Finding the Jordan form for a $4\times4$ matrix

Question

$$A:= \begin{bmatrix}4 & -4 & -11 & 11\\3 & -12 & -42 & 42\\ -2 & 12 & 37 & -34 \\ -1 & 7 & 20 & -17 \end{bmatrix}$$

I'm struggling with this matrix: it has $p_A(x) = (x-3)^4 $, yet $\ker(A - 3I)^3 $ is already the whole space $ \mathbb C^4 $. I read on another post that (for the $ \mathbb C^3 $ case of a similar matrix) I should take this to be the size of the biggest Jordan block, thus leaving me with $1$ standard eigenvector and a Jordan chain (image of the image of $\ldots$ ) of a vector $v$ not in $\ker(A - 3I)^2$. But this doesn't work for any $v$.

What steps should I follow to find the Jordan form of $A$?

Answer 1

The issue here is that the method you're trying to apply is not quite right. My guess is that the result that is attempted to being applied is that "the algebraic multiplicity of $3$ in the minimal polynomial of your matrix is the size of the largest Jordan block". (you're looking at the geometric multiplicity of your eigenvalue in the minimal polynomial instead).

This is true, and it's also true that the minimal polynomial is $(x-3)^{3}$. This tells me that the size of the largest Jordan block is $3$.

Once you've applied this, we're done, we only have one eigenvalue, the largest Jordan block is of size $3$, the dimension of our space is $4$, we only have room for one more block, so we know what the Jordan Canonical form is (right?).

Answer 2

It maybe of interest to have the explicit decomposition

$A=PJP^{-1}$ with

$$J=\left(\begin{array}{rrr|r}3 & 1 & 0 & 0\\ 0 & 3 & 1 & 0\\ 0 & 0 & 3 & 0\\ \hline 0 & 0 & 0 & 3\end{array}\right) \ \ \text{and} \ \ P=\begin{pmatrix} \ \ 3 & 3 & 14 & 0\\ \ \ 9 & 0 & 0 & 0\\ -6 & 6 & 1 & 1\\ -3 & 6 & 0 & 1\end{pmatrix}.$$

Let $B=A-3I_4$. We have $\ker(B) \subsetneqq \ker(B^2) \subsetneqq \ker(B^3)$ with:

$\ker(B)=\langle \begin{pmatrix} 0\\0\\1\\1\end{pmatrix},\begin{pmatrix} 2\\6\\-1\\1\end{pmatrix}\rangle, \ \ \ker(B^2)=\langle \begin{pmatrix} 0\\0\\1\\1\end{pmatrix},\begin{pmatrix} 2\\6\\-1\\1\end{pmatrix},\begin{pmatrix} 1\\0\\0\\0\end{pmatrix}\rangle, \ \ \ker(B^3)=\mathbb{R}^4.$