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Let $P \in \mathbb C^{n \times n}$ be a projection matrix.

Is it true that if $\lVert P \rVert_\mathrm{F} = \sqrt{\mathrm{rank}(P)}$, then $P$ is Hermitian (and, thus, an orthogonal projection matrix)?

I know that the converse of the above statement is true and I was wondering if the above holds as well, since I couldn't find a counterexample.

entechnic
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    Related result – Ben Grossmann Oct 31 '24 at 15:06
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    @BenGrossmann Just to record this, here is an easier variational argument for the linked question: Assume $P$ has spectral norm $1$. Assume to the contrary that it is not orthogonal, then there exists $y\in\text{range}(P)^\perp$ s.t. $x=Py\neq0$. Note that $x\perp y$. Thus, for any $t>0$,$$(t+1)^2|x|^2=|(t+1)x|^2=|P(tx+y)|^2\leq|tx+y|^2=t^2|x|^2+|y|^2$$Hence,$$2t|x|^2+|x|^2\leq|y|^2$$for all $t>0$, which is impossible as $x\neq0$. – David Gao Oct 31 '24 at 15:19
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    @DavidGao You're blowing my mind today, thank you – Ben Grossmann Oct 31 '24 at 15:47

2 Answers2

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This is indeed true. Let $\{e_i\}_{i = 1}^{\text{rank}(P)}$ be an ONB of $\text{range}(P)$ and extend it to an ONB $\{e_i\}_{i = 1}^n$ of $\mathbb{C}^n$. Note that the Frobenius norm can be calculated by,

$$\begin{split} \text{rank}(P) = \|P\|_F^2 &= \sum_{i = 1}^n \|Pe_i\|^2\\ &= \sum_{i = 1}^{\text{rank}(P)} \|Pe_i\|^2 + \sum_{i = \text{rank}(P) + 1}^n \|Pe_i\|^2\\ &= \text{rank}(P) + \sum_{i = \text{rank}(P) + 1}^n \|Pe_i\|^2 \end{split}$$

Thus, $Pe_i = 0$ for all $\text{rank}(P) < i \leq n$. But this means $P$ is zero on $\text{range}(P)^\perp$, i.e., $P$ is an orthogonal projection.

David Gao
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Write out Schur's Inequality:
$\big\Vert P\big \Vert_F^2 \geq \sum_{k=1}^n \vert \lambda_k\vert^2\geq \text{trace}\big(P^2\big)=\text{trace}\big(P\big)=\text{rank}\big(P\big)=\big\Vert P\big \Vert_F^2$
The second inequality is triangle inequality, and the the first inequality is due to Schur, which is met with equality iff $P$ is normal. Conclude: $P$ is unitarilly diagonalizable with all eigenvalues $\in \big\{0,1\big\}$, i.e. $P$ is Hermitian.

Another argument:
directly compute $\big\Vert P-P^*\big \Vert_F^2 = 2\cdot \big\Vert P\big \Vert_F^2 - 2\cdot \text{rank}\big(P\big) =0$

user8675309
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