Hilbert space subspace of "equally projected elements"

Question

Apologies for the title, bit of a struggle to come up with something non-generic.

Let $H$ be a Hilbert space and $p:H\rightarrow H$ an orthogonal projection.

Suppose $h_1,\,h_2\in H\backslash\{0\}$ and $$\alpha:=\frac{\|p(h_1)\|^2}{\|h_1\|^2}=\frac{\|p(h_2)\|^2}{\|h_2\|^2}.$$

Does it follow that:

$$\frac{\|p(h_1+h_2)\|^2}{\|h_1+h_2\|^2}=\alpha?$$

Answer 1

No. Let $H=\mathbb{C}^2$, $p$ the projection onto the first coordinate and $h_1=(1,1)$, $h_2=(1,-1)$. Then $$ \frac{\lVert p(h_1)\rVert^2}{\lVert h_1\rVert^2}=\frac{\lVert p(h_2)\rVert^2}{\lVert h_2\rVert^2}=\frac1 2, $$ yet $$ \frac{\lVert p(h_1+h_2)\rVert^2}{\lVert h_1+h_2\rVert^2}=1. $$