)(T))
Remark: This also can be applied to square matrices thus expressing the matrix inverse of a matrix M as a polynomial applied to M.
Example:  Let `M=( (3, 5),(0,2))` then M has minimal polynomial: `m = (X-3)(X-2) = X^2 -5X +6`.  Since m(0)= 6, we have that `M^{-1} = -1/6 (M-5I) = -1/6((-2,5),(0,-3)) `.  Check!

11-30
Invariant Subspaces: V, T as usual.
Def'n:W is called an invariant subspace for T if for all w in W, T(w) is also in W... i.e. T(W)<W.
If W is an invariant subspace of T, then T:W->W is a linear operator as well, denoted T|W.
Fact: If W is an invariant subspace of T, then the minimal polyonmial of
T|W is a factor of the minimal polynomial of T.

Invariant Subspaces, Diagonal and Block Matrices:
If
V is a fdvs / F and T is in L(V)  with W1 ,W2 ,...,Wk invariant subspaces for T and V = W1 `oplus`W2 `oplus`...`oplus` Wk .
and if the basis for V is B is composed of bases for each
W1 ,W2 ,... ,Wk in order, then M(T)  is composed of matrix blocks - each of which is M(T|Wi). Furthermore if m1 ,m2 ,... , mk is the minimal polynomial for T restricted to W1 ,W2 ,... ,Wk then the minimal polynomial for T is the lowest common multiple of the polynomials m1 ,m2 ,... , mk.

Examples of Matrices, characteristics values and minimal polynomials.
[Some references here to the characteristic polynomial from previous course work in Linear algebra.]
Example 1. `((3,0,0),(0,3,0), (0,0,2))` is a diagonal matrix that has "characteristic polynomial"  `(x-3)^2 (x-2)` but minimal polynomial `(x-3)(x-2)`.
Example 2.
`((3,1),(0,3))` is not diagonalizable. It has characteristic polynomial and minimal polynomial `(x-3)^2`.
These examples are relevant because of the following

Prop. Suppose m in F[X] is the mininal polynomial for T.
T is diagonalizable
if and only if there are distinct c1,...,cm in F where m =  (X-c1)... (X-cm)

Proof :
=> Suppose T is diagonalizable and T has eigenvalues c1,...,cm. By the preceeding Proposition,  for each c1,...,cm each (X-c1),...,(X-cm) is a factor of the minimal polynomial, and by our previous work, S=(X-c1)*...*(X-cm) is in Ann(T) so m = S.

<=
[ Modified from proof in Hoffman and Kunze- Linear algebra 2nd Ed]
Suppose
there are distinct c1,...,cm in F where m = (X-c1)*...*(X-cm).
Let W be the subspace spanned by all the characteristic vectors of  T. So W =
W1 +W2 +...+ Wm  where Wk = Null(T-ck).
We will show that V = W indirectly. Suppose V is not W.

Lemma: There is a vector v* not in W and a characteristic value cj  where w*=(T-cj Id)(v*) is in W. [Proof is below.]

Now express w* in terms of vectors in
Wk
Then for any polynomial h,
h(T)(w*) = h(c1)w1 +  ... +h(ck) wk.
Now
m = (X-cj )q and q-q(cj ) = (X-cj )k.

THUS...
q(T)(v*)-q(cj )(v*) = k(T)(T-cj Id)(v*)= k(T)(w*) which is in W!

BUT 0 = m(T)(v*)=(T-cj Id)(qT)(v*).... so q(T)(v*) is in W.

Thus q(cj )(v*)=q(T)(v*)- k(T)(w*) is in W.

But we assumed v* is not in W, so q(cj ) = 0. So the factor (X-cj ) appeared twice in m!  A Contradiction!
Proof of Lemma: We must find a vector v* not in W and a characteristic value cj  where w*=(T-cj Id)(v*) is in W.

Suppose b is in V but not in W.
Consider C={ all polynomials f where f (T) (
b) is in W}.
[There are some non-trivial elements of C since m(T)(b) = 0, so is in C.]
Of all the elements of C, there is a unique non-zero monic polynomial of least degree which we will call g.
[Proof left as exercise for Friday]
Then g is a factor of m. [Proof left as exercise for Friday.] Since b is not in W,  g is not a constant, and so deg( g ) >0.
Since we know all the factors of m, for some
cj ,
(X- cj  ) is a  factor of g.

So g= (X- cj  ) * h, and because g was of minimal degree for polynomials where  f (T) (b) is in W,
h(T)(b)=v* is not in W.

But  w* = g(T)(b) = (T-cj Id)h(T)(b) =  (T-cj Id)(v*) is in W.
End of lemma's proof.

Remark: Every monic polynomial is the minimal polynomial for some linear operator:
Suppose `f` is the polynomial and degree of `f` is `n`, `f = a_0 + a_1 X + a_2 X^2 + ... + a_{n-1}X^{n-1} + X^n`.
Let T: `R^n -> R^n` be defined by  T(`e_i`) = `e_{i+1}` for `i = 1,2,...n-1` and
T(`e_n`) =
`- (a_0e_1 + a_1e_2 + a_2 e_3 + ... + a_{n-1}e_n)`. Then the minimal polynomial of T is `f`.

Example:
`f = (X-2)(X-1)^2 = (X-2) (X^2 -2X +1) = X^3  -4X^2  -3X -2 `. Then using the standard basis T has matrix:
`((0,0, 2),(1,0,3), (0,1, 4))` and  `f` is the minimal polynomial for T.

Nilpotent Operators and Jordan Canonical Form.

Now what about operators where the minimal polynomial splits into powers of linears? or where the minimal polynomial has non-linear irreducible factors?

First Consider the case of powers of linear factors.
The simplest is just a power of X.  (or (X-c)).

Example: D: P3(R) -> P3(R), the derivative. Then the minimal polynomial for D is X4.

Definition:
In general, an operator N is called nilpotent if  for some k>0, Nk =0. The smallest such k is called the index of nilpotency for N, and  if  k is the index of nilpotency for N, then the minimal polynomial for N is Xk .

Proposition: If N is nilpotent of index k and dim V = k, then there is a basis for V , {b1,b2,...bk} where N(bk)= 0 and N(bi) = bi+1 for all i <k.
Proof: Since
the minimal polynomial for N is Xk, there is some vector v* in V where Nk-1 (v*) is not zero but Nk (v*)=0. Let b1 = v* and b2 = N(b1), b3 = N(b2), ...,bk = N(bk-1).
Then clearly N(
bk )= Nk (v*)=0. It suffices to show that (b1,b2,...bk) is linearly independent. Suppose a1b1 + a2b2+...+ akbk= 0. Then a1v* + a2N(v*)+...+ akNk-1 (v*)= 0. Now apply N to obtain Nk-1(a1b1 + a2b2+...+ akbk)= 0 or a1Nk-1b1 + a2Nk-1b2+...+ akNk-1bk= 0. But Nk-1bj =0 for  j>1, so a1Nk-1b1=0. But Nk-1 (v*) is not zero, so a1 = 0.  Now by using Nk-2(a2b2+...+ akbk)= 0 a similar analysis shows that a2 =0. Continuing we can show that a3 =0, ..., ak-1
= 0. But that leaves ak bk = 0 and thus ak = 0, so {b1,b2,...bk} is linearly independent.

Alternatively: Let f = (
a1, a2,...,ak) the polynomial of degree k-1 with a1, a2,...,afor coefficients. If is not the zero polynomial and f(N)(v*)=0 then  f  must be a factor of  Xk . But the assumption is that Nk-1 (v*) is not zero, so f must be the zero polynomial and all of the coefficients are 0. Hence, (b1,b2,...bk) is linearly independent.

Example: Find the basis for D: P3(R) -> P3(R).
• Using the basis (b1,b2,...bk) the matrix for N, M(N) has 0 everywhere except for 1's that are below the main diagonal.
• If  T is an operator on a vector space V with dim V= k and the minimal polynomial for T is (X-c)k .
Then let N = T-cId and  so
Nk =0. So N is nilpotent with index of nilpotency k. Using the basis for V determined in the last proposition we have M(T-cId ) = M(T) - cM(Id), So the matrix of T  , M(T) = M(T-cId) + cM(Id). This matrix is called the Jordan Block matrix of dimension k for the characteristic value c. J(c;k)
• The BIG Picture.
• Theorem (Jordan Canonical Form): Suppose T is a linear operator on V and m is the minimal polynomial for T.
If m =
(X-c1)p1... (X-cm)pm then
there are
T- invariant subspaces W1,....,Wq with V = W1`oplus`... `oplus`Wq. Furthermore the minimal polynomial for T restricted to each subspace Wi is of the form (X-cj)riwhere dim(Wi) =ri and ri=<pj and ri= pj for at least one i and each j.

• Thus there is a basis for V so that using that basis, M(T) is a block matrix where each block has the form of the J(cj ; ri). This matrix is called the Jordan Canonical Form for T.
Fill in examples of possible matrices for T when m =
(X- 2)3(X+3)2X3 (X-1)
where dim (V) =12. Discuss uniqueness.

• By the Fundamental Theorem of Algebra, If V is a finite dimensional vector space over C,  then any linear operator on V has a basis for which M(T) is a Jordan Canonical Form.

• Corollary: For T a linear operator on V, a finite dimensional vs over R or C,  the degree of the minimal polynomial for T is not greater than the dimension of V.
Proof: By the Jordan Theorem, the degree of m is p1 + p2+ ....+ pm which is no greater than the r1+r2+ ... +rq = dim(V).

• Application to powers of matrices:
Suppose A is a n by n matrix with complex coefficients. Then there is a linear operator T on C
n that has A for it's matrix using the standard basis for Cn . By the Jordan Theorem, there is a basis for Cn where M(T)=J is in Jordan form. Thus there is an inverible matrix S where A =S-1 J S.
Now consider A
n = (S-1 J S )n = S-1J n S.
So we can determine the behavior of
An by studying powers of the Jordan blocks.
Do an example  with J( 1/2;3),
J( 1;3), J(2;3) , and J( i ;3).
What conclusions can we infer about the convergence of
An  for large n based on the eigenvalues of A?
If any eigenvalue ,c, has  |c|>1 then the sequence diverges. If |c| = 1 and c is not 1, then the sequence will not converge, c= 1 and the block is J(1;k) with k > 1, then the sequence diverges.
Otherwise, the sequence will converge!
• Look at J(c;3) and J(c;4) to see with examples... using technology.

• Application to Markov Chains:
A markov chain consist of a finite number, n, of states and a matrix T where T(i,j) = the probability that something currently in state j will move to state i after one period of transition. Thus 0 <= T(i,j)  <= 1 for all 1 <= i,j <= n. It is assumed the only possible changes of state from j to i are listed, so T(1,j) + T(2,j) +... +T(n,j) = 1 for any j. If v = (v1, v
2, ... ,vn) gives the initial distribution of objects in the various states, then T(v) is the likely distribution of objects after one period  of transition.
• T(v) i = T(i,1)v1 + T(i,2)v2 +... +T(i,n)vn , so T gives a linear operator on Rn and Cn. Notice that using the standard basis, M(T) = T. If we assume that the probabilities of T remain the same for every period of transition, then T is called a  Markov transition matrix. Given the initial distribution v, the distribution after k periods of transition is T(...(T(v))...) = Tk(v).  Treating T as a linear operator on Cn, there is a basis for Cn that is composed of Jordan blocks. Thus, the question of what happens to the distribution between the states in the long run is determined by the eigen values of T. T is called a regular Markov process, if  for some k, all the entries of Tk are positive.
• DO example to illustrate how this works on graph.

Theorem: If T is a regular Markov process, then 1 is an eigenvalue for T and all other eigenvalues of T have magnitude less than 1. Furthermore: Dim Null(T-Id)=1 and the power of X-1 in the minimal polynomial of T is 1.
• Corollary: For k large, Tk(v) is close to the unique vector v*in Cn where
T(v*)=v* and v1 + v2 +... +vn =v*1 + v*2 +... +v*n.
Illustrate this with example... and technology.
• Use the result to find long run for a 5 state markov process using graph  of transitions and technology to find the limit as the solution to T(v*)=v* or (T-Id)(v*)=0.

• Proof (outline ..based in part on Friedberg, Insel, Spence): Suppose T is a regular Markov process and let J be the Jordan matrix for T.
Fact: If c is an eigenvalue for T, then c is an eigenvalue for the transpose of T, Tt.
Proof of fact (outline): check: f (T)t = f(Tt), so T and Tt have the same minimal polynomial, hence the same eigenvalues.

Lemma 1: If c is an eigenvalue for T, then |c| ≤ 1.
Proof of Lemma 1:  Consider c as an eigenvalue for Tt. Suppose x be an eigenvector with eigenvalue c for Tt.
• Then (Tt(x))i =T(1,i)x1 + T(2,i)x2 +... +T(n,i)xn = cxi for each i. Let b = Max{|x1|,|x2|,...,|xn|}
Then for some k,
|c| b = |c xk| = |T(1,k)x1 + T(2,k)x2 +... +T(n,k)xn|
|T(1,k)x1|+ |T(2,k)x2 |+... +|T(n,k)xn|   [triangle inequality}
|T(1,k)||x1| + |T(2,k)||x2| +... +|T(n,k)||xn|          [  |ab|=|a| |b|  ]
|T(1,k)|b + |T(2,k)|b +... +|T(n,k)|b    [ b =Max{|x1|,|x2|,...,|xn|}]
≤ [|T(1,k)| + |T(2,k)| +... +|T(n,k)|]b = b
since T(1,j) + T(2,j) +... +T(n,j) = 1  and 0 T(i,j).
• Lemma 2: 1 is an eigenvalue for T.
Proof: 1 an eigenvalue for
Tt: T(1,j)1 + T(2,j)1 +... +T(n,j)1 = 1 so Tt has an eigenvector of (1,...,1) = u.
• Lemma 3: J has single blocks for c=1, of the form J(1;1).
Powers of T have real number entries that are never larger than 1. [Prove as exercise.]
But if there is a Jordan block in J  with J(1;k) with k>1, the powers will get much larger than 1.

• Lemma 4: If c is an eigenvalue for T with |c|=1, then c = 1 and Dim(Null(T-Id))= 1.
Proof: Since T is a regular Markov process, we may assume that all entries of T are positive.
First, since |c| = 1, all the previous inequalities in Lemma 1 are actually equalities for any eigenvector x for the eigenvalue c.

b =|c| b = |c xk| = |T(1,k)x1 + T(2,k)x2 +... +T(n,k)xn|
=|T(1,k)x1|+ |T(2,k)x2 |+... +|T(n,k)xn|     (*)
=|T(1,k)||x1| + |T(2,k)||x2| +... +|T(n,k)||xn|
=|T(1,k)|b + |T(2,k)|b +... +|T(n,k)|b   (**)
=[|T(1,k)| + |T(2,k)| +... +|T(n,k)|]b = b .
T(1,k) + T(2,k) +... +T(n,k) = 1
• Notice that for any complex numbers a and b,  |a+b| = |a| + |b|  only if a and b are "colinear", that is, there is some complex number z with |z| = 1 and non-negative real numbers r and where a = rz and b = sz.
Using this fact for  (*)  there must be some complex number z*  with |z*| = 1  and real numbers rj where

rj z* =T(j,k)xfor all j.
• From (**) ,  since b > 0, either T(j,k) = 0  or |xj|=b, but the assumption of regularity is that T(j,k)>0, so |xj|= b > 0  for all j. But then rj z*/T(j,k)=xso
b
=
|xj| = |rj z* /T(j,k) | =  rj |z*| /T(j,k) = rj/T(j,k) for all j. Thus b z* =xfor all j.
SO.... x = bz*(1,1,...,1) and c = 1.
Thus if If c is an eigenvalue for T with |c|=1, then c = 1 and if x is an eigenvector with eigenvalue 1 then x is a scalar multiple of (1,1,...,1), so Dim(Null(T-Id))= 1
EOP for Lemma.

• To Finish the Proof of the Theorem- note that since Dim(Null(T-Id))= 1, the factor of the minimal polynomial that corresponds to the eigenvalue 1 can only be (X-1), since all blocks in the Jordan matrix from the eigenvalu 1 are of the form J(1,1).

Proof of the Corollary:
By the Jordan Theorem, there is a basis for Cn where M(T)=J is in Jordan form. Thus there is an inverible matrix S where T =S-1 J S.
Now consider T
n = (S-1 J S )n = S-1J n S.
Since T has only 1 for a complex eigenvalue of magnitude 1, and all other eigenvalues have magnitude less than 1, and since there is only one block ( J(1,1) ) for the eigenvalue 1. For k large Tk, is close to the matrix  S-1 K S where K(1,1) = 1 and K(i,j) = 0 for (i,j) different from (1,1).
Now one can show that
S-1 K S is the square matrix which has each column equal to the eigenvector with eigenvalue  1and with the sum of its components equal to one.
From this it can be shown that for any v,
Tk(v) is close to the unique vector v*in Cn where
T(v*)=v* and v1 + v2 +... +vn =v*1 + v*2 +... +v*n.

• Do Example. We looked at a 3 by 3 example of a regular Markov chain matrix modelling the movement of rental cars between SF, LA, and Vegas. We assumed there are 400 cars in the fleet. To find the v* suggested by the Theorem, we need to solve Tv* = v* and 400=v*1 + v*2 +v*3 . This is equivalent to solving the system of equations (T-I)v* = 0 with 400=v*1 + v*2 +v*3 . Since the columns of T add to 1, the matrix T-I has linearly dependent rows (they add to 0). replacing the final row with 1 1 1  and  0 with (0,0,400) the equations had a unique solution, 400(3/11,6/11, 2/11). Obviously, if the total number of cars were N, then the result would be N(3/11,6/11, 2/11).  The proportions  are fixed then by  the characteristic vector  (3/11,6/11, 2/11).  Furthermore, this result is independent of the starting distribution, so  if all cars started in SF, the result would be the same, and similarly for LA or Vegas. So for n large Tn is approximately the matrix with each column equal to the transpose of (3/11,6/11, 2/11).

• Notice that T n -> T* where each column of T* is v* with  v*1+... + v*n = 1

Proof of JCF Theorem:  ?? Outline