Matrix manipulation (inverse, lin. alg.)

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  • #1
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Homework Statement



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.





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.
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement



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

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.
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.


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.
 
  • #3
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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
 
  • #4
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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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