$\|P\|_2 \ge1$ if and only if $P$ is an orthogonal projector.

Question

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!

Answer 1

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.