# Diagonalization of non-orthogonal projection

I feel confused. Consider the statement.

Matrices are normal iff they are similar to diagonal via multiplication by unitary matrices.

As I was explained, projection doesn't belong to the class of normal matrices. However, projection can be diagonalized.

Any projection has a full set of linearly independent eigenvectors, which we can turn into orthonormal set via Gram-Schmidt process. Therefore, we can express a projection as a product of orthogonal (which are, of course, unitary) matrices and a diagonal one. Nevertheless, it need not be a normal matrix. For me, it looks like a contradiction to the result in quoting marks above.

• But are the diagonalized by unitary matrices? – Squirtle Jul 10 '15 at 18:14
• When you do Gram-Schmidt on your set of eigenvectors it can very easily be that the new, orthonormal basis is no longer of eigenvectors ... – Timbuc Jul 10 '15 at 18:15
• @Timbuc, so this is where my reasoning went wrong! Thanks! – push Jul 10 '15 at 18:40

$$A = \begin{bmatrix} 1 & -1 \\ 0 & 0\end{bmatrix}.$$ which projects onto the $y = 0$ axis (i.e., the $x$-axis). Thus $(1, 0)$ is an eigenvector for $1$. And $(1,1)$ is an eigenvector for $0$. But these, taken together, are not perpendicular, so you cannot form a unitary matrix to conjugate your matrix to Diag(1, 0).
When you have a projection $P$ you can write the space as direct sum of its kernel $\ker P$ and its image $\operatorname{im} P$. Every eigenvector is in $\ker P$ (eigenvalue $0$) or $\operatorname{im} P$ (eigenvalue $1$). If you want a basis of eigenvectors you must take the vectors from $\ker P$ and $\operatorname{im} P$ only.
It is true that for $\ker P$ and $\operatorname{im} P$ separately you could chose orthonormal basis. But if these two subspaces, $\ker P$ and $\operatorname{im} P$, are not orthogonal you can never get an orthogonal basis of eigenvectors.