# $2$-norm of idempotent matrix

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

• Nov 26, 2021 at 13:31
– abx
Nov 27, 2021 at 9:26
• @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. Nov 27, 2021 at 11:14
• @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.
– abx
Nov 27, 2021 at 13:04
• 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.) Nov 28, 2021 at 1:01

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 $$\|I-P\|\leqslant C$$.