$(Px,x)=\|Px\|^2$ implies that $P$ is a projection.

Question

Let $H$ be a Hilbert space over $\mathbb{R}$ and $P$ is a bounded linear operator over it. If $$ (Px,x)=\|Px\|^2,\quad \forall x\in H, $$ I want to show $P$ is a projection.

If $H$ is a Hilbert space over $\mathbb{C}$, I know how to prove it. Note that $$ (Px,(I-P)x)=0,\quad\forall x\in H. $$ For any $x,y\in H$, we have $$ (P(x+y),(I-P)(x+y))=0. $$ Using $(Px,(I-P)x)=(Py,(I-P)y)=0$ we can get $$ (Px,(I-P)y)+(Py,(I-P)x)=0. $$ Substitute $x$ by $\mathrm{i}x$ and we can get $$ (Px,(I-P)y)-(Py,(I-P)x)=0 $$ and $$ (Px,(I-P)y)=0,\quad\forall x,y\in H. $$ Thus, $P$ is a projection.

But for $H$ over $\mathbb{R}$, I have no idea. If the symmetry of $P$ could be proven, then $$ (Px,(I-P)y)=((I-P)x,Py)=(Py,(I-P)x) $$ and $P$ is a projection.

Any help is welcome!

Answer 1

This is not true. On $\mathbb{R}^2$, the matrix,

$$P = \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}$$

satisfies the required conditions, but is not even symmetric.

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Since this matrix may be a bit mysterious, here is the process that led to it. Let $R$ be the rotation by $90^\circ$ on $\mathbb{R}^2$. We want to find $P$ of the form $\alpha I + \beta R$, $\alpha, \beta > 0$. Then,

$$\alpha \|x\|^2 = (Px, x) = \|Px\|^2 = \alpha^2 \|x\|^2 + \beta^2 \|x\|^2$$

So $\alpha = \alpha^2 + \beta^2$. One solution to this is $\alpha = \beta = \frac{1}{2}$, which yields the matrix $P$ in the answer.