Calculate the Jordan normal form

Question

I have the matrix $A=\begin{bmatrix} -2 & -3 & 6 \\ 1 & 2 & -2\\ -1 & -1 &3 \end{bmatrix}$ and I have to find the transformation matrix and its Jordan normal form.

This is what I did so far:

Char. polynomial: $p_A=(\lambda-1)^3$ so I have eigenvalue $\lambda=1$.

Then I calculated the kernel: $$\ker(A-I_3)=\ker\begin{pmatrix} -3 & -3 & 6 \\ 1 & 1 & -2 \\ -1 & -1 &2 \end{pmatrix} = \operatorname{span}\left\{\begin{pmatrix} -1\\1\\0\end{pmatrix};\begin{pmatrix} 2\\0\\1\end{pmatrix}\right\}$$

Then I have to calculate a third vector $v_3$, such that: $(A-I_3)v_3=v_2$ but the system doesn't give me a solution for this vector, am I missing something?

Answer 1

The eigenspace corresponding to $\ker(A-I)$ is two dimensional and we have one eigenvalue therefore there must be a generalized eigenvector in the kernel of $(A-I)^2$.
As if $(A-I)^2v=0$, then $(A-I)v$ must be one of your eigenvectors (as there are only 2 corresponding to the eigenvalue and we would have $(A-I)[(A-I)v]=0$.
Thus in fact you are looking for the basis of the kernel of $(A-I)^2$. This will give you the $v_3$ you are looking for.

For a great reference on computing and understanding Jordan-Normal form at a basic level please see https://www.maths.tcd.ie/~pete/ma1212/chapter2.pdf.

Answer 2

Let $v_3=\begin{bmatrix}1\\0\\0\end{bmatrix}\in\ker(A-I)^2=O.$

Then $v_2=(A-I)v_3=\begin{bmatrix}-3\\1\\-1\end{bmatrix}\in\ker(A-I)$. And we may take $v_1=\begin{bmatrix}-1\\1\\0\end{bmatrix}\in\ker(A-I).$

Hence, $P=\begin{bmatrix}v_1|v_2|v_3\end{bmatrix}$ gives $P=\begin{bmatrix}-1&-3&1\\1&1&0\\0&-1&0\end{bmatrix}.$

Check: $P^{-1}AP=$ is Jordan form of $A$!

I followed the algorithm given in Wikipedia.