# Thread: Linear algebra: Jordan forms & eigenvectors

1. ## Linear algebra: Jordan forms & eigenvectors

Dear MHF members,

I have the following problem.
If the Jordan form of $A$ is a single Jordan block of size $n \times n$,
then how many independent eigenvectors does A have and what is the dimension of $\ker (A-\lambda I)^2$?

bkarpuz

2. Originally Posted by bkarpuz
Dear MHF members,

I have the following problem.
If the Jordan form of $A$ is a single Jordan block of size $n \times n$,
then how many independent eigenvectors does A have and what is the dimension of $\ker (A-\lambda I)^2$?

bkarpuz

As the number of Jordan blocks corr. to an eigenvalue equals the dimension of the eigenspace corr. to that eigenvalue, we have in this

case that there's one unique eigenvalue for that matrix and its eigenspace is of dimension 1...

Tonio

3. Originally Posted by bkarpuz
and what is the dimension of $\ker (A-\lambda I)^2$?

One way for this question (not the most elegant): if J is the Jordan form of A, then

$(J-\lambda I)^2=\begin{bmatrix} 0 & 1 & 0 & & \ldots &0 & 0 & 0\\ 0 & 0 & 1 & &\ldots &0 & 0 & 0 \\0 & 0 & 0 & & \ldots & 0 & 0 & 0\\ \vdots&&&&&&&\vdots \\ 0 & 0 & 0 & & \ldots & 0 & 1 & 0\\ 0 & 0 & 0 & & \ldots & 0 & 0 & 1\\0 & 0 & 0 & &\ldots & 0 & 0 & 0\end{bmatrix}^2=\begin{bmatrix} 0 & 0 & 1 & & \ldots &0 & 0 & 0\\ 0 & 0 & 0 & &\ldots &0 & 0 & 0 \\0 & 0 & 0 & & \ldots & 0 & 0 & 0\\ \vdots&&&&&&&\vdots \\ 0 & 0 & 0 & & \ldots & 0 & 0 & 1\\ 0 & 0 & 0 & & \ldots & 0 & 0 & 0\\0 & 0 & 0 & &\ldots & 0 & 0 & 0\end{bmatrix}$

As a consequence,

$\dim \ker (A-\lambda I)^2=\dim \ker (J-\lambda I)^2=n-\textrm{rank}(J-\lambda I)^2=n-(n-2)=2$