Hi!

i need to prove that if matrix $\displaystyle A^n$ is invertible, then A is also invertible.

TIA!

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- Dec 4th 2012, 11:54 PMStormeyprove that matrix is invertible
Hi!

i need to prove that if matrix $\displaystyle A^n$ is invertible, then A is also invertible.

TIA! - Dec 5th 2012, 01:00 AMchiroRe: prove that matrix is invertible
Hey Stormey.

Recall that det(AB) = det(A)*det(B) for operators A and B. - Dec 5th 2012, 01:04 AMDevenoRe: prove that matrix is invertible
it's easier to use chiro's hint if you prove the contrapositive:

if A is singular, so is A^{n} - Dec 5th 2012, 03:04 AMStormeyRe: prove that matrix is invertible
we didn't learn determinants yet, so i'm guessing it can be solved without using it.

nevertheless, from what i read in wiki, if matrix A is invetrtible, then its determinant is not 0.

and since i want to prove the contrapositive, i need to assume it's singular, and the its det is equal 0.

so:

$\displaystyle det(A)=0$

i can multiply it by det(A) $\displaystyle n-1$ times, and then say that $\displaystyle det(A^n)=0$

but what conclusion can i draw from it?

does that prove it?

BTW, i have no idea what determinant is, so bare with me... (Giggle) - Dec 5th 2012, 07:10 AMDevenoRe: prove that matrix is invertible
you don't need to use determinants, really.

suppose that A is singular. this means there is some non-zero vector v, with Av = 0.

hence A^{n}v = A^{n-1}(Av) = A^{n-1}(0) = 0.

so A^{n}is singular, as well. - Dec 5th 2012, 07:41 AMHallsofIvyRe: prove that matrix is invertible
To show that $\displaystyle A^n$ is invertible, you must show that there exist matrix B such that $\displaystyle A^nB= BA^n= I$ where I is the identity matrix.

Since A is invertible, there exist a matrix C such that AC= CA= I.

Now, what can you say about $\displaystyle A^nC^n= A^{n-1}(AC)C^{n-1}$ and $\displaystyle C^nA^n= C^{n-1}(CA)(A^{n-1})$? - Dec 5th 2012, 09:27 AMStormeyRe: prove that matrix is invertible
thanks, Deveno.

HallsoiIvy, i need to prove that if**p**then**q**, not if**q**then**p**.

i managed to prove the other direction, but thanks anyway.