Nonzero eigenvalue and its eigenspace for a projector

A square matrix $A$ is idempotent when $A^2=A$. Suppose $A$ is a $10\times10$ idempotent matrix that is diagonalisable and has nullity $4$. Let $\lambda$ be its only possible non-zero eigenvalue and $r$ the dimension of the corresponding eigenspace. Find $\lambda^2+r^2$.

$37$. If $Av=\lambda v$ with $v\ne0$, then $\lambda v=A^2v=\lambda^2v$, so $\lambda\in\{0,1\}$ and the non-zero one is $\lambda=1$. Diagonalisability splits $\mathbb{R}^{10}$ into the $0$-eigenspace (dimension $4$) and the $1$-eigenspace, so $r=6$ and $1+36=37$.