- #1
Ted123
- 446
- 0
Let [itex]A =\begin{bmatrix} -14 & 5 & -2 & 1 \\ -38 & 13 & -4 & 5 \\ 25 & -10 & 7 & 5 \\ -17 & 5 & -2 & 4 \end{bmatrix}[/itex] .
The Jordan Normal Form of A is [itex]J =\begin{bmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{bmatrix}[/itex] .
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 [itex]\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}[/itex] and for the eigenvalue 3, an eigenvector is [itex]\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}[/itex] .
But [itex]\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\}[/itex]
but clearly [itex]\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \notin \text{Ker} (A-2I)[/itex]
All eigenvectors should be in the kernel so why isn't this - isn't it an eigenvector?
Similarly [itex]\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\}[/itex]
but clearly [itex]\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} \notin \text{Ker} (A-3I)[/itex]
The Jordan Normal Form of A is [itex]J =\begin{bmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{bmatrix}[/itex] .
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 [itex]\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}[/itex] and for the eigenvalue 3, an eigenvector is [itex]\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}[/itex] .
But [itex]\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\}[/itex]
but clearly [itex]\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \notin \text{Ker} (A-2I)[/itex]
All eigenvectors should be in the kernel so why isn't this - isn't it an eigenvector?
Similarly [itex]\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\}[/itex]
but clearly [itex]\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} \notin \text{Ker} (A-3I)[/itex]