# Finding Jordan normal form when $(A-\lambda I)$ is singular and nilpotent

I'm having trouble finding generalised eigenvectors of the following matrix:

$$A= \begin{bmatrix} 7 &1 \\ -4 &3 \end{bmatrix}$$ This matrix has a double eigenvalue $\lambda=5$, with one corresponding eigenvector $\vec{v_1}=\left[\matrix{-1\\2}\right]$. Now if I try to use the Jordan technique, I get: $$(A-\lambda I)^2\vec{v_2}=\begin{bmatrix} 2 &1 \\ -4 &-2 \end{bmatrix}^2\vec{v_2}=\vec{0}$$ but since $$\begin{bmatrix} 2 &1 \\ -4 &-2 \end{bmatrix}^2=0$$ this doesn't get me very far. Now I also tried $$(A-\lambda I)\vec{v_2}=\begin{bmatrix} 2 &1 \\ -4 &-2 \end{bmatrix}\vec{v_2}=\vec{v_1}$$ But again, since $\begin{bmatrix} 2 &1 \\ -4 &-2 \end{bmatrix}$ is singular, this doesn't get me anywhere either.

Now how do I proceed in finding the Jordan normal form of $A$?

• Has double eigenvalue corresponding to the eigenvector. Commented Nov 8, 2015 at 8:24

As you get $(A-\lambda I)^2=0$, the equation $$0\cdot \vec v_2=0$$ has lots of solutions to pick - you can take any vector you like. Then to set $\vec v_1=(A-\lambda A)\vec v_2$ and you are done.
In your second attempt: there is a solution, namely, $$\vec v_2=\left[\matrix{0\\-1}\right]$$ even if the matrix is singular.