Martin Flashman's Courses
MATH 344 Linear Algebra Fall, 2003
Class Notes and Summaries

week1 8-25 8-27 8-29
week 2 9-1 9-3 9-5
week 3 9-8 9-10 9-12
week 4
9-15 9-17
9-19
week 5
9-22
9-24
9-26
week 6
9-29
10-1
10-3
week 7
10-6
10-8
10-10
week 8
10-13
10-15
10-17
Exam

week 9
10-20
10-22
10-24
week 10
10-27
10-29
10-31
week 11
11-3
11-5
11-7
week 12
11-10
11-12
11-14
week 13
11-17
11-19
11-21
week 14
NoClasses

11-24
11-26
11-28
week 15
12-1
12-3
12-5
week 16
12-8
12-10
12-12


8-27-03







  • Abstract Vector Spaces


  • 9-5-03




    9-8-03
  • We began class with a discussion of two problems from the assignment relating to subspaces.

  • In particular we showed that if U and W are subspaces of V and U union W is also a subspace of V, then either U < W or W < U.
  • Comment: Make correction on remarks made about countability last class.
  • Do Examples
  • F[X] = { f in F, where f(n) = 0 for all but a finite number of n.} < F


  • 9-10-03
  • (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}.

  • 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) = akxk  where ak is in F} then Uj^Uk = {0} for j not equal to k.

    ...



    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+  ... + 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 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+  ... + un for  uk in Uk implies u1=u2=...=un=0.
    Theorem: (Prop 1.9) V = U#W if and only if V = U+W and U^W={0}.



    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 Wc = { v in Fn where vA = cv}.
    9-15-03
    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=(

    5/61/6
    1/43/4
     
    )
     
      Example B.  
      vB = cv where


      B=(

      2/31/3
      1/43/4
       
      )
       
    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)(

    5/61/6
    1/43/4
     
    )= c(x,y)
     
      Example B


      (x,y)(

      2/31/3
      1/43/4
       
      )= 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  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 W5/12 .


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

    9-17-03
    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 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,  ... , un in S  and scalars a1, a2,  ..., an NOT ALL 0 in F where 0 = a1u1+ a2u2+  ... + anun .
    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 a1, a2,  ..., an in F where 0 = a1u1+ a2u2+  ... + anun , the scalars are all 0, i.e. a1 = a2 = ... =an = 0 .


    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 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, the R*(A) = the column space of A.

    "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.


    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.


    Comments:

    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.


    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.


    Problem 2.12: Suppose p0,...,pm are in Pm(F) 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(F).

    But SPAN ( 1,x, ..., xm) = Pm(F) .
    By repeatedly applying Lemma 2.4 to these two sets of m+1 polynomials as in Theorem 2.6, we have SPAN (p0,...,pm)=P
    m(F)
    , a contradiction
    . So {p0,...,pm} is not linearly independent.
    End of proof.



    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.


    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.


    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,  ... , un) .  If v is in V, then there are unique vectors  scalars a1, a2,  ..., an in F where v = a1u1+ a2u2+  ... + anun . The scalars are called the coordinates of v w.r.t. B, and we will write v = [a1, a2,  ..., an]B.


    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)(

      2/31/3
      1/43/4
       
      )= c(x,y)
       


    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 W5/12 .
    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.


    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


    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.



    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 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 :  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.
     


    Monday 9-29.

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


    More notes on Chapter 1 and 2


    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


    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+  ... + un for  uk in Uk. Then each ui =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=...=un=0 and V = U1 #...# UnEOP


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

    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.


    Visualize with Winplot?


    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).


    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.


    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.


    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+  ... + anun
      ... then let T(v)  = a1S(u1)+ a2S(u2)+  ... + anS(un) ....
    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.


    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.


    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.


    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.


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

     

    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).


    Oct.8
    Footnote on notation for Matrices: If the basis for V is B and for W is C and T:V->W,
    the matrix of T with respect to those bases can be denoted MBC(T). The matrix for a vector V is denoted
    MB(v).  Thus MC(T(v))=MBC(T)MB(v).

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

    Invertibility of Linear Transformations:
    Def'n: T: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 (3.17): T is invertible iff  T is 1:1 and onto  (injective and surjective).
    (i) =>  Assume S... (a)show T is 1:1. (b) show T is onto.
    (ii) <=  Assume T is 1:1 and onto. Define S. Show S 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)
    Comments:
    (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.
    [Start 10-10 here!]
    Theorem (3.18) Suppose V and W are finite dimensional v.s.
    Then
    V=T W if and only if dim(V) = dim(W).
    Proof: Done! See text!
    Theorem (3.19) Suppose B= (v1,...,vn) and C=(w
    1,...,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.

    Cor. Prop 3.20: Dim L(V,W) = Dim(V) Dim(W).


    Prop 3.21. 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.


    End of material for midterm exam

    10-13 Connection to square matrices:
    A is invertible  is equivalent to....Systems of equations statements.

    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 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 W5/12 .
    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)=v1  while 5/12 had an eigenvector (1,-1)=v2. These two vectors are a basis for R2.
    B=(v1,v2)
    What is the matrix of T using this basis.

    MBB(T)=(

    1 0
    0
    5/12
     
    )
    Using this basis and matrix makes it easy to see what happens when the transfomation is applied repeatedly:
    MBB(Tn)=[MBB(T)]n=(

    1 0
    0
    5/12
     
    )n=

    1 0
    0
    (5/12)n
     


    10-15 Change of basis:
    So ... What is the rel
    ationship between this very nice matrix for T that results from using the basis B of eigenvectors and the matrix for T that uses the standard basis, E = (e1,e2)?

    The key to understanding the relationship between these is the identity map! 
    We consider the matrix for the identity operator using B for the source and
    E for the target.

    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:
    MBE(Id)MBB(T)MEB(Id)=MEE(IdTId)=MEE(T)
    and
    MEB(Id)MEE(T)MBE(Id)=MBB(IdTId)=MBB(T).

    If we let P =MEB(Id) and L = MBE(Id) = P-1, then we have

    LMBB(T)P =P-1MBB(T)P=MEE(T)
     or

    M
    BB
    (T)P= PMEE(T)
    and
    PMEE(T)L=MBB(T).

    10-20
    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)MBB(T)MEB(Id)=MEE(T)
    and
    MEB(Id)MEE(T)MBE(Id)=MBB(T).

    If we let P =MEB(Id) and L = MBE(Id) = P-1
    [
    LP =MBE(Id)MEB(Id) = MEE(Id)= In ]
    then we have
    LMBB(T)P =P-1MBB(T)P=MEE(T)
     or
    MBB(T)P= PMEE(T)
    and likewise  PMEE(T)L=MBB(T).

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

    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: If P is an invertible n by n matrix, then TP:Fn ->
    Fn defined by the matrix P where MEE(TP) =P maps every basis B of Fn to a basis, TP(B)= B' .


    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 eignevector for T. 
    Proposition: a is an eignevalue 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.
    10-22
    Fact: If T is diagonalizable with distinct eigenvalues  a1,...,ak , then S = (T-
    a1Id)(T-a2Id).... (T-akId) = 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-akId)(v) = (T-a1Id)(T-a2Id).... (T-ajId)... (T-akId)(T-ajId)(v) = 0.
    What about the Converse? If  there are distinct scalars a1,...,ak where S(v) = (T-a1Id)(T-a2Id).... (T-akId)(v) = 0 for any v in V, is T diagonalizable? we will return to this later....!


    10-24
    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... 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] the 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).
    If A is any algebra over F with unity, and f is in F[X],
    f= (a1,a2,...an,0,0,...), then we have a function, f :A->A defined by
    f
    (t) =
    a1 I + a2t+ ... +antn  where I  is the unity for A.

    In particular (i)A can be the field F itself, so f is in P(F).
    Example: F = Z2. f  = X2 + X in F[X]. Then f is not (0,0,0....) but f(t) = 0 for all t in F.



    (ii) A can be L(V) where Vis 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].

    10-27
    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":
    10-29
    Theorem: Suppose V is nontrivial f.d.v.s / C.  T in L(V). Then T has an eigenvalue.
    Comment: First we considered this with the
    Coke/Pepsi example B:
    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(v) + a1T(v) + a2T2(v)=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) (1/3,1/3)= 0, so (1/3,1/3) is an eignvector for T with eigemvalue 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, T as usual. 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 C[X] where f (T) = 0, i.e., f(T)(v) = 0 for all v  in V.
     


    10-31
    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).

    Theorem: 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!

    Prop. Suppose m in F[X] is the mininal polynomial for T.  Then T has eigenvalues 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

    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))=
    11-3
    Invariant Subspaces: V, T as usual.
    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.
    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 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 #W2 #...# 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.
    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 )h.

    THUS...
    q(T)(v*)-q(cj )(v*) = h(T)(T-cj Id)(v*)= h(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*)- h(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!
     11-5
    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 polynomial is the minimal polynomial for some linear operator: STILL To be filled in here.
    Example: f = (X-2)(X-1)^2. = (X-2) (X^2 -2X +1) = X^3  -4x^2 ...
    Define

    Nilpotent Operators.

    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 .

    11-7

    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).
    Comments:
    12-10