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In the book "The Symmetric Eigenvalue Problem" by BN Parlett, Exercise 1.4.5 asks

Show that $\operatorname{rank}(\boldsymbol E_j) = \operatorname{dimension}(\mathcal N_j)$. Note that $\operatorname{rank}(\boldsymbol E_j) = \operatorname{trace}(\boldsymbol E_j)$. Use the fact that $\boldsymbol E_j^2 = \boldsymbol E_j$ and $\operatorname{trace}(\boldsymbol B\boldsymbol C) = \operatorname{trace}(\boldsymbol C\boldsymbol B)$.''

This matrix $\boldsymbol E_j$ is defined by

A matrix $\boldsymbol E$ is a projector if $\boldsymbol E^2= \boldsymbol E$. It is an orthogonal projector if $\boldsymbol E\boldsymbol y = \boldsymbol 0$ for all $\boldsymbol y$ orthogonal to $\boldsymbol E$'s range, meaning $\boldsymbol E^{\mathsf{H}}= \boldsymbol H$. or any square matrix $\boldsymbol B$ the spectral projector $\boldsymbol E_\lambda$ for an eigenvalue $\lambda$ satisfies $\boldsymbol B\boldsymbol E_\lambda = \boldsymbol E_\lambda\boldsymbol B = \lambda\boldsymbol E_\lambda$.

I'm not sure how to take the definition to prove the required statement regarding $\operatorname{rank}(\boldsymbol E_j)$ and $\operatorname{dimension}(\mathcal N_j)$. I thought about starting from $(\boldsymbol B - \lambda\boldsymbol I)\boldsymbol y = \boldsymbol 0$ which should all for all $\boldsymbol y \in (\boldsymbol B - \lambda\boldsymbol I)^\perp$, I believe, but this doesn't seem like it would back to what it's telling me to use with the trace and idempotence of $\boldsymbol E_j$.

LaguerreGroup
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    Some interesting answers here: https://math.stackexchange.com/questions/101512/proving-the-trace-of-an-idempotent-matrix-equals-the-rank-of-the-matrix – Emil Vatai Sep 25 '23 at 11:47

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I don't think you need any knowledge of the trace in this case, since the spectral projector $E_j$ is by definition the orthogonal projection onto $N_j = \mathrm{ker}(B - \lambda_j I)$. Hence, the range of $E_j$ is precisely $N_j$, so $$ \mathrm{rank}(E_j):= \mathrm{dim} (\mathrm{ran}(E_j)) = \dim(N_j). $$

Wizact
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