So I have the following question. I have to prove if this statement is true or false. If it's true, I need to explain why. If it's false, I need a counterexample.
If a square matrix $A$ satisfies $A^k=5{\rm{Id}}_k$ for some $k \geq 1$, then $A$ is invertible.
I can easily do this with determinants.
$\det(A^k)=5^k \det({\rm{Id}}_k)$
$\det(A)^k=5^k (1)$
Since $k \geq 1$ and $\det(A)^k \neq 0$, we have:
$\det(A)^k \neq 0$
$\det(A) \neq 0$
This implies that the statement is true.
However, is there a way to do this without determinants? Supposedly you have to use the properties of the invertibility of a square matrix.
For some background, the concept of span, linear combinations, RREF and transpose has been covered. Also, the conditions of invertibility $AB=BA={\rm{Id}}_n$ has been covered too.
Is there another way I could show this?
Best Answer
$$A \left(\frac15 A^{k-1}\right)=\left(\frac15 A^{k-1}\right)A=\rm{Id}.$$