If A is a matrix satisfying $A^3 = A^2$ and the determinant of $A$ cannot equal $0$, must $A=I$.

Question

$I$ is an identity matrix. How would one go about conducting a proof for something like this?

I know that this doesn't work when $A$ is $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ , are there any other cases where this statement is true and $A$ does not equal $I$? Any hints are appreciated!

Answer 1

If the determinant of A is not 0, then you can multiply both sides of the equation by $(A^{-1})^2 $ to get $A=I$

Answer 2

Note that the statement is not always true if the entries of $A$ are taken from a commutative ring with zero divisors. E.g. in $\mathbb Z/4\mathbb Z$ (i.e. with modulo $4$ arithmetic), we have $2^3=2^2$ and $\det(2)=2\ne0$ but $2\ne1$.

So, for counterexamples with non-diagonal matrices, you may consider $A=\pmatrix{2&\ast\\ 0&1}$ over $\mathbb Z/4\mathbb Z$.