The question is:
If A is a square matrix such that $A^2=A$ then $A^n =A$ for all natural numbers $n$ greater than one. What is $A$ if $A \ne 0$ and $A \ne I$.
I figured out an answer but I can't tell if that's the only answer. Let's say that $a_{kk}$ is a value in $A$. Every value in $A$ is $0$ except for $a_{kk}$ and $a_{00}$, they're $1$. I haven't gotten this answer mathematically.
I've tried a few approaches but ended up with the identity matrix.
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$\begingroup$Perhaps one more example, besides $A=0$ and $A=I$ may be insightful: take the block matrix $$ A=\begin{pmatrix} I & 0 \cr 0 & 0 \end{pmatrix} $$ Of course, $A=A^2$, but $A\neq 0,I$. The (square) blocks can be of any size, so we obtain several examples. Up to similarity, these are the only ones, too. See "canonical forms" in the wikipedia article.
$\endgroup$ 3 $\begingroup$Such a matrix is called idempotent. Here are some examples and properties.
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