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Let $A$ be an $n \times n$ matrix. Show that $A$ is invertible if and only if any power $A^k$ (with $k\geqslant1$) of $A$ is invertible.

I've been looking over the Theorem of Invertible Matrices but I can't seem to find anything in reference to how $k$ might affect $A$. I was also thinking of using the inverse, but I'm not sure I can since my teacher wants $k$ to either be one or more.

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    $\begingroup$ Suppose that $B$ is an inverse of $A^2$. Can you find an inverse of $A$ in terms of $A$ and $B$? $\endgroup$ – Lord Shark the Unknown Oct 25 '17 at 20:46
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If $B$ is an inverse of $A$, then $A^kB^k=\operatorname{Id}$.

On the other hand, if $C$ is such that $C.A^k=\operatorname{Id}$, then $(C.A^{k-1})A=\operatorname{Id}$, and therefore $A$ has an inverse.

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$\det(A) = 0$ iff $\det(A^k) = \det(A)^k = 0$. So $A$ is non-invertible iff $A^k$ is.

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$\textbf{$A$ is invertible $\Rightarrow$ $A^k$ is invertible :}$

$A$ is invertible $\Rightarrow$ there exists an unique matrix $B$ such that $AB=BA=I$.

Now multiply the above on left (right) equation by $A^{k-1}$ $\Rightarrow$ to get $A^kB=BA^k=A^{k-1}$.

Multiply the above equation on left (right) by $B^{k-1}$ $\Rightarrow$ $A^kB^k=B^kA^k=I$.

Hence $B^k$ is the inverse of $A^k$ $\Rightarrow$ $(A^k)^{-1}=(A^{-1})^{k}$.

$\textbf{$A^k$ is invertible $\Rightarrow$ $A$ is invertible :}$

$A^k$ is invertible $\Rightarrow$ there exists an unique matrix $B$ such that $A^kB=BA^k=I$.

Hence $A[A^{k-1}B]=[BA^{k-1}]A=I$.

Note $[A^{k-1}B]=[BA^{k-1}]$ (which can be seen by multiplying on left (right) with $A^k$).

Hence $A$ is invertible and $A^{-1}=A^{k-1}B$.

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