# Eigenvectors of a matrix in Jordan Normal Form

1. Dec 19, 2011

### Ted123

Let $A =\begin{bmatrix} -14 & 5 & -2 & 1 \\ -38 & 13 & -4 & 5 \\ 25 & -10 & 7 & 5 \\ -17 & 5 & -2 & 4 \end{bmatrix}$ .

The Jordan Normal Form of A is $J =\begin{bmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{bmatrix}$ .

Now, I've been told that eigenvectors can be read off as the columns with no 1's on the superdiagonal.

So for the eigenvalue 2, an eigenvector is $\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$ and for the eigenvalue 3, an eigenvector is $\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$ .

But $\text{Ker} (A-2I) = \left\{ \begin{bmatrix} w \\ 3w \\ 0 \\ w \end{bmatrix} : w\in\mathbb{R} \right\} = \text{span} \left\{ \begin{bmatrix} 1 \\ 3 \\ 0 \\ 1 \end{bmatrix} \right\}$

but clearly $\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \notin \text{Ker} (A-2I)$

All eigenvectors should be in the kernel so why isn't this - isn't it an eigenvector?

Similarly $\text{Ker} (A-3I) = \left\{ \begin{bmatrix} 0 \\ \frac{2}{5}z \\ z \\ 0 \end{bmatrix} : z\in\mathbb{R} \right\} = \text{span} \left\{ \begin{bmatrix} 0 \\ 2 \\ 5 \\ 0 \end{bmatrix} \right\}$

but clearly $\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} \notin \text{Ker} (A-3I)$

2. Dec 19, 2011

### Ray Vickson

The vectors (1,0,0,0)^T and (0,0,1,0)^T are eigenvectors of J, not of A.

RGV

3. Dec 19, 2011

### Ted123

Of course, $J=P^{-1}AP \neq A$ for some change of basis matrix $P$.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook