Orthogonal Projection
Definition 1. Let $T \in \mathcal{L}(V)$ be a projection, where $V$ is an inner product space. We say that $T$ is an orthogonal projection if $R(T)^{\perp} = N(T)$ and $N(T)^{\perp} = R(T)$.
Remark
Remark. (a) If $V$ is finite-dimensional, by Theorem 2, we need only assume that one of the preceding conditions holds.
(b) If $W \leq V$ is finite-dimensional, then there exists exactly one orthogonal projection on $W$.
($\because$) Since $V = W \bigoplus W^{\perp}$, $\forall y \in V, y = u + z$ for some $u \in W, z \in W^{\perp}$ by Theorem 1. Define a function $T: V \longrightarrow V$ by $T(y) = u, \forall y \in V$. Then $T$ is an orthogonal projection on $W$.
If $T$ and $U$ are orthogonal projection on $W$, then $R(T) = W = R(U)$ and $N(T) = R(T)^{\perp} = R(U)^{\perp} = N(U)$. Thus $T = U$.
Theorem 1
Theorem 1. Let $T \in \mathcal{L}(V)$ where $V$ is an inner product space. Then $T$ is an orthogonal projection $\iff$ $T$ has an adjoint $T^*$ and $T^2 = T = T^*$.
Proof. $(\Longrightarrow)$
Since $T$ is a projection, $T = T^2$ by Theorem 2. Let $x, y \in V$ denote $x = x_1 + x_2$ and $y = y_1 + y_2$ for some $x_1, y_1 \in R(T)$ and $x_2, y_2 \in N(T)$. Then we have $\langle T(x), y \rangle = \langle x_1, y_1 + y_2 \rangle = \langle x_1, y_1 \rangle + \langle x_1, y_2 \rangle = \langle x_1, y_1 \rangle$ and $\langle x, T(y) \rangle = \langle x_1 + x_2, y_1 \rangle = \langle x_1, y_1 \rangle + \langle x_2, y_1 \rangle = \langle x_1, y_1 \rangle$ because $T$ is an orthogonal projection. Thus $\langle T(x), y \rangle = \langle x, T(y) \rangle, \forall x, y \in V$, so there exists $T^*$ and $T = T^*$.
($\Longleftarrow$)
We need only to show that $R(T) = N(T)^{\perp}$ and $R(T)^{\perp} = N(T)$.
Let $y \in R(T)$. Then $\forall z \in N(T)$, $\langle y, z \rangle = \langle T(y), z \rangle = \langle y, T(z) \rangle = \langle y, \mathbf{0} \rangle = 0$. Thus $y \in N(T)^{\perp}$.
Let $y \in N(T)^{\perp}$. Then we have $||y - T(y)||^2 = \langle y - T(y), y - T(y) \rangle = \langle y, y - T(y) \rangle - \langle T(y), y - T(y) \rangle \\ = - \langle T(y), y - T(y) \rangle = - \langle y, T(y) - T^2(y) \rangle = 0$ because $y - T(y) \in N(T)$. Thus $T(y) = y$, so $y \in R(T)$. Thus $R(T) = N(T)^{\perp}$.
Since $R(T) \subseteq N(T)^{\perp}$, $N(T) \subseteq (N(T)^{\perp})^{\perp} \subseteq R(T)^{\perp}$. Thus $N(T) \subseteq R(T)^{\perp}$.
Note that $\forall x \in R(T)^{\perp}$, $||T(x)||^2 = \langle T(x), T(x) \rangle = \langle x, T^2(x) \rangle = \langle x, T(x) \rangle = 0$. Thus $x \in N(T)$. Then we have $N(T) \subseteq R(T)^{\perp}$, so $N(T) = R(T)^{\perp}$. Hence $T$ is an orthogonal projection. $\blacksquare$
