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Let $P$ be a nonzero projector.

$\|P\|_2 \geq 1$ with equality iff $P$ is an orthogonal projector.

I already proved $(\Leftarrow)$ way.

However, It is a sticking point to solve $(\Rightarrow$)

I saw that there is a similar post, but I don't undertstand with this post, since it is about equality, but I want to solve this with inequality.

I tried $P^2=P$, since $P$ is a projector, and doing some inequality but it doesn't work.

How do I solve this problem?

Thanks!

Sammy Black
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JAEMTO
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  • See also this post – Ben Grossmann Oct 04 '23 at 21:51
  • Notation: use \|P\| for better spacing between norm bars: $|P|$ vs. $||P||$. – Sammy Black Oct 04 '23 at 21:57
  • you can streamline the argument by using an inequality of Schur i.e. if $P$ has rank $r$, it is diagonalizable with eigenvalues of 0 and 1 and $\big \Vert P\big \Vert_2\ = 1 \implies r = \big \Vert P\big \Vert_2^2\cdot r \geq \big \Vert P\big \Vert_F^2\geq \sum_{k=1}^n \vert \lambda_k\vert^2 = r$ -- is met with equality so $P$ is normal, i.e. unitarily diagonalizable with real spectrum, hence Hermitian, i.e. an orthogonal projector – user8675309 Oct 05 '23 at 16:19

1 Answers1

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It is generally true that for a non-zero projector $P$, $\|P\|_2 \geq 1$. This can be seen, for instance, by noting that for any non-zero vector $x$ in the image of $P$, we have $Px = x$, hence $$ \|x\| = \|Px\| \leq \|P\|_2 \|x\| \implies \|P\|_2 \geq 1. $$ As the linked post shows, $\|P\|_2 = 1$ if and only if $P$ is an orthogonal projection. For all other cases, we can conclude that $\|P\|_2 > 1$ using the above observation.

Ben Grossmann
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  • I read your answer several times. Overall, I understood ∥P∥2=1 if and only if P is an orthogonal projection, but how can I show that $|P|_2 > 1$ implies $P$ is orthogonal projector? Can you be more specific, please? – JAEMTO Oct 06 '23 at 08:27
  • Come to think about it, "$|P|_2 \geq 1$ iff $P$ is an orthogonal projection" is weird, because $|P|_2 = 1$ iff $P$ is an orthogonal projection. – JAEMTO Oct 06 '23 at 09:26
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    @JAEMTO Your comment is confusing. The goal here is to conclude that $|P|_2 > 1$ implies that $P$ is not an orthogonal projector. Also, I never say that "$|P|_2 \geq 1$ iff $P$ is an orthogonal projection." In any case, maybe the following explanation will help: in general, it is true (for a non-zero projector $P$) that either $|P|_2 = 1$ or $|P|_2 > 1$ (i.e. $|P|_2 \geq 1$). If $P$ is not orthogonal, then it cannot be true that $|P|_2 = 1$, so it must be true that $|P|_2 > 1$. – Ben Grossmann Oct 06 '23 at 16:37
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    @JAEMTO I assumed that your title was meant to be a shortening of the correct statement from your question, "$|P|_2 \geq 1$ with equality iff $P$ is an orthogonal projector." Your title as it currently stands is an incorrect statement. – Ben Grossmann Oct 06 '23 at 16:41