= Solution
Suppose $E^2=E$. Every <vector> has the decomposition
$$
v=Ev+(v-Ev),\qquad Ev\in\operatorname{im}E,\quad v-Ev\in\ker E.
$$
If $w=Ez$ also belongs to $\ker E$, then $w=E^2z=Ew=0$. Hence
$$
\boxed{V=\operatorname{im}E\oplus\ker E.}
$$
On its <image of a linear map>, $E$ acts as the identity; on its <kernel of a linear map>, it acts as zero. Concatenate a <basis> of the image with a <basis> of the kernel. The resulting <matrix> is $\operatorname{diag}(I_r,0)$, where $r=\operatorname{rank}E$.
Conversely, this block <matrix> squares to itself, so any <linear map> having that matrix in a <basis> satisfies $E^2=E$. Thus \b[<projection operators> are precisely the <linear maps> with this block form]. This is an algebraic projection; no <inner product> or orthogonality is required.
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