Hi!
i need to prove that if matrix $\displaystyle A^n$ is invertible, then A is also invertible.
TIA!
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...
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 $\displaystyle C^nA^n= C^{n-1}(CA)(A^{n-1})$?