Jordan form and canonical basis

Question

Let
$$
A = \begin{bmatrix}
2 & 1 & 2 & 0 & 0 \\
0 & 4 & 0 & 3 & 4 \\
0 & 0 & 2 & 1 & 2 \\
0 & 0 & 0 & 4 & 1 \\
0 & 0 & 0 & 0 & 2
\end{bmatrix}.
$$
I have to find a Jordan form of a matrix and canonical basis, can someone help me with canonical basis (what is this and how can Ifind it). I have already found Jordan form of a matrix.
Thanks in advance
$$
J = \begin{bmatrix}
2 & 1 & 0 & 0 & 0 \\
0 & 2 & 1 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 4 & 1 \\
0 & 0 & 0 & 0 & 4
\end{bmatrix}.
$$
First of all, I found characteristic polynomial $(2 – \lambda)^3(4-\lambda)^2)$. It implies that the algebraic multiplicity of $2$ is $3$ and the algebraic multiplicity of $4$ is $2$ Then I have found $\dim\operatorname{Ker}(A-2I)$, that equals $1$ and $\dim\operatorname{Ker}(A-4I)=1$. That means that I have only $2$ Jordan boxes and I put $1$ above the main diagonal

Answer 1

Let $f$ be the linear transformation defined by the matrix $J$. We start to examine the bottom right block. The canonical basis $\mathcal B=\{v_1,\ldots,v_5\}$ must satisfy $f(v_4)=4v_4$ and $f(v_5)=v_4+4v_5$, i.e. $g(v_4)=0$ and $g(v_5)=v_4$, where $g:=f-4I$. So $v_5\in\ker g^2\setminus\ker g$, $v_4=g(v_5)\neq 0$ and $v_4\in\ker g$. With some calculations it's easy to see that $\ker g^2=\mathcal L\{\left(1,2,0,0,0\right),\left(-1,0,2,4,0\right)\}$. We can choose $v_5=(-1,0,2,4,0)$ and $v_4=g(v_5)=(6,12,0,0,0)$.

Now, we have to determine $v_1,v_2,v_3$. If $h:=f-2I$, the first three vectors of the base must satisfy: $h(v_1)=0$, $h(v_2)=v_1$ and $h(v_3)=v_2$. Proceeding as before:
$\ker h^3=\mathcal L\{(1,0,0,0,0),(0,0,1,0,0),(0,5,0,2,-4)\}$
$\ker h^2=\mathcal L\{(1,0,0,0,0),(0,0,1,0,0)\}$
$\ker h=\mathcal L\{(1,0,0,0,0)\}$
Let's choose $v_3=(0,5,0,2,-4)$, $v_2=h(v_3)=(5,0,-6,0,0)$ and $v_1=h(v_2)=(-12,0,0,0,0)$.

You can directly verify that the matrix $$P=\begin{bmatrix} -12 & 5 & 0 & 6 & -1\\ 0 & 0 & 5 & 12 & 0\\ 0 & -6 & 0 & 0 & 2\\ 0 & 0 & 2 & 0 & 4\\ 0 & 0 & -4 & 0 & 0 \end{bmatrix}$$ satisfies the relation $$J=P^{-1}AP$$