Proof: adjoint map of projection is a projection and ...

Question

Let $V$ be a pre hilbert space and $\pi \in \mathrm{End}(V)$.

Show: the adjoint map $\pi^+$ of a projection (meaning: $\pi^2 = \pi$) is a projection itself. Show then: a projection $\pi$ is orthogonal projection if and only if $\pi = \pi^+$.

What does that mean and how can I proof it?

Answer 1

• For all $x,y\in V$ and if $u,v\in \operatorname{End}(V)$ then $$\langle uv x,y\rangle=\langle vx,u^+y\rangle=\langle x,v^+u^+y\rangle$$ so $$(uv)^+=v^+u^+$$ Now if $\pi$ is a projection then $\pi^2=\pi$ and then $$(\pi^2)^+=(\pi^+)^+=\pi^+$$ and then $\pi^+$ is a projection.

• If $\pi$ is an orthogonal projection so let $x=x_1+x_2$ and $y=y_1+y_2$ where $x_1,y_1\in\operatorname{Im}\pi$ and $x_2,y_2\in (\operatorname{Im}\pi)^\perp$ then $$\langle \pi x,y\rangle=\langle x_1,y_1+y_2\rangle=\langle x_1,y_1\rangle=\langle x_1+x_2,y_1\rangle=\langle x,\pi y\rangle$$ hence we have $$\pi^+=\pi$$

Answer 2

In general $(\alpha\beta)^{+}=\beta^{+}\alpha^{+}$, so if $\pi=\pi^2$ then $\pi^+=(\pi^2)^+=(\pi^+)^2$. So $\pi^+$ is a projection. $\pi$ is orthogonal if $\pi\pi^+=1$, multiplying by $\pi$ on the left we get $\pi^2\pi^+=\pi$ or $\pi\pi^+=\pi$. now multiplying $\pi\pi^+=1$ by $\pi^+$ on the right we have $\pi(\pi^+)^2=\pi^+$ or $\pi\pi^+=\pi^+$ so now $$\pi=\pi\pi^+=\pi^+.$$