I'm highly doubtful that the answer is "yes," but I fail to see what's incorrect about this very basic proof I've thought of. If someone could point out my error, I'd appreciate it. My logic is as follows:
Claim: Every projection on a Hilbert space is orthogonal.
"Proof":
1. For any linear space $X$, it is true that given a projection $P: X \rightarrow X$ (where $P$ is a projection iff $P$ is linear and satisfies $P^2 = P$), we have $X = \text{ran}(P) \oplus \text{ker}(P)$.
2. Assume X is a Hilbert space (which is, by definition, linear). Since $\text{ker}(P)$ is a closed linear subspace of $X$, then by the projection theorem, $X = \text{ker}(P) \oplus \text{ker}(P)^\perp$ is an orthogonal direct sum.
3. Therefore, $\text{ker}(P)^\perp = \text{ran}(P)$, so $X = \text{ran}(P) \oplus \text{ker}(P)$ is an orthogonal direct sum. Thus, $P$ is an orthogonal projection.
Best Answer
Your argument is wrong because I guess you have in mind a orthogonal projector, namely a map $P: X \to X$ such that $P^2 = P$ and also $P^{\top} = P$ (or $P^* = P$). Here is a simple example. Decompose $\mathbb{R}^2 = \mathbb{R}e_1 \oplus \mathbb{R}(e_1 + e_2)$ where $e_1,e_2$ is the canonical basis i.e. $e_1=(1,0), e_2=(0,1)$. Then you have a projection $P: \mathbb{R}^2 \to \mathbb{R}^2$ given by taking any vector $v$ to its component in $\mathbb{R}(e_1 + e_2)$ along $\mathbb{R}e_1$ e.g. $P(e_1) = 0 , P(e_2) = e_1 + e_2$, such $P$ is a projector i.e. $P^2 = P$ by definition. But it is not orthogonal since the vectors $e_1$ and $e_1 + e_2$ are not perpendicular.