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Proving eigenvalues and diagonalizability

  1. Nov 12, 2015 #1
    < Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

    Let A be a square matrix of order n such that A^2 = I
    a) Prove that if -1 is the only eigenvalue of A, then A= -I
    b) Prove that if 1 is the only eigenvalue of A, then A= I
    c) Prove that A is diagonalizable.

    For part a and b, I consider A=-I and I then showed that the eigenvalues are -1 and 1 respectively. Is it valid by assuming A=-I and I then proving the eigenvalues are -1 and 1?
    If A=-I, Av=λv
    -v=λv
    therefore, λ=-1.

    For part c, I have no clue how to even start. I did some research, and I haven't learn minimal polynomial so I can't use it to prove.
     
    Last edited by a moderator: Nov 12, 2015
  2. jcsd
  3. Nov 12, 2015 #2

    fresh_42

    Staff: Mentor

    Consider the expression ##(t \cdot E_n - A) \cdot (t \cdot E_n + A)## and the zeros of the characteristic polynomial of ##A## to prove c).
    a) and b) are direct consequences of c).
     
  4. Nov 12, 2015 #3

    Mark44

    Staff: Mentor

    No. You are proving the converse of what you're asked for. For a), assume that -1 is the only eigenvalue of A, then show that A must be -I. Same idea for b).
     
  5. Nov 12, 2015 #4
    Av=λv
    Av=-v
    Av=-Iv
    A=-I
    Can I deduce A=-I from that?
     
  6. Nov 12, 2015 #5
    I didn't learn anything about that in my course. Can you explain that more explicitly? I know the eigenvalue of A are -1 and 1.
     
  7. Nov 12, 2015 #6

    Mark44

    Staff: Mentor

    You've omitted what's given, and a step or two. How do you go from equation 2 above to equation 3?
     
  8. Nov 12, 2015 #7

    fresh_42

    Staff: Mentor

    ##A## is diagonalizable iff there is a matrix ##S## so that ##S^{-1} A S = diag (λ_1, ... , λ_n)##
    ##diag (λ_1, ... , λ_n)## denotes a diagonal matrix with entries ##λ_1, ... , λ_n## on its diagonal and 0 elsewhere. ##E_n = diag(1, ... , 1)##

    The characteristic polynomial ##char_A (t) ## of ##A## in ##t## is defined as ##det ( t \cdot E_n - A )##. Now
    ## char_A (t) \cdot det ( t \cdot E_n + A )##
    ##= det ( t \cdot E_n - A ) \cdot det ( t \cdot E_n + A )##
    ##= det [( t \cdot E_n - A ) \cdot ( t \cdot E_n + A )]##
    ##= det (t^2 \cdot E_n - A^2)##
    ##= det (t^2 \cdot E_n - E_n)##
    ##= (t^2 - 1)^n##
    ##= (t-1)^n (t+1)^n##

    The ##λ_i## are exactly the zeros of ##char_A (t)##. The equation above shows that only ##(t ± 1)## are divisors of ##char_A (t)## and therefore all ##λ_i## must be 1 or -1, proving c) and then a) and b).

    I admit there's eventually an easier way to prove this without determinant and characteristic polynomial.
    Maybe I'm too used to them that I didn't see it. Which rules about matrices are you expected to use?
     
    Last edited: Nov 12, 2015
  9. Nov 12, 2015 #8
    I multiply both sides by identity matrix so it IAv is still Av.
     
  10. Nov 12, 2015 #9

    Mark44

    Staff: Mentor

    I don't believe that the OP is this far along in his/her studies.
    Try to keep your hints aligned with what the person you're attempting to help can handle.
     
  11. Nov 12, 2015 #10

    Mark44

    Staff: Mentor

    Yes, that's clear. What I really meant was how did you go from equation 3 to equation 4?
     
  12. Nov 12, 2015 #11
    I'm not sure whether it is right. But I compared them like coefficients.
     
  13. Nov 12, 2015 #12

    Mark44

    Staff: Mentor

    The trouble is, that isn't guaranteed to work, when you're doing matrix multiplication. For example, it's possible for AB = 0, where neither A nor B is the zero matrix.
    For example:
    ##A = \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}## and ##B = \begin{bmatrix} 0 & 2 \\ 0 & 0 \end{bmatrix}##

    So starting here: Av=-Iv, do you have any other ideas? You know the v is an eigenvector, right? Is there a property of eigenvectors you can use?
     
  14. Nov 12, 2015 #13
    The only properties I know are that they are linearly independent and they form a basis for R^n.
     
  15. Nov 12, 2015 #14

    Mark44

    Staff: Mentor

    "They" - for each problem part, you have only one eigenvector.
    Can any old vector be an eigenvector, or are there any restrictions?
     
    Last edited: Nov 12, 2015
  16. Nov 12, 2015 #15
    If I know that the rank(I-A) + rank(I+A) = n, then does it mean that the dimension of eigenspace of A is n, therefore there are n eigenvectors so A is diagonalizable.
     
  17. Nov 12, 2015 #16
    Eigenvector cannot be the zero vector?
     
    Last edited by a moderator: Nov 12, 2015
  18. Nov 12, 2015 #17

    Mark44

    Staff: Mentor

    Yes!
    OK, now starting from here -- Av = -Iv, can you get another equation and conclude that A = -I?
     
  19. Nov 12, 2015 #18
    Another equation? (A+I)v=0?
     
  20. Nov 12, 2015 #19

    Mark44

    Staff: Mentor

    That's what I was steering you toward. With what you know about v, and what you know about A (from post #1), why can you conclude that A + I must be the zero matrix?

    Note that because of the way matrix multiplication works, it's possible for Bx = 0, with ##\vec{x} \ne \vec{0}## and B not equal to the zero matrix. For example, with ##A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}## and ##\vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}##.
     
  21. Nov 12, 2015 #20

    fresh_42

    Staff: Mentor

    mea culpa. You're right.
     
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