A projection $P$ is orthogonal if and only if its spectral norm is 1 [closed]

I have to show what the title says.

A projection $P$ is orthogonal if and only if its spectral norm is $1$.

I suppose I have to use the following identity: $$\|P\|_{2}=(\lambda_{\max}(P^*P))^{\frac{1}{2}}.$$

Fact 1 If $P\neq 0$ is idempotent, then $\|P\|_2\geq 1$.

Proof: Since $P^2=P\neq 0$, $0\neq\|P\|_2=\|P^2\|_2\leq\|P\|_2^2$. Divide by $\|P\|_2$. $\Box$

Fact 2 If $P\neq 0$ is idempotent and Hermitian, then $\|P\|_2=1$.

Proof: Take a nonzero $x$ and consider the splitting $x=Px+(I-P)x$. We have using $P^2=P$ and $P^*=P$ that $$ \|x\|_2^2=\|Px\|_2^2+\|(I-P)x\|_2^2 $$ and hence $\|Px\|_2\leq\|x\|_2$. It follows that $\|P\|_2\leq 1$ and together with Fact 1, we have $\|P\|_2=1$. $\Box$

Fact 3 If $P$ is an idempotent matrix, then $P^*$ is idempotent too. The range of $P$ has the form $$\mathrm{Im}(P)=\{x:\;x=Px\}.$$

Proof: For the first statement, $(P^*)^2=P^*P^*=(PP)^*=P^*$. If $x=Px$, then obviously $x\in\mathrm{Im}(P)$. On the other hand, if $x\in\mathrm{Im}(P)$, then $x=Py$ for some $y$. But $Px=P^2y=Py$ so $x=Py=Px$. $\Box$

Fact 4 An idempotent matrix $P$ is Hermitian if and only if $\mathrm{Im}(P)=\mathrm{Im}(P^*)$.

Proof: One direction is obvious. If $P$ is Hermitian, then $\mathrm{Im}(P)=\mathrm{Im}(P^*)$. On the other hand, assume $\mathrm{Im}(P)=\mathrm{Im}(P^*)$ and let us show that $Px=P^*x$ for all $x$. Since $\mathrm{Im}(P)=\mathrm{Im}(P^*)$, we have $\mathrm{Ker}(P)=\mathrm{Ker}(P^*)$. The fundamental theorem of linear algebra gives that $\mathrm{Im}(P)=\mathrm{Ker}(P^*)^{\perp}=\mathrm{Ker}(P)^{\perp}$. Take any $x$ and consider the splitting $x=y+z$, where $y\in\mathrm{Im}(P)$ and $z\in\mathrm{Ker}(P)$. We have hence $y=Py=P^*y$, which follows from Fact 3, and $Pz=P^*z=0$. Consequently, $Px-P^*x=(Py+Pz)-(P^*y+P^*z)=y-y=0$. $\Box$

Fact 5 If $P$ is idempotent and $\|P\|_2\leq 1$, then $P^*=P$.

Proof: In view of Fact 4, we just need to show that $\mathrm{Im}(P)\perp\mathrm{Ker}(P)$. Let $x$ be a vector such that $x\perp\mathrm{Ker}(P)$ and consider $y:=Px-x$. It is easy to see that $Py=0$ and hence $y\in\mathrm{Ker}(P)$. Note that $x\perp y$ (so $\|x\|_2^2+\|y\|_2^2=\|x+y\|_2^2$). Since $\|P\|_2\leq 1$, we have $\|Px\|_2\leq\|x\|_2$. It follows that $$ \|x\|_2^2\leq\|x\|_2^2+\|y\|_2^2=\|x+y\|_2^2=\|Px\|_2^2\leq\|x\|_2^2 $$ and hence $y=0$, which implies that $Px=x$ and thus $x\in\mathrm{Im}(P)$. Therefore, $\mathrm{Ker}(P)^{\perp}\subset\mathrm{Im}(P)$. On the other hand, if $z\in\mathrm{Im}(P)$, $Pz=z$. Let $x\in\mathrm{Ker}(P)$ and $y\perp\mathrm{Ker}(P)$ be such that $z=x+y$, so $Pz=Px+Py=Py$. Since we already know that $\mathrm{Ker}(P)^{\perp}\subset\mathrm{Im}(P)$, $Py=y$ and hence $z=y$. Therefore, $\mathrm{Im}(P)\subset\mathrm{Ker}(P)^{\perp}$. From, $\mathrm{Ker}(P)^{\perp}\subset\mathrm{Im}(P)$ and $\mathrm{Im}(P)\subset\mathrm{Ker}(P)^{\perp}$, it follows that $\mathrm{Im}(P)=\mathrm{Ker}(P)^{\perp}$. $\Box$

Your statement follows from Fact 2 and Fact 5 (by substituting "projector" for "idempotent matrix" and "orthogonal projector" for "Hermitian idempotent matrix"). You need to add an assumption that $P\neq 0$! (Obviously, $P=0$ is an orthogonal projector too.)

We begin with the following:

Fact: $P$ is orthogonal iff its eigenvalues are all $1$ and $0$ and $P = P^*$

This can be proven using the spectral theorem: note that "$P$ is orthogonal" is true iff $P$'s eigenvectors are mutually orthogonal, and perhaps you can deduce the statement about the eigenvalues.

Fact: if $P$ is a projection, $\|P\|_2 \geq 1$

stealing Pavel's proof for this one: $\|P\|_2 = \|P^2\|_2 \leq \|P\|_2^2$

Now, note that $P$ has a Schur decomposition. That is, there is an upper triangular matrix $T = U^*PU$ where $U$ is unitary.

Verify that $T$ is a projection, and that $\|T\|_2 = \|P\|_2$. That is, "we assume without loss of generality that $P$ is upper triangular".

Now, $T$ can be written as follows: $$ T = \pmatrix{ 0&&&&&\star\\ &\ddots\\ &&0\\ &&&1\\ &&&&\ddots\\ &&&&&1 } $$ Now, remember that by definition, $$ \|T\|_2 = \max_{x \neq 0} \frac{\|Tx\|}{\|x\|} $$ The sketch of the rest of the proof is as follows: Suppose that $T^2 = T$ and $\|T\| = 1$.

First, we write $T$ as a block matrix: $$ T = \pmatrix{A & B\\0&C} $$ We have $T^2 = T$, which is to say $$ \pmatrix{A^2 & AB + BC\\0&C^2} = \pmatrix{A & B\\0&C} $$ $A$ is nilpotent with $A^2 = A$, so $A = 0$ (and $C-I$ is nilpotent with $C^2 = C$, so $C = I$). So, we have $$ T = \pmatrix{0 & B\\0&I} $$ And we now want to show that $B = 0$.

Suppose there is a non-zero entry in the $i$th column, above one of the $1$s on the diagonal. Verify that $\|Te_i\|_2 > 1$. Thus, if $\|T\|_2 = 1$, $B = 0$ (and $C = I$).

Thus, we conclude that if $T$ is an upper-triangular projection with $\|T\|_2 = 1$, then $T$ is diagonal. The conclusion follows.