Let $A$ be a diagonalizable real matrix such as $A^3=A$. Prove that $\mbox{rank}(A) = \mbox{tr}(A^2)$.
My way so far:
I know that if $B$ is simillar to $A$ then $B^3=B$.
$A$ is diagonalizable so there is a diagonal matrix $D=P^{-1}AP$
$rank(A)=rank(D)$ and $trace(A)=trace(D)$
$rank(A^2)=rank(D^2)$ and $trace(A^2)=trace(D^2)$
$rank(A^2)\le rank(A)$
$D^3=D$
But how can I continue from here?
Best Answer
Since $D^3=D$ so the eigenvalues are either $0$ or $-1$ or $1$. If we denote the number of nonzero eigenvalues by $p$ and the dimension of matrix by $n$ then since $A^2$ is equivalent to $D^2$ it has $p$ eigenvalues with value $1$ and $n-p$ zeros. therefore $$tr(A^2)=\sum{\lambda_i}=p$$ also the rank of a matrix is number of nonzero eigenvalues since the zero ones make linear dependence between columns or rows. Therefore $$rank(A)=p$$which leads us to $$rank(A)=tr(A^2)$$