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Given a projector $P (\text{e.g}\; P^2 =P )$, prove that if $ ||P||_2 = 1$ then $P^* = P$ namely $P$ is symmetric matrix and an orthogonal projector.

I tried taking $x = u+v $ where $ u \in Col(P) $ and $v \in Null(P)$ and then, given that $Pu=u$ and $Pv=0$ I have

$ ||P||_2 = \sup_{u,v} \frac{||P(u+v)||_2}{||u+v||_2} = \sup_{u,v}\frac {||u||_2}{||u +v||_2} =1 $

But even if this mandates that $v=0$,I don't think it proves that $Col(P)$ and $ Null(P)$ are orthogonal sub spaces and therefore $P$ is a symmetric matrix and an orthogonal projector.

A rural reader
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Tomer
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  • It's slightly preferable to use something like $\lVert x \rVert$, which produces $\lVert x \rVert$, in place of just a double vertical line. – Brian Tung Dec 09 '24 at 01:44
  • See here – Igor Dec 09 '24 at 02:05
  • I saw but I could not understand this convoluted thread. – Tomer Dec 09 '24 at 02:08
  • See my comment under this question where I provided an easy variational argument for the result. – David Gao Dec 09 '24 at 02:54
  • Maybe there is a more standard proof ? – Tomer Dec 10 '24 at 01:58
  • @Tomer I added an accessible proof. – copper.hat Dec 10 '24 at 04:30
  • why aren't you able to pursue one of the arguments in the link? In particular I suggest writing $0\leq \big\Vert P-P^*\big \Vert_F^2 = 2\cdot \big\Vert P\big \Vert_F^2 - 2\cdot \text{rank}\big(P\big) $. Then argue $\big\Vert P\big \Vert_2 =1 \implies \big\Vert P\big \Vert_F^2\leq \text{rank}\big(P\big)$ – user8675309 Dec 10 '24 at 19:49
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    This question has been asked and answered before, see e.g. https://math.stackexchange.com/questions/21668/orthogonal-projection, https://math.stackexchange.com/questions/4464608/idempotent-operator-on-hilbert-space-with-operator-norm-1-is-projection, https://math.stackexchange.com/questions/1109755/a-projection-p-is-orthogonal-if-and-only-if-its-spectral-norm-is-1, https://math.stackexchange.com/questions/3622683/show-that-if-q-1-then-q-is-the-orthogonal-projection-of-h-onto-rq, https://math.stackexchange.com/questions/2738473/norm-1-projection-in-a-c-algebra (generalized to $C^*$-algebras) – leslie townes Dec 12 '24 at 03:55
  • @leslietownes perhaps none were right for 'adaption'? – copper.hat Dec 12 '24 at 19:14

2 Answers2

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Suppose $\langle Px, (I-P)y \rangle \neq 0 $ for some $x,y$. Multiplying by a scalar as necessary, we can assume that it is real and negative. Consider the point $Px+t(I-P)y$ for positive $t$. Note that $P( Px+t(I-P)y ) = Px$ and $\|Px+t(I-P)y\|^2 = \|Px\|^2 + 2t \langle Px, (I-P)y \rangle + t^2 \|(I-P)y\|^2$. In particular, since the inner product is negative, we can choose the $t>0$ sufficiently small such that $\|Px+t(I-P)y\| < \|Px\|$, or in other words, $\|P\| > 1$. (To reiterate, if we let $w=Px+t(I-P)y$, then $Pw = Px$ and $\|w\| < \|Pw\|$, hence $\|P\| > 1$.)

Hence if $\|P\|=1$, we must have $\langle Px, (I-P)y \rangle = 0 $ for all $x,y$, or $P= P P^*$, taking the Hermitian conjugate shows that $P=P^*$.

copper.hat
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Based on copper.hat's answer, I have reformulated the idea slightly differently:

Suppose $P^* \ne P$. This implies that there exist unit vectors $x \in \text{Col}(P)$ and $y \in \text{Null}(P)$ such that $x^* y \ne 0$. Define $w = - (x^* y) y$ and observe that

$$ \| x + w \|^2 = \|x\|^2 + 2w^*x + \|w\|^2 = \|x\|^2 - 2(x^* y) y^*x + \|(-x^* y) y\|^2 = \|x\|^2 - (x^* y)^2 < \|x\|^2 \implies \| x + w \| < \|x\|. $$

Next, consider

$$ \frac{ \| P(x+w) \| }{\|x+w\|} = \frac{ \| P(x) + P(w) \| }{\|x+w\|} = \frac{ \| x \| }{\|x+w\|} > 1 \implies \|P\| > 1. $$

This contradicts the assumption that $\|P\| = 1$. Therefore, it follows that $P^* = P$, and $P$ is an orthogonal projection.

Tomer
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