• 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?



A= ( 0 1 1 0   )

A= ( 1 1 0 1   )

A= ( r 0 0 s   )

 
1. r = 1/2 s= 1/2
2. r= 2, s = 1/2
3. r=1, s= -1

• Complex numbers: C = { z = a+bi, where a and b are real numbers}

The arithmetic of C. ||z|| = sqrt(a^2 + b^2)
z = ||z||(cos(t) +i sin(t))    [called the "polar form" of the complex number]

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 you use the product of the magnitudes of z and w to determine the magnitude of the product and use the sum of the angles to determine the angle of the product.

8-29-03
Powers of complex numbers.
Based on the previous result on products, zⁿ =  ||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, the the powers of Z will oscillate around the circle of radius 1 without having any limit.

Notation: cos(t) + i sin(t) is somtimes written as cis(t).
Note: If we consider the series for e^x = 1 + x + x²/2! +x³/3! + ...
then e^ix = 1 + ix + (ix)²/2! +(ix)³/3! + ... = 1 + ix -  x²/2! - ix³/3! + ...
... = cos(x) + i sin(x)
Thus e^pi = cos(p) + i sin(p)= -1.  Thus ln(-1) = p i.

Matrices with complex number entries.
If r and s are complex numbers in the matrix A, then as n get 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,  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, making 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, or 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.

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

Definition- Axioms See Fields.
Examples: R, Q, C, Z₂, Z_p, where p is a prime number, F₄, {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.

9-3-03
• The class began with a discussion of the properties of a field with 4 elements- from the homework assignment.
• 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, Z_p, 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.
• Motivation Question II

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

A. First resample: 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 Resample: 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.  v0TA = v1 where TA= ( 5/6 1/6 1/4 3/4   )

Example B.  v0TB = v1 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 v_n  when n is large?
Does the distribution of v_n 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 = v_n.

Vectors and vector spaces.
• Traditional vectors: vectors in R², R², and Rⁿ.
• Lists. Definition: A list is an ordered collection of a finite number of n-objects, separated by comma and surrounded by parentheses.

A list is also called an n-tuple.
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 = (v₁  , v₂  , ..., v_n  ), w = (w₁  ,w₂  , ..., w_n  ) and v_i = w_i  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 = (v₁  , v₂  , ..., v_n  ) where  v_i  is 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 =  (v₁  , v₂  , ..., v_n  ) and w =  (w₁  ,w₂  , ..., w_n  ) are in Fⁿ  and a is in F then v + w = (v₁  + w1, v2 + w₂  , ..., v_n+w_n  ) and
a*v =  (av₁  , av₂  , ..., av_n  ) using the addition and multiplication of F for the operations on the individual elements of F.
• Proposition: With the operations just defined on Fⁿ  the following properties are true:
• If v, w, and z are in Fⁿ  and a and b are in F, then v+w and a*v are in Fⁿ  [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
• 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]

• 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]

• Vectors 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: When S = {1,2,...,n} we can give a 1 to 1 correspondence between V(S,F) and Fⁿ by letting  f (k) = v_k when v=(v₁  , v₂  , ..., v_n  ) 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)  and
a*v is the function defined by (a*v)(s) =  a v(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)  the following properties are true:
• 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 s  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 coresponds to the function f  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 + ...+ a_nxⁿ} where a_i is in F for each i} = the polynomial functions from F to F.
• 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}.

• Abstract Vector Spaces (Again)
• Definition: Suppose V is a nonempty 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]

• Simple properties of Vector spaces
• 0  is unique.
• -v is unique.
• 0v= 0.
• a*0 = 0.
• (-1)v = -v.
• If v+z = w + z then v=w. [cancellation]
• If a*v = 0, then either a = 0 or v = 0.

• 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 0v=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.

Do 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 F^k } <  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 F^k where for some v in Fⁿ, vA= w  } <  F^k
• 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 F^k 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).
• Fact: If U<W and W <V, then U<V.

Proof:The operation on U is the same operation for W and 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) = a_kx^k  where a_k is in F} . Then U0+ U1+ U2+  ... + Un = {f : f (x) = a₀ + a₁x + a₂x² + ...+ a_nxⁿ where a₀ ,a₁ ,a₂,...,a_n are in F}.

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}= ...
Let Uk = {f in P(F): f(x) = a_kx^k  where a_k is in F} then Uj^Uk = {0} for j not equal to k.

...

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9-12-03
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+  ... + u_n for  u_k in Uk is unique, i.e., if v = u₁'+ u₂'+  ... + u_n' for  u_k' in Uk then u₁ = u₁', u₂=u₂', ... , u_n=u_n'. In these notes we will write V =  U1# U2#  ... # Un.
Examples:Uk = {v in Fⁿ: 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: (Prop 1.8) V =  U1# U2#  ... # Un if and only if (i)U1+ U2+  ... + Un = V AND 0=u₁+ u₂+  ... + u_n for  u_k in Uk implies u₁=u₂=...=u_n=0.
Theorem: (Prop 1.9) V = U#W if and only if V = U+W and U^W={0}.

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Examples using subspaces and direct sums in appplications:
Suppose A is a square matrix (n by n) with entries in the field F.
For c in F, let W_c = { v in Fⁿ where vA = cv}.
9-15-03
Fact: For any A and any c,  W_c< Fⁿ . [Comment: for most c, W_c= {0}. ]
Definition: If W_c is not the trivial subspace, then c is called an eigenvalue or characteristic value for the matrix A and nonzero elements of W_c  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 W_c 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 = W_c1 + W_c2 for these subspaces? Is this sum direct?

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Focusing on Example B we consider for which c  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
(8-12c)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  or 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 W_5/12 .

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General Claim: If c is different from k, then W_c ^ W_k = {0}
Proof:?
generalize?
What does this mean for  v_n  when n is large?
Does the distribution of v_n when n is large depend on v₀?

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9-17-03
Application 2: For c a real number let
W_c = {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!

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If c is different for k, then W_c ^ W_k = {0}
Proof:...

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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,  ... , u_n in S  and scalars a₁, a₂,  ..., a_n in F where v = a₁u₁+ a₂u₂+  ... + a_nu_n .
Comment: For Axler: S is a finite set.
 
Def'n. Span (S) = {v in V where v is a linear combination of vectors in S}
Span (S) = V we say that S spans V. "finite dimensional" v.s.

Linear Independence.
Def'n. A set of vectors S is linearly dependent if there are vectors u₁, u₂,  ... , u_n in S  and scalars a₁, a₂,  ..., a_n NOT ALL 0 in F where 0 = a₁u₁+ a₂u₂+  ... + a_nu_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,  ... , u_n in S  and scalars a₁, a₂,  ..., a_n in F where 0 = a₁u₁+ a₂u₂+  ... + a_nu_n , the scalars are all 0, i.e. a₁ = a₂ = ... =a_n = 0 .

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9-19-03
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 F^k where for some v in Fⁿ, vA= w  } <  F^k.
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, the R*(A) = the column space of A.

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"Infinite dimensional" v.s. examples: P(F), F^∞, C^∞ (R)
P(F) 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 P(F) cannot equal SPAN(p1,...,pn) for any finite set of polynomials- i.e, P(F) is NOT finite dimensional.

Some Standard examples.

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2.4 Linear dependence Lemma : Suppose S is a  finite linearly dependent set indexed by 1,2,.. , n and v1  is not 0, then for some index j,  vj is in the span(v1,...v(j-1)) and Span (S) = Span(S -{vj}).
Proof:  See LA p25.

2.6 Theorem: Suppose S is a finite set of vectors with V = Span (S) and T is a linearly independent set of vectors in V. Then T is also finite and n( T)< = n(S)
Proof:  See LA p25-26.

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Comments:

• (2.4) shows how to constuct a basis for a non trivial finite dimensional v.s., V. Start with a finite set of vectors  S that span V. We can assume S has some non-zero vector in it. Put that element first.

• If S is linearly independent you are done. If not, apply (2.4) repeatedly until the resulting set of vectors is linearly independent. This must happen since at worst you will be left with v1 which was not 0. Thus we have proven

Theorem 2.10: Every finite spanning list in a vector space can be reduced to a basis.
and the Corollary (2.11). Every finite dimensional vector space has a basis.

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Comment:The proof of Theorem 2.6 also shows that given T, a  linearly independent subset of V and S, a finite set where SPAN(S) = V, one can step by step replace the elements of S with elements of T at the beginning of the list of vectors, so that eventually you have a new set S' where Span(S') = V and T  contained in  S'. Now by applying repeatedly the Lemma to S', one will arrive at a set B that is a basis for V with T contained in B.  This proves

Theorem 2.12: Every Linearly independent subset of a finite dimansional vector space can be extended to a basis of the vector space.

Prop: A Subspace of a finite dimensional vs is finite dimensional.

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Problem 2.12: Suppose p₀,...,p_m are in P_m(F) and p_i(2) = 0 for all i.
Prove {p₀,...,p_m} is linearly dependent.

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

(a₀p₀+..+a_mp_m )(2) = a₀p₀(2)+..+a_mp_m(2) = 0

and since u(x)= 1 has u(2) = 1, u (= 1) is not in the SPAN(p₀,...,p_m).
Thus SPAN(p₀,...,p_m) is not P_m(F).
But SPAN ( 1,x, ..., x^m) = P_m(F) .
By repeatedly applying Lemma 2.4 to these two sets of m+1 polynomials as in Theorem 2.6, we have SPAN (p₀,...,p_m)=P_m(F), a contradiction. So {p₀,...,p_m} is not linearly independent.
End of proof.

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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.

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

Proof: fill in ... based on 2.6.

Define dimension of a finite dimensional v.s. over F.

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9-22: Filled in much above on Bases and the proofs of theorems about bases.

9-24
Discuss dim({0}).
What is Span of the empty set? Characterize 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!

2.8: bases and representation of vectors in a f.d.v.s.

Suppose B is a finite basis for V with its elements in a list, (u₁, u₂,  ... , u_n) .  If v is in V, then there are unique vectors  scalars a₁, a₂,  ..., a_n in F where v = a₁u₁+ a₂u₂+  ... + a_nu_n . The scalars are called the coordinates of v w.r.t. B, and we will write v = [a₁, a₂,  ..., a_n]_B.

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Examples: In R², P₄(R).

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)

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We considerd 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 W_5/12 .
Now we can use technology to find eigenvectors in each of these subspaces.
Matrix calculator, gave as a result that the eignevalue 1 had an eigenvector (1,4/3)  while 5/12 had an eigenvector (1,-1). These two vectors are a basis for R².

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Dimension Results: 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) S contains a basis, B.  If B is smaller than S then B has less than n elements, which contradicts that fact that any basis for V has exactly n elements. So B = S and S is a basis.
IRMC

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9-26
2.18: 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.

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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).

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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 trasnformation.
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 matric 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 :  T_A  (left argument) and _AT (right) are linear transformations on F^k and Fⁿ.
T_A  (x) = x A for x in F^k 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.
 

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Monday 9-29.

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

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More notes on Chapter 1 and 2

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1.9:V = U # W if and only if V = U+W  and U^W={0}.
Proof: => suppose v is in U^W, then v=u in U and v=w in W, so 0 = u-v. But since V= U#W, this means u=w = 0 so v=0, so U^W={0}.
Note: This argument extends to V as the direct sum of any family of subspaces.

<= Suppose u is in U and w is in W and u+w = 0. Then, u = -w so u is also in W, and thus u is is U^W={0}. So u=0 and then w= 0 . Since V=U+W, we have by 1.8, V=U#W. EOP

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2.19 If V is f.d.v.s.  and U1, ...Un are subspaces with  V = U1 +...+ Un and
dim(V) = dim(U1)+...+ dim(Un) then V = U1 #...# Un

Proof outline: Choose bases for U1, ..., Un and let B be the union of these setes. Since V = U1 +...+ Un every vector in v is a linear combination of elements from B. But B has exactly dim(U1)+...+ dim(Un) = dim(V) elements in it, B is a basis for V. Now suppose  0=u₁+ u₂+  ... + u_n for  u_k in Uk. Then each u_i =can be expressed as a linear combination of the basis vectors for Ui, and the entire linear combination is 0 implies that each coefficient is 0 because B is a basis. So u₁=...=u_n=0 and V = U1 #...# Un.  EOP

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How do you find a basis for the SPAN(S) in Rⁿ?
Outline of use of row operations...

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Back to linear transformations:
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.

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Visualize with Winplot?

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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).

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Coke/Pepsi example B: T(x,y) =(2/3 x +1/4 y, 1/3 x+3/4 y)
T(v₀) = v₁, T(v₁) = v₂.... T(v_k)=T(v_k+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.

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D... Differentiation: on polynomials, on ...

Example: (D(f))(x) = f' (x) or D(f) = f'.

T(f)(x) = f''(x) - f(x) or T(f) = DD(f) - f = (DD-Id) f.

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Wednesday 10-1
Theorem: T : V->W  linear, B a basis, gives S(T):B ->W.
Suppose S:B -> W, then there is a unique linear transformation T(S):V->W such that S(T(S))=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+  ... + a_nu_n  ... then let T(v)  = a₁S(u₁)+ a₂S(u₂)+  ... + a_nS(u_n) ....
This well defined since the representation of v is unique. Left to show T is linear.  Clearly... if u is in B then S(T(S))(u) = S(u).
Example: T: P(F) -> P(F).... S(xⁿ) = nx ^n-1.
Or another example: S(xⁿ) = 1/(n+1) x ⁿ⁺¹.

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Algebraic stucture 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: L(V,W) is a vector space over F.

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(TU) = (ST)U.

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Key focus: L(V,V) , the set of linear "operators" on V.... also called L(V).
If T and U are in L(V) then UT is also in L(V).  This is the key example of what is called a "Linear Algebra"... a vector space with an extra internal operation usually described as the product. That satisfies the distributive and associative properties.

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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.

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

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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.

Friday 10-3
More details to follow on this lecture:
he first part of the lecture discussed the importance of the Null Space of T, N(T) is undertanding 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)  the 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 trasnformation 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.

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.

Major result of the day: 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: Done in class- see text: 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).

Next: Monday. Oct.6. Matrices  and Linear transformations. (with Dr. B).

MEE(T)=

(

2/3 1/4 1/3 3/4

)

 

MBE(Id)=

(

1 1 4/3 -1

)

 

MBB(T)=

(

1 0 0 5/12

)

 

• Using the basis (b₁,b₂,...b_k) 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 N^k =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)

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• 11-10
  
• 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-c_m)^pm then
  there are T- invariant subspaces W1,....,Wq with V = W1#... #Wq. Furthermore the minimal polynomial for T restricted to each subspace Wi is of the form (X-c_j)^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(c_j ; 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).
  

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• 11-12
  
• Application to powers of matrices:
  Suppose A is a n by n matrix with complex coefficients. Then there is a linear operator T on Cⁿ 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 Aⁿ = (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.
  
  

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• 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, v₂, ... ,v_n) 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)v_n , 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))...) = T^k(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 T^k are positive.
• DO example to illustrate how this works on graph. 

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• Proof on Monday 11-17 for the following
  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, T^k(v) is close to the unique vector v*in Cⁿ where
  
  T(v*)=v* and v₁ + v₂ +... +v_n =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.
  Notice: If c is an eigenvalue for T, then c is an eigenvalue for the transpose of T, T^t.
  [ f (T)^t = f(T^t), so T and T^t have the same minimal polynomial, hence the same eigenvalues.   Write up the details of this as an exercise!]
  Lemma 1: If c is an eigenvalue for T, then |c| ≤ 1.
  Proof of Lemma 1:  Consider c as an eigenvalue for T^t. Suppose x be an eigenvector with eigenvalue c for T^t.
• Then T^t(x)_i =T(1,i)x₁ + T(2,i)x₂ +... +T(n,i)x_n = cx_i for each i. Let b = Max{|x₁|,|x₂|,...,|x_n|}
  Then for some k,
  
  |c| b = |c x_k| = |T(1,k)x₁ + T(2,k)x₂ +... +T(n,k)x_n|
  ≤|T(1,k)x₁|+ |T(2,k)x₂ |+... +|T(n,k)x_n|
  ≤|T(1,k)||x₁| + |T(2,k)||x₂| +... +|T(n,k)||x_n|
  ≤|T(1,k)|b + |T(2,k)|b +... +|T(n,k)|b
  ≤ [|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 T^t: T(1,j)1 + T(2,j)1 +... +T(n,j)1 = 1 so T^t has an eigenvector of (1,...,1) = u.

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• 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.
•  Start here on 11-19.
• 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 x_k| = |T(1,k)x₁ + T(2,k)x₂ +... +T(n,k)x_n|
  =|T(1,k)x₁|+ |T(2,k)x₂ |+... +|T(n,k)x_n|     (*)
  =|T(1,k)||x₁| + |T(2,k)||x₂| +... +|T(n,k)||x_n|
  =|T(1,k)|b + |T(2,k)|b +... +|T(n,k)|b   (**)
    =[|T(1,k)| + |T(2,k)| +... +|T(n,k)|]b = b .
  And we already have that 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 real numbers r and s  where a = rz and b = sz.
  Using this fact for  (*)  there must be some complex number z*  with |z*| = 1  and real numbers r_j where
  
  r_j z* =T(j,k)x_j for all j.
• From (**) ,  since b > 0, either T(j,k) = 0  or |x_j|=b, but the assumption of regularity is that T(j,k)>0, so |x_j|= b > 0  for all j. But then r_j z*/T(j,k)=x_j  so
  b =|x_j| = |r_j z* /T(j,k) | =  r_j |z*| /T(j,k) = r_j/T(j,k) for all j. Thus b z* =x_j for all j.
  
  SO.... x = bz*(1,1,...,1) and c = 1.   EOP
  
• Finish Proof of Theorem and corollary.
  

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  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
  

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  12-1
  Inner Products:  Unit vectors ("normal"). Orthogonal vectors.
• V... finite dimensional i.p.s.  Notation: If  w = x+ yi then use ~w to denote the conjugate of w, so ~w = x - yi.
  Orthonormal basis: ||v_j||=<v_k,v_k> = 1. <v_k,v_j> = 0 for  k and j different.
  
• Suppose B = (v₁,v₂ ... v_n) is an orthonormal basis for Fⁿ.  Let A be the matrix that has each row  determined as the coordinates of the basis B in terms of the standard basis. Let A* be the matrix that is the conjugate of the transpose of A. Then AA* = I.
  
  
• If  B = (v₁,v₂ ... v_n) is an orthonormal basis for V then v = sum <v,v_j>v_j .
  Proof: v = Σ a_jv_j so <v,v_i> = <Σ a_jv_j , v_i> = Σ a_j <v_j,v_i> = a_i . 
  
• Proposition: If S is an orthonormal set of vectors then S is linearly independent.
  Proof: If 0= Σ a_jv_j then 0=<0,v_i> = <Σ a_jv_j , v_i> = Σ a_j <v_j,v_i> = a_i for each i, so the vectors are linearly independent..

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• How to find an orthonormal basis for a finite dimensional vector space.
  Convert a basis C= (w₁,w₂ ... w_n). [Called the Gram-Schmidt process]: Start: v₁= w₁ / ||w₁||.
  
  Next: Let  z₂= w₂ - <w₂,v₁>v₁, so <z₂,v₁ > =0. Now let v₂= z₂ / ||z₂||. Then <v₁,v₂ > = 0 and Span(v₁, v₂ ) = Span (w₁,w₂ ).
• Continue....Let  z₃= w₃ - (<w₃,v₁>v₁ + <w₃,v₂>v₂), so <z₃,v₁ >=<z₃,v₂>  =0. Now let v₃= z₃ / ||z₃||. Then <v₃,v₁ >=<v₃,v₂ > = 0 and Span(v₁,v₂,v₃>) = Span (w₁,w₂ ,w₃).  and so on.... until we are done.
  
• An Alternative: v₃= w₃ - (<w₃,v₁>/ ||v₁||² v₁ + <w₃,v₂>/||v₂||² v₂) , etc.
• Do an example  in R ³; discuss the geometry of this. 

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• 12-3
  Miscellaneous topics:
• Orthogonal Linear Operators:
  Def. T:V-> V in L(V) is called an orthogonal if ||T(v)|| = ||v|| for all v in V.
  Relation to Inner Products, angles:
  Prop: T is orthogonal if and only if <T(v),T(w)> =<v,w> for all v and w in V.
  An Orthonormal operator is 1:1 and "preserves the angles between vectors", in particular it transforms a set of orthonormal vectors to a set of orthonormal vectors.
  An Orthonormal operator will transform an orthonormal basis to an orthonormal basis.
  O(n): If V = Rn with the usual inner product, then O(n) = { T in L(V): T is orthonormal}
  Prop: If T and S are in O(n) then , so TS and T^-1 are also in O(n).
  Comment: Since Multiplication of operators is associative, and clearly, Id is in O(n), O(n) forms a "group".... called the "orthogonal group."
  12-5
  Relation to matrices: A square matrix defines a linear operator by multiplication.  If T is the operator from the matrix M then using the standard basis, M(T) = M. Then T is orthogonal if and only if M^t*M = I.  Thus O(n) = {n by n matrices M | M^t*M = I}
  
  

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• The determinant: An axiomatic approach with multilinearity.
  
  Definition: V a fdvs  over F. A Function T: V ⁿ -> F is called a multilinear function on V if
  T(v₁,..., v_n) is a linear function for variable vector, i.e. for all i,
   T(v₁,. ,av_i+v'_i,.., v_n)= aT(v₁,.., v_i,..., v_n) + T(v₁,..,v'_i,..., v_n)
  Prop.: If T is multilinear then T(v₁,...,0,..., v_n)=0.
  
• T is called alternating if  
  T(v₁,...,v_i ,..., v_j, ..., v_n) = -T(v₁,... ,v_j,...,v_i,..., v_n)

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• Prop: If 1+1 is not 0, then T is alternating if and only if T(v₁,..,v_i ,...,v_i,.., v_n)= 0 
• Assume for the remainder of this topic that  1+1 is not 0 in F.

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• 12-8
  
• If T is alternating and multilinear then   (v₁,. ,v_i,...,av_i+v_j,.., v_n)= T(v₁,..,v_i ,...,v_j,..., v_n)
  
• Theorem: There is a unique alternating multilinear function D on Fⁿ  satisfying the property that D(e1,...,en) = 1.
  Furthermore when considered as a function on the n by n matrices with coefficients in F,
  D(M) = 0 if and only if M is singular.
  
• Start of Proof: Every square matrix is row equivalent to an upper triangular matrix. [Math 241.]
  Lemma: If M is an upper triangular matrix, then D(M) is the product of the entries along the main diagonal.
  Proof:Let the product be denoted p. Then D(M) = pD(M') where M' is upper triangular with 1's along the main diagonal. Then since D is alternating and multilinear, D(M') = D(I) = 1.
• D(MP) = D(M)D(P) ..... Proof : If D(P) = 0 then P is singular and so is MP, so D(MP) = D(M)*D(P).
  Suppose D(P) is not 0. Let D* be defined on M by D*(M) =1/D(P)* D(MP). Show D* is multilinear, alternating and D*(I)=1.
  
• Cor. If M is invertible with inverse P then D(P) = 1/D(M).
• Cor. If T is a linear operator on a fdvs / F then D(T) = D(M(T)) is well defined using any basis.