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Proving and disproving matrix properties?

  1. Oct 5, 2009 #1
    I have no idea how to approach these.. any help would be greatly appreciated.
    I can't seem to find this in the book either so if there are any links on the web that relate to these questions please let me know.. Thanks!

    Prove or disprove the following statements. I and 0 represent the identity
    and zero matrix of the same size as A.

    1. If A is a matrix such that A^2 - 3A + 2I = 0, then A - 2I is invertible.
    2. If A is a matrix such that A^2 - 5A + 5I = 0, then A - 3I is invertible.
    3. If A is invertible, then A^2 + I is invertible.
    4. Prove that there is no 5 by 5 matrix such at A^2 + I = 0.
     
  2. jcsd
  3. Oct 6, 2009 #2

    jbunniii

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    Hint for (2): note that

    A^2 - 5A + 5I = (A - 3I)(A - 2I) - I
     
  4. Oct 6, 2009 #3

    jbunniii

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    Hint for (1): note that

    A^2 - 3A + 2I = (A - 2I)(A - I)

    Think about what

    (A - 2I)(A - I) = 0 means. Right away you can see two possible solutions for A. What is (A - 2I) in each case?
     
  5. Oct 6, 2009 #4

    Dick

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    For 2. I would try to prove the contrapositive. If (A-3I) is NOT invertible then A^2 - 3A + 2I is NOT zero. Hint: A-3I not invertible tells you something the existence of a eigenvector of A. For the last two you must mean A to be a real matrix. Otherwise there's a diagonal counterexample for both. Hint: det(A^2)=det(-I).
     
  6. Oct 6, 2009 #5
    hey guys thanks for your help.. so far for one i have
    A = I then A-2I is invertible so the statement is true.
    for two i have
    A = 3I or A = 4I but neither of these work when i plug it back into the original equation?
    the last two are real matrices..
    for four i'm trying to prove that A^2 can not = -I ?
     
  7. Oct 6, 2009 #6

    Mark44

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    For 4, don't forget that your working with 5 x 5 matrices. I'm sure that is significant.
     
  8. Oct 6, 2009 #7

    Hurkyl

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    P.S. keep in mind matrices do not have the cancellation property. While
    A=I and A=2I​
    are both solutions to
    (A-I)(A-2I)=0,​
    there is no reason to think those are the only solutions.
     
  9. Oct 6, 2009 #8

    Hurkyl

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    Really, we cannot help effectively without know what you know about polynomial equations involving matrices. This is part of why our homework guidelines insist that you post what you have done on the problem, and what facts or approaches you believe may be relevant. At the very least, it would help to know what class (and at what level) the question comes from.
     
  10. Oct 6, 2009 #9

    Dick

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    I agree with Hurkyl. You got 1 wrong by picking half of a hint and then jumping to a wrong conclusion. Why don't you concentrate on one of these problems at a time and fully explain your reasoning? Then we can figure out what you know and what you don't know.
     
  11. Oct 6, 2009 #10
    hey in response to your message hurkyl, i'm taking "elementary linear algebra"
    and right now we have covered basic matrix stuff like inverse, determinants, solving systems of linear equations, multiplying/adding matrices etc..
    i think these questions relate to chapter two which is about cofactors, adjoint, cramer's rule, eigenvalues/eigenvectors and diagonilization; however, i missed the past two classes so right now i'm trying to self-learn a lot of this stuff from the book...
     
  12. Oct 6, 2009 #11

    Dick

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    You probably do need chapter 2 for problem 2. It's about eigenvectors. You can handle 1 and 3 without it if you are clever with finding counterexamples. You can also do 4 if you learned determinants.
     
  13. Oct 6, 2009 #12

    jbunniii

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    A = I is ONE solution. The factorization I gave in my earlier hint shows that there is (at least) one more solution, namely A = 2I. Is A - 2I invertible in this case?
     
  14. Oct 6, 2009 #13

    jbunniii

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    Actually problem 2 is trivial with the hint

    A^2 - 5A + 5I = (A - 3I)(A - 2I) - I

    although this doesn't illustrate any sort of general method for solving similar problems.

    Without more context it's hard to know what machinery is expected to be available for these problems.
     
  15. Oct 7, 2009 #14

    HallsofIvy

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    As jbunniii said originally, "A^2 - 5A + 5I = (A - 3I)(A - 2I) - I = 0" so (A- 3I)(A-2I)= I. Yes, it is trivial!
     
  16. Oct 7, 2009 #15

    Dick

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    That is clever. I didn't quite see where that hint was going.
     
  17. Oct 7, 2009 #16
    thanks everyone for your input it really helped me out..
    for one i have A= I or 2I but if A=2I then A-2I is not invertible so the statement is false

    for two i don't really understand Jbunniii's hint.. i originally thought A equaled 4I and 3I but when i put these back into the original A^2 - 5A + 5I equation, these matrices do not equal 0.. i decided to use the quadratic formula so now i have A = (5+/- sqroot5)/2 and these answers both give 0 when plugged back into the original equation and they are both invertible for A-3I so for two i have the statement is true?
     
  18. Oct 7, 2009 #17

    jbunniii

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    A^2 - 5A + 5I = (A - 3I)(A - 2I) - I = 0 <==> (A - 3I)(A - 2I) = I

    What does the equation

    (A - 3I)(A - 2I) = I

    mean? What does it say about the matrices (A - 3I) and (A - 2I)?
     
  19. Oct 7, 2009 #18
    .. they are inverses?
     
  20. Oct 7, 2009 #19

    jbunniii

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    Correct. The inverse of (A-3I) is (A-2I) and vice versa. Therefore you can answer "yes" to the original question (is A - 3I invertible?), and you got the bonus information that (A - 2I) is invertible as well, and the even more bonus information that they are inverses of each other!
     
  21. Oct 7, 2009 #20
    Anyone has any ideas how to do 3 and 4?
     
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