Proving the sum of orthogonal projections is an orthogonal projection Q: Let $P_1, \dots, P_B$ be a set of orthogonal projections with $P_iP_j = 0$ for $i \neq j$. Show that $Q = P_1 + \dots +P_m$ is an orthogonal projection.
I was able to show it's a projection pretty easily. Here's my work so far on trying to show that it's orthogonal...

A matrix $P$ be a projection matrix. It is an orthogonal projection matrix iff $\mathcal R(P) \perp \mathcal N(P)$.
We've already shown that $Q$ is a projection matrix.
$Q$ is an orthogonal projection if we can show that $x - Qx$ is orthogonal to the column space of $Q$ for any vector $Qc$ in the column space of $Q$.
Let $y = (cQ)^H = Q^Hc^H$ represent an arbitrary vector $y$ in the row space of $Q$.
Let $0 = Qz$ represent arbitary vector $z$ that is in the null space of $Q$.
We need to show that $\langle y, z \rangle = 0$ and $\overline{\langle z, y \rangle} =0$ .
$$
\begin{align*}
\langle y, z \rangle &= \overline y^H z \\
          &= ((\bar c \bar Q)^H)^H z \\
          &= \bar c(\bar Qz) \\
          &= 0
\end{align*}
$$
$$
\begin{align*}
\overline{\langle z, y \rangle} &= \overline{ \overline{z}^H y} \\
      &= z^H \bar y \\
      &= z^H \bar Q^H \bar c^H \\
      &= (\bar Q z)^H \bar c^H \\
      &= 0
\end{align*}
$$
I realized that I don't actually know that $\bar Q z =0$ which makes the proof erroneous.
I'm not sure how to find an element in the null space of $P_1 + \dots + P_m$ and then invoke this property because I'd need to show that if $(P_1 + \dots + P_m)x = 0$ then we'd need $(\bar P_1 + \dots + \bar P_m)x = 0$ which doesn't seem to follow...

EDIT
I reworked it and think I'm a little closer:
Consider matrix $P_i$ in the sum that generates $Q$. For $P_i$ to a be an orthogonal projection, it must be the case that $\langle (cP_i)^H, z \rangle = 0$ where $P_i z = 0$ because $\mathcal R(P_i) \perp \mathcal N(P_i)$. A symmetrical argument must hold for $\overline{\langle z, (cP_i)^H \rangle}$ which will evaluate to $0$ by definition of othogonal.
Now we consider arbitrary $y = (cQ)^H$ and arbitrary $Qz = 0$. Note that $Qz= (P_1z + \dots + P_m z) = 0$.
We must show that  $\langle y, z \rangle = 0$ and $\overline{\langle z, y \rangle} =0$.
$$
\begin{align*}
\langle y, z \rangle &= \langle (cQ)^H, z \rangle \\
                     &= \langle (c(A_1+ \dots + A_m))^H, z\rangle \\
                     &= \langle (cA_1+ \dots + cA_m)^H, z\rangle \\
                     &= \langle (cA_1)^H+ \dots + (cA_m)^H, z\rangle \\
                     &= \langle (cA_1)^H, z\rangle + \dots + \langle (cA_m)^H, z\rangle \\
                     &= 0
\end{align*}
$$
The other case holds by symmetry argument stated for $P_i$.
But now the problem with the argument is that the null space changes when we add matrices and not in a predictable way (as far as I can tell), also the column space can change dramatically as well.
Maybe I can try getting the professor to explain the answer for this one at the start of class tomorrow.
 A: For each $i \in \{1, \dots, B\}$, let $U_i = R(P_i)$. The condition $i \neq j \implies P_iP_j = 0$ implies that if $i \neq j$, then $U_j \subset U_i^{\perp}$. In other words, $U_1, \dots, U_B$ are all orthogonal to each other. Thus intuitivley, it would make sense that $P_1 + \dots + P_B$ is the orthogonal projection onto $U_1 + \dots + U_B$. This turns out to be true. I recommend you try to prove this, but I left a proof below.

! To prove this we need to show that for any vector $v$, $(P_1 + \dots + P_B)v \in U_1 + \dots + U_B$ and $v - (P_1 + \dots + P_B)v \in (U_1 + \dots + U_B)^{\perp}$. The first assertion is trivial. For the second assertion, by linearity of the inner product, it suffices to show that $v - (P_1 + \dots + P_B)v \in U_i^{\perp}$ for each $i$. But this is true because if $u \in U_i$,
\begin{align}
(v - (P_1 + \dots + P_B)v, u) &= (v, u) - ((P_1 + \dots + P_B)v, u) \\
&= (v, u) - ((P_1v, u) + \dots + (P_Bv, u)) \\
&= (v, u) - (P_iv, u) \\
&= (v - P_iv, u) \\
&= 0.
\end{align}

For a quicker but less intuitive argument, you can use the fact that in a finite dimensional space, $P$ is an orthogonal projection if and only if $P^2 = P$ and $P^* = P$.
A: Here's what I came up with after searching around and stewing on it for a bit. I wish I didn't have the define the inner product but...

Let $P$ be an orthogonal projection. We show that must be symmetric.
Consider vectors $v$ and $w$. then by definition of projection we have
$$
\begin{align*}
v = v_p + v_n\\
w = w_p + w_n
\end{align*}
$$
By definition of orthogonal projection, it's the case that the range of $P$ is perpendicular to the null space of $P$. (i.e. $\mathcal R(P) \perp \mathcal N(P)$).
$$
\begin{align*}
\langle Pv, w \rangle &\qquad \langle v, Pw\rangle\\
\langle v_p, w \rangle &\qquad \langle v, w_p\rangle \\
\langle v_p, w_p + w_n \rangle &\qquad \langle v_p + v_n, w_p\rangle \\
\langle v_p, w_p \rangle + \langle v_p, w_n \rangle &\qquad \langle v_p , w_p\rangle + \langle v_n, w_p \rangle \\
\langle v_p, w_p \rangle &\qquad \langle  v_p, w_p\rangle\\
\end{align*}
$$
Thus
$$
\begin{align*}
\langle Pv, w \rangle = \langle v, Pw\rangle
\end{align*}
$$
Now taking $\langle x, y \rangle = x^H y$ we have:
$$
\begin{align*}
\langle Pv, w \rangle &= \langle v, Pw\rangle \\
v^HP^Hw &= v^HPw \\
\implies P^H &= P
\end{align*}
$$
Therefore all orthogonal projection matrices are (Hermitian) symmetric.
We now show that $Q = P_1 + \dots + P_m$ is an orthogonal projection. We need to show that $\langle Qv, y - Qy\rangle = 0$. Well
$$
\begin{align*}
\langle Qv, y - Qy\rangle &= \langle (P_1 + \dots + P_m)v, y - (P_1 + \dots + P_m)y \rangle \\
 &=\langle P_1v, y - P_1y\rangle + \dots + \langle P_1v, y - P_m y\rangle \\
 &\quad \langle P_2v,y - P_1y\rangle + \langle P_2v,y - P_2y\rangle + \dots + \langle P_2v, y - P_m y\rangle \\
 &\quad \dots \\
 &\quad \langle P_m v, y- P_1y\rangle + \dots + \langle P_m v, y- P_my\rangle
 \end{align*}
$$
First note the cases where we have $\langle P_iv, y - P_iy\rangle$ . Then
$\langle P_iv, y - P_iy\rangle = 0$ because $P_i$ is an orthogonal projection. This is the "diagonal" of foiling.
Now now the cases $\langle P_i v, y - P_jy\rangle$ where $i \neq j$. Fix $i$ and fix $j$. We have
$$
\begin{align*}
\langle P_i v, y - P_jy\rangle &= v^HP_i^H(y - P_j y) \\
                              &= v^HP_i(y - P_j y) \\
                              &= v^HP_iy - v^HP_j y
\end{align*}
$$
Which does not sum to zero unless we also have the case where $v^HP_jy - v^HP_i y$, but we do because when $(P_1+\dots+P_m)(P_m + \dots P_1)$ We have exactly one pair of $P_i P_j = P_j P_i$ by symmetry of foiling. In our case, we're "foiling" the inner product and for every $\langle P_i v, y - P_jy\rangle$ we have  $\langle P_j v, y - P_i y\rangle$ when $i \neq j$.
Therefore we have that the sum of these "off-diagonal" elements sum to zero.
Therefore we have that $\langle Qv, y - Qy\rangle = 0 $ which completes our proof that $Q$ is an orthogonal projection.
