 Solution

The identity $T^2=T$ gives the decompositionThe first summand lies in the kernel of a linear map $\ker T$, and the second in the image of a linear map $\mathrm{im}T$. If $w$ belongs to their intersection, write $w=Tu$; then $Tw=T^{2}u=Tu=w$, while $Tw=0$. Thus $w=0$, proving the direct sumOn these two invariant subspaces, $T$ is respectively zero and the identity map. Joining a basis of each produces a basis of eigenvectors, with matrixHence every linear projection is diagonalizable, over the original field, including in characteristic two.