 Solution

Suppose $E^2=E$. Every vector has the decompositionIf $w=Ez$ also belongs to $\ker E$, then $w=E^{2}z=Ew=0$. HenceOn 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 $\mathrm{diag}(I_r,0)$, where $r=\mathrm{rank}E$.

Conversely, this block matrix squares to itself, so any linear map having that matrix in a basis satisfies $E^2=E$. Thus projection operators are precisely the linear maps with this block form. This is an algebraic projection; no inner product or orthogonality is required.