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Square matrix proof

  1. Feb 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Show that if B is a square matrix that satisfies the equation
    B^2 - 2B + I = 0
    then
    B^-1 = 2I - B


    2. Relevant equations



    3. The attempt at a solution

    I've never before done a proof involving actual unknown matrices. Basically i know what a square matrix is (n x n matrix) im also pretty sure that I is the identity matrix. Ive tried algebraic operations on the equations to make them equal, however im rusty on what can and cant be done with matrices. Any help even a starting location will be much appreciated
     
  2. jcsd
  3. Feb 8, 2007 #2

    Tom Mattson

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    1. First deduce the possible values for the determinant of B. If B is not singular, then B-1 exists.

    2. If B-1 exists (hint: it does exist), then you can multiply both sides of the equation by it.
     
  4. Feb 8, 2007 #3
    AHHH thank you! I was able to get the first equation to equal what it was suppose to be. If i'm faced with questions like these what do you recommend i do first? Also i did not really understand what you ment by "First deduce the possible values for the determinant of B. If B is not singular, then B-1 exists"

    thanks for the help!
     
  5. Feb 8, 2007 #4

    radou

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    A recommendation is to look up the definition of a regular matrix. Further on, this problem is nothing else than adjusting the equation to fit into your definition.
     
  6. Feb 8, 2007 #5
    no no, radou i was able to do the problem after the first hint, im just wondering how i should approach different proofs (with matrices) in the future).
     
  7. Feb 8, 2007 #6

    radou

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    Just do a few more proofs and your question will be answered automatically.
     
  8. Feb 8, 2007 #7
    Allright well how about: If A is a square matrix such that A^3 = 0, show that (I-A)^-1 = I + A + A^2

    The books answer is as follows:
    (I-A)(I + A + A^2) = I - A^3 = I - 0 = I, therefore (I-A)^-1 = I + A + A^2

    i mean i understand how they got (I-A)(I + A + A^2) = I, what i dont understand is why they did that and how that prooves (I-A)^-1 = I + A + A^2

    once again any help is appreciated
     
  9. Feb 8, 2007 #8

    radou

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    These are just steps which require some experience, which you'll definitely gain after solving a considerable amount of problems. My replies may not seem very helpful, but I hope you get my point. :smile:
     
  10. Feb 8, 2007 #9
    AHHH, nevermind i figured out the question... well i think i did... its actually quite similar to the first one...

    multiplie both sides of (I-A)^-1 = I + A + A^2
    by (I-A) thus getting I = (I + A + A^2) x (I-A)
    then just do the thing the book shower and you get I=I , anyway i think thats right
     
  11. Feb 8, 2007 #10
    yah i do radou i just get frustrated when i cant figure out problems :P. Answers at the back of books usualy anger me as they dont really show steps. I was suprised when my book came with no solutions manual... speaking of which why do textbooks cost 110$ and not even have a solutions manual...
     
  12. Feb 8, 2007 #11

    radou

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    You can't use the solution to the problem to solve the problem, that's what you just did.
     
  13. Feb 8, 2007 #12

    robphy

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    Basically, you have to learn some aspects of the matrix algebra of square matrices, which is very similar to (but not exactly the same) as the algebra of real variables. You can add and take linear combinations, etc... However, something like (1/A) doesn't make sense. Instead, one has the inverse A-1, if it exists, such that AA-1=I=A-1A. In addition, one has to be aware that matrix multiplication is not generally commutative: AB =/= BA in general. So, "multiplying on the left side" is different from "multiplying on the right side".
     
    Last edited: Feb 8, 2007
  14. Feb 8, 2007 #13
    I dont think i quite understand what you mean. The question is If A is a square matrix such that A^3 = 0, show that (I-A)^-1 = I + A + A^2

    All i have to proove is that (I-A)^-1 = (I + A + A^2)
    or that LS = RS
    LS = (I-A)^-1, i then multiply it by (I-A) getting I
    then for RS
    (I + A + A^2) x (I-A) , because what you do to one side you gotta do to the other you get I... therefore LS = RS and the proof is prooven.. no?
     
  15. Feb 8, 2007 #14

    radou

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    And another useful fact is that AB = 0 does not imply that either A or B (or both) equal zero (i.e. the zero matrix).

    Edit: not related to your problem, but could be to some other one you'll encounter.
     
  16. Feb 8, 2007 #15
    Bah, now i dont know what to do, haha
     
  17. Feb 8, 2007 #16

    radou

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    They want to make you think on your own for 110$. :wink:
     
  18. Feb 8, 2007 #17

    Tom Mattson

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    You can't multiply both sides of the equation by B-1 if that matrix doesn't exist, right? So before you get to that step, you need to make sure that B even has an inverse. There's a theorem that says that a square matrix B has an inverse if and only if its determinant is not zero (that is, if and only if the B is nonsingular). You should be able to deduce from your equation that the only possible value of det(B) is 1, so the inverse of B exists, and you can proceed.
     
  19. Feb 8, 2007 #18

    mathwonk

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    to show something is the inverse of [ ], just show it gives the identity when you multiply it by [ ].

    i.e. a proof should start from a definition,a nd thats the definition on inverse.

    in this case the definition pretty much solves all your questions immediately.
     
    Last edited: Feb 8, 2007
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