Thread: prove that matrix is invertible

Hi!

i need to prove that if matrix is invertible, then A is also invertible.

TIA!

Hey Stormey.

Recall that det(AB) = det(A)*det(B) for operators A and B.

it's easier to use chiro's hint if you prove the contrapositive:

if A is singular, so is Aⁿ

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:



i can multiply it by det(A) times, and then say that
but what conclusion can i draw from it?
does that prove it?

BTW, i have no idea what determinant is, so bare with me...

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 Aⁿv = A^n-1(Av) = A^n-1(0) = 0.

so Aⁿ is singular, as well.

To show that is invertible, you must show that there exist matrix B such that 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 and ?

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.