Jordan Normal Form and eigenvalue 0

Question

I understand the processes of putting a matrix into Jordan normal form and forming the transformation matrix associated to "diagonalizing" the matrix. So here's my question:

Why is it that when you have an eigenvalue x=0 with algebraic multiplicity greater than 1, that you don't put a 1 in the superdiagonal of the JNF matrix but when the eigenvalue is non-zero and satisfies the same properties, we put a 1 in the superdiagonal of the Jordan normal form?

My professor posted solutions to an assignment involving finding a matrix exponential, but the JNF of a matrix had eigenvalue x=0 with algebraic multiplicity of 3,yet had no entries of 1 along the superdiagonal.

In advance, I would like to thank you for your help.

Answer 1

We have a single eigenvalue of $\lambda_1 = 1$ and a triple eigenvalue of $\lambda_{2,3,4} = 0$.

For $\lambda=0$, we need to find three linearly independent eigenvectors and can just use the null space of $A$ for this. We have:

$$NS(A) = NS \left(\begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}\right)$$

This produces $v_{2,3,4} = (0,0,0,1), (0,0,1,0), (0,1,0,0)$ as three linearly independent eigenvectors, thus this matrix is diagonalizable and we can write the Jordan block using the eigenvalues down the main diagonal as:

$$J = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$$

Answer 2

I can make some counter-example about $1$.

Let's see two matrices: $A=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$.

In both cases eigenvalues are $1$ with multiplicity $2$, but $A$ has two eigenvectors such that $Ae=e (\{1,0\},\{0,1\})$ and $B$ has only one such eigenvector$(\{1,0\})$.

We put a $1$ into subdiagonal only then the corresponding block has only one eigenvector with such eigenvalue.