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!
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 $\dim H=\infty$.)
Let $U$, $V$ be the image and kernel of $P$, respectively. Then $\mathbb{C}^n=U\oplus V$ and for fixed $C>1$ we have $$ \|P\|\leqslant C\\\Leftrightarrow \forall u\in U, v\in V\colon\,\|u\|^2\leqslant C^2 \|u+v\|^2 \\ \Leftrightarrow \forall u\in U, v\in V\colon\,0\leqslant (C^2-1) \|u\|^2+2C^2 \Re \langle u,v\rangle+C^2 \|v^2\|\\ \Leftrightarrow\forall u\in U, v\in V,t\in \mathbb{R}\colon 0\leqslant (C^2-1) t^2\|u\|^2+2C^2 t\Re \langle u,v\rangle+C^2 \|v^2\|\\ \Leftrightarrow \forall u\in U, v\in V\colon\,C^4 \Re \langle u,v\rangle^2\leqslant C^2(C^2-1) \|v^2\|\cdot \|u\|^2, $$ here $U$ and $V$ come in symmetric fashion, so $\|P\|\leqslant C$ if and only if $\|I-P\|\leqslant C$.
