Let $V$ be an inner product space that is an inner direct sum $U_1 \oplus U_2$ of its subspaces $U_1, U_2$, and let $T: V → V$ be the projection to $U_1$ along $U_2$.
Prove that $T$ is self-adjoint if and only if $U_1 ⊥ U_2$.
Honestly, I have no clue how to approach this question. help me out please
Since $V$ is a direct sum, every element $u \in V$ has a unique expression $u=u_1+u_2$ with $u_1 \in U_1$ and $u_2 \in U_2$. The map $T$ is the one which satisfies $T(u_1+u_2) = u_1$.
Suppose $T$ is self-adjoint. We want to show that $\langle u_1, u_2\rangle = 0$ for any $u_1 \in U_1$ and $u_2 \in U_2$. Compute: $$\langle u_1, u_2\rangle = \langle T(u_1), u_2\rangle = \langle u_1, T(u_2) \rangle.$$ But what is $T(u_2)$?
Conversely, suppose $U_1 \perp U_2$. If $u_1, w_1 \in U_1$ and $u_2, w_2 \in U_2$, then we want to compute: $$\langle T(u_1+u_2), w_1+w_2\rangle = \langle u_1, w_1+w_2\rangle = \langle u_1, w_1\rangle + \langle u_1, w_2 \rangle = \langle u_1, w_1 \rangle.$$
Similarly, $\langle u_1+u_2, T(w_1+w_2) \rangle = \langle u_1, w_1 \rangle.$ What can you conclude about $T$?