# Calculate the Jordan normal form

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

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?

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.
• Thanks, but actually $(A-I)^2$ gives a zero matrix, does that mean that there is no third vector? – Dada Mar 5 at 9:23