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?