Nilpotent blocks of matrices

Question

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!

Answer 1

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.

Also, the form that the book is talking about is much more general than just for nilpotent operators. This type of decomposition is called the Jordan Normal Form. The fact that the geometric multiplicity of an eigenvalue is the number of Jordan blocks is a fundamental fact of the Jordan Normal Form and your result is an immediate consequence.

Answer 2

i found answer of your question in matrix analysis and applied linear algebra (carld.meyer)