# Is the composition of a self-adjoint map $f$ with its projection also self-adjoint?

This is probably obvious. But if $V = V_1 \oplus V_2$ where $V_i$ are irreducible subspace such that they are inequivalent (meaning it is not possible to find an isomorphism $A : V_1 \to V_2$ such that $A(\phi_1(g)(v)) = \phi_2(g)(A(v))$ where $\phi_i : G \to Aut(V_i)$ for all $g \in G$ and $v \in V_1$, in other words $A\phi_1 = \phi_2A$ or it is not possible to find similarity transform between them).

Define $f : V \to V$ to be a self-adjoint map $f^* = f$ and define $\pi_2 : V \to V_2$ to be the projection of $V$ onto $V_2$. Must, $\pi \circ f$ be selfad-joint?

This reduces to asking if the projection is self-adjoint. and if $(\pi fv, w) = v, f\pi^*w)$

EDIT: actually I got it; if we further furnish that $V_1$ and $V_2$ are orthogonal with respect to some inner product $(,)$, then we know the projection is self-adjoint

Yes. Because $f$ maps $V_2$ into itself, so when $\pi$ is selfadjoint, $$f\,\pi=\pi\,f\,\pi.$$ And then $$\pi\,f=(f^*\,\pi^*)^*=(f\,\pi)^*=(\pi\,f\,\pi)^*=\pi^*\,f^*\,\pi^*=\pi\,f\,\pi=f\,\pi.$$ In particular, $$(\pi\,f)^*=f^*\,\pi^*=f\,\pi=\pi\,f.$$
• Lol so it has nothing with my conjecture, would you mind explaining the second equality $(f^* \pi^*)^* = (f\pi)^*$. You stated $f$ maps into $V_2 \to V_2$ using the invariance of $f$ since it is self-adjoint right? Not because of my stupid unnecessary comment about orthogonal complements. – jacob smith Sep 27 '16 at 3:33
• Yes. As for the second equality, I'm just using that both $f$ and $\pi$ are selfadjoint, so $f^*=f$ and $\pi^*=\pi$. – Martin Argerami Sep 27 '16 at 3:34
• Also for your 5th equality, don't you have to justify that $\pi$ is self-adjoint to get $\pi^* f^* \pi^* = \pi f \pi$? – jacob smith Sep 27 '16 at 3:35