 Solution

If $Av=\lambda v$ for a nonzero eigenvector, then $A^{2}v={\lambda }^{2}v$. Since $A^2=A$, $\lambda (\lambda -1)=0$: every eigenvalue is zero or one. Let $m$ be the number of ones, counted with algebraic multiplicity. The trace is $m$ and the determinant is the product of these eigenvalues. ThereforeEvery output $y=Ax$ is fixed by a second application:The image of a linear map is thus the eigenspace for the eigenvalue one. To establish its dimension without assuming diagonalizability, decompose every input asThe first term lies in the image of a linear map $A$ and the second in its kernel. Their intersection is zero, because a vector in the image is fixed by $A$ whereas a vector in the kernel is killed by $A$. Thus the space is their direct sum, and an adapted basis puts $A$ into the form $\mathrm{diag}(I_r,0)$, where $r=\mathrm{rank}A$. Comparing traces gives $r=m$:This is an idempotent linear map, and trace of an idempotent equals its rank. If $m=0$, no nonzero output exists. The dimension in the question refers to the image subspace, including its zero vector.