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Suppose $n > 1$ is an integer. Let $P \in \mathbb C^{n \times n}$ be a matrix such that $P^2=P$ and $1\leqslant {\rm rank}(P)<n$. Prove that $\Vert P \Vert_2 = \Vert I - P \Vert_2$.

I have been working on the problem for hours. Please let me know if any can help. Thanks!

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    $\begingroup$ See here: math.nthu.edu.tw/~amen/2014/131012(final).pdf $\endgroup$ Nov 26 '21 at 13:31
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    $\begingroup$ I don't understand why this question is closed. This is nontrivial, see the nice reference given by Christian Remling. $\endgroup$
    – abx
    Nov 27 '21 at 9:26
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    $\begingroup$ @abx: I beg to disagree: the proof in that reference is ingenious, but entirely elementary. Furthermore, things are easier in this post because the Hilbert space is finite-dimensional, so in principle one has more ways of attacking the problem. This is definitely a good question for MSE, undoubtedly, but maybe less so for MO. $\endgroup$
    – Alex M.
    Nov 27 '21 at 11:14
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    $\begingroup$ @Alex M.: I think the OP meant the operator norm w.r.t. ${\lVert\ \rVert}_2$ on $\Bbb{C}^n$ — otherwise this is trivially false. $\endgroup$
    – abx
    Nov 27 '21 at 13:04
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    $\begingroup$ Also @AlexM. I agree with abx (my first interpretation of the question was yours, but then once I realized there are diagonal counterexamples I assumed that -- as is sometimes commonplace in matrix analysis -- $\Vert \cdot\Vert_p$ is used for the $\ell_p^n\to \ell_p^n$ norm. (The Hilbert-Schmidt norm is then often denoted, in those sources, by "F" for "Frobenius.) $\endgroup$
    – Yemon Choi
    Nov 28 '21 at 1:01
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Let $U$, $V$ be the image and kernel of $P$, respectively. Then $\mathbb{C}^n=U\oplus V$ and $$ \|P\|\leqslant C\,\, \text{for some}\,\,C>1\Leftrightarrow \forall u\in U, v\in V\colon\,\|u\|^2\leqslant C^2 \|u+v\|^2 \\ \Leftrightarrow \forall u\in U, v\in V\colon\,0\leqslant (C^2-1) \|u\|^2+2C^2 \Re \langle u,v\rangle+C^2 \|v^2\|\\ \forall u\in U, v\in V,t\in \mathbb{R}\colon 0\leqslant (C^2-1) t^2\|u\|^2+2C^2 t\Re \langle u,v\rangle+C^2 \|v^2\|\\ \Leftrightarrow \forall u\in U, v\in V\colon\,C^4 \Re \langle u,v\rangle^2\leqslant C^2(C^2-1) \|v^2\|\cdot \|u\|^2, $$ here $U$ and $V$ come in symmetric fashion, so $\|P\|\leqslant C$ if and only if $\|I-P\|\leqslant C$.

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This is very easy to confirm when $n=2$: We can then also assume that $Pe_1=e_1$, so $$ P=\begin{pmatrix} 1 & a \\ 0 & 0 \end{pmatrix} , $$ and $\|P\|=\| 1-P\| = \sqrt{1+|a|^2}$.

In general, pick an $x$ with $\|x\|=1, \|Px\| = \|P\|$ and restrict $P$ to the invariant subspace $V$ spanned by $x,Px$. Then $\dim V=2$ and $P\not= 0,1$ also on $V$, unless we are in the trivial case $\|P\|=1$, so the first part shows that $\|(1-P)\bigr|_V\|=\|P\|$. Thus $\|1-P\|\ge \|P\|$ and then also $\|1-P\|=\| P\|$ by symmetry.

(This argument, very slightly modified, also works in general, when $\dim H=\infty$.)

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