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Matrix manipulation (inverse, lin. alg.)

  1. Jan 4, 2013 #1
    1. The problem statement, all variables and given/known data

    Let A[itex]\in[/itex]M[itex]_{n}[/itex]([itex]\Re[/itex]) a matrix verifying

    A[itex]^{3}[/itex]-A[itex]^{2}[/itex]-I[itex]_{n}[/itex]=0

    a) Show that A is inversible and calculate it
    b) Show that the solution X[itex]\subset[/itex]M[itex]_{n}[/itex]([itex]\Re[/itex]) of the equation

    A[itex]^{k}[/itex](A-I[itex]_{n}[/itex])X=I[itex]_{n}[/itex]

    has a unique solution.





    3. The attempt at a solution

    I'm having trouble with starting this one. I'm quite rubbish with these matrices in linear algebra, but I have exams in a few days and this question was on it, so i need help!

    I know the criteria for matrix inverse (AB=BA=I). However there's too much going on... help me dissect it? thanks a lot to anyone for any help, much appreciated.
     
  2. jcsd
  3. Jan 4, 2013 #2

    HallsofIvy

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    This is kind of trivial! [itex]A^3- A^2= A(A^2- A)= (A^2- A)A= I[/itex].

    Again, from [itex]A^3- A^2- I= 0[/itex], we have [itex]A^2(A- I)= I[/itex] so for any [itex]k\ge 2[/itex], [itex]A^k(I- I)X= A^{k-2}(A^2(A- I)X= A^{k- 2}X= I[/itex]. And since A has an inverse, you just multiply both sides by [itex]A^{-1}[/itex] k- 2 times. The cases where k= 0 or k= 1 are simple.


    There was really no "matrix" algebra involved here, just the definitions and basic algebraic manipulation.
     
  4. Jan 4, 2013 #3
    Oh I see, that was easier than I thought. I always get tripped up when they mention matrices and think that I have to tread very cautiously.

    thank you
     
  5. Jan 5, 2013 #4
    Actually, can you help me out with the last steps for solving? I'm still a bit caught up. I attached my attempt where you left off.

    thanks
     

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