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I am trying to do the same exercise as in this question:

Let $P\in C^{m×m}$ be a non-zero projector. Show that $||P||_2=1$ iff $P$ is an orthogonal projector.

I managed to prove everything but the implication that if the non-zero projector $P$ has one as its 2-norm then it is an orthogonal projector. The linked answer says this conclusion is easy after showing the result that the 2-norm of a matrix is equal to the 2-norm of its SVD, but I am not getting it.

I tried using the SVD to try and show the projector has the form

$$P = Q \Sigma Q^*,$$

but I cannot figure it out. I also attempted to try and calculate the SVD of $P^*$ to show it is equal to that of $P$, but also to no avail.

tigre200
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1 Answers1

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For any non-zero vector $x\in\operatorname{range}(P)$, write $x=u+v$ where $u\in\ker(P)$ and $v\in\ker(P)^\perp$. Note that $v\ne0$ or else $x\in\ker(P)\cap\operatorname{range}(P)=0$. Since $u+v=x=Px=Pv$, we get $$ \|P\|_2\ge\frac{\|Pv\|_2}{\|v\|_2}=\frac{\|u+v\|_2}{\|v\|_2}=\frac{\sqrt{\|u\|_2^2+\|v\|_2^2}}{\|v\|_2}=\sqrt{\frac{\|u\|_2^2}{\|v\|_2^2}+1}. $$ So, when $\|P\|_2=1$, we must have $u=0$ and $x=v\in\ker(P)^\perp$. As $x$ is an arbitrary non-zero vector in $\operatorname{range}(P)$, we obtain $\operatorname{range}(P)\subseteq\ker(P)^\perp$. Therefore $P$ is an orthogonal projector.

tigre200
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user1551
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