Matrix manipulation (inverse, lin. alg.)

boings
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Homework Statement



Let A\inM_{n}(\Re) a matrix verifying

A^{3}-A^{2}-I_{n}=0

a) Show that A is inversible and calculate it
b) Show that the solution X\subsetM_{n}(\Re) of the equation

A^{k}(A-I_{n})X=I_{n}

has a unique solution.





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.
 
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boings said:

Homework Statement



Let A\inM_{n}(\Re) a matrix verifying

A^{3}-A^{2}-I_{n}=0

a) Show that A is inversible and calculate it
This is kind of trivial! A^3- A^2= A(A^2- A)= (A^2- A)A= I.

b) Show that the solution X\subsetM_{n}(\Re) of the equation

A^{k}(A-I_{n})X=I_{n}

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

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.

There was really no "matrix" algebra involved here, just the definitions and basic algebraic manipulation.
 
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
 
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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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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