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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.

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    $\begingroup$ By a projection you mean a linear map $P:X \to X$ such that $P^2 = P$ ? $\endgroup$ – Holonomia Jun 18 '15 at 16:53
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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.

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  • $\begingroup$ Thanks! The example helped make it particularly clear. $\endgroup$ – user218389 Jun 19 '15 at 3:25
  • $\begingroup$ you are welcome. $\endgroup$ – Holonomia Jun 19 '15 at 8:26
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Careful, $E=F\oplus G=F \oplus H$ does not imply $G=H$.

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You state that $X = \ker(P) \oplus \ker(P)^\perp $ and $X = \ker(P) \oplus \text{ran}(P)$ ; but then, your implication (3) that "therefore $\ker(P)^\perp = \text{ran}(P)$" rests on the assumption that the supplement space of $\ker(P)$ is unique, which is not true in general.

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A projection $P$ should be thought of as being onto a subspace $A$ relative to a subspace $B$.

As a simple example, look at $\mathbb{R}^{2}$ with standard basis $\{ e_1, e_2\}$. Another basis for $\mathbb{R}^{2}$ is $\{ e_1, e_1+e_2 \}$. The orthogonal projection of $\mathbb{R}^{2}$ onto $[\{e_1\}]$ is $Px = (x,e_1)e_1$. However, if you want to write $x$ in terms of the second basis, \begin{align} x & = (x,e_1)e_1+(x,e_2)e_2 \\ & = (x,e_1)e_1+(x,e_2)(e_1+e_2)-(x,e_2)e_1 \\ & = \{ (x,e_1) - (x,e_2) \}e_1 + (x,e_2)(e_1+e_2). \end{align} This gives rise to a second projection onto $[\{e_1\}]$: $$ Qx = (x,e_1-e_2)e_1. $$ You can check that this is a projection onto $[\{e_1\}]$ as well: \begin{align} Q^{2}x & = (Qx,e_1-e_2)e_1 \\ & = ((x,e_1-e_2)e_1,e_1-e_2)e_1) \\ & = (x,e_1-e_2)(e_1,e_1-e_2)e_1 \\ & = (x,e_1-e_2)e_1 = Qx. \end{align} Both projections are onto $[\{e_1\}]$, but the first is relative to $[\{e_2\}]$, while the second is relative to $[\{e_2-e_1\}]$; these are different even though $[\{e_1\}]\oplus[\{e_2\}] = [\{e_1\}]\oplus[\{e_2-e_1\}]$.

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