# Nilpotent blocks of matrices

I'm trying to read through Hirsch and Smale's "Differential Equations, Dynamical Systems, and Linear Algebra", and I don't understand how this theorem follows from this other theorem.

The first theorem says

Theorem 1. Let $N$ be a nilpotent operator on a real or complex vector space $E$. Then $E$ has a basis giving $N$ a matrix of the form $$A = diag\{A_1, \dots, A_r\}$$ where $A_j$ is an elementary nilpotent block, and the size of $A_k$ is a nonincreasing function of $k$. The matrices $A_1, \dots, A_r$ are uniquely determined by the operator $N$.

They go on to say that if $A$ is an elementary nilpotent matrix, then the dimension of $Ker A = 1$, since the rank of $A$ is $n-1$. That part I understand.

What I don't understand is how Theorem 2 follows from Theorem 1:

Theorem 2. In Theorem 1 the number $r$ of blocks is equal to the dimension of $Ker A$.

Can you help me?

Thanks!

-
The block decomposition of your matrix $A$ corresponds to a decomposition of your vector space into a direct sum of $A$-invariant subspaces, say $V=V_1\oplus\cdots\oplus V_r$, where each $V_i$ is $A_i$-invariant. Therefore $\ker A=\ker A_1\oplus\cdots\oplus\ker A_r$. Now the claim easily follows. – Matemáticos Chibchas Jan 15 '13 at 3:35

Each elementary nilpotent block has rank $n-1$ and the rank the matrix is the sum of the ranks of the block diagonal matrices. Equivalently the nullity of the matrix is the sum of the nullities of the block diagonal matrices. If the nullity of $A$ is $k$ it follows that there must be precisely $k$ blocks, each contributing $1$ to the nullity.