On page 21 of Matrix Differential Calculus by Magnus and Neudecker (3rd ed, ISBN:0-471-98632-1), the book states, without any apparent justification, that to prove the statement:
If $A$ has $r$ non-zero eigenvalues, then rank($A$) $\geq r$.
We start by
...using [Schur Decomposition], $S^*AS=M$. We partition $M = \left( \begin{array}{cc}M_1 && M_2 \\ 0 && M_3 \end{array} \right)$ where $M_1$ is a non-singular upper triangular $r \times r$ matrix and $M_3$ is strictly upper triangular.
where $M$ is the upper diagonal matrix and $S$ is the unitary matrix resulting from Schur Decomposition. This statement is later repeated for Theorem 21.
Just looking at an arbitrary upper triangular matrix such as
$\left(\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5\\ 0 & 0 & 1 & 2 & 3\\ 0 & 0 & 4 & 5 & 6\\ 0 & 0 & 0 & 7 & 8\\ 0 & 0 & 0 & 0 & 9 \end{array}\right)$
I can't see how this is generally possible. Is there a theorem regarding Schur Decomposition that I'm missing?
You can do as follows. Let $E=\bigoplus_{i|\lambda_i\not=0}\ker((A-\lambda I)^n),F=\ker(A^n)$; note that $A_{|E}\in L(E),A_{|F}\in L(F)$ and $E\bigoplus F=\mathbb{C}^n$. We choose an orthonormal basis of $E$ that triangularizes $A_{|E}$ and an orthonormal basis of $E^{\perp}$. The matrix $A$ becomes $B=\begin{pmatrix}M_1&N_2\\0&N_3\end{pmatrix}$ where $M_1$ is invertible triangular and $N_3$ is nilpotent. It remains to choose another orthonormal basis of $E^{\perp}$ that triangularizes $N_3$.