Order of vectors in $P$ for the Jordan Canonical form.

Question

Given is the Matrix:

$A = \begin{bmatrix} 2&0&0&0 \\ 0&2&0&0\\ 0&0&2&1 \\ 1&0&0&2 \end{bmatrix}$

Minimal and characteristic polynomial are:

$p(\lambda) = (\lambda-2)^4 \\ m(\lambda)= (\lambda-2)^3 $

The Jordan "matrix" is:

$ J = \begin{bmatrix} 2&1&0&0 \\ 0&2&1&0\\ 0&0&2&0 \\ 0&0&0&2 \end{bmatrix}$

Now I have some problems finding $P$ such that $A = PJP^{-1}$.

Here are the kernels:

$null(A-2I) = span\{\begin{bmatrix}0\\1\\0\\0\end{bmatrix};\begin{bmatrix}0\\0\\1\\0\end{bmatrix}\}$

$null((A-2I)^2) = span\{\begin{bmatrix}0\\1\\0\\0\end{bmatrix};\begin{bmatrix}0\\0\\1\\0\end{bmatrix};\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\}$

and of course:

$null((A-2I)^3) = span\{\begin{bmatrix}1\\0\\0\\0\end{bmatrix};\begin{bmatrix}0\\1\\0\\0\end{bmatrix};\begin{bmatrix}0\\0\\1\\0\end{bmatrix};\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\}$

I know now that I should take this vector $\begin{bmatrix}1\\0\\0\\0\end{bmatrix}$ from $null((A-2I)^3)$ then I take $\begin{bmatrix}0\\0\\0\\1\end{bmatrix}$ from $null((A-2I)^2)$ and finally the whole $null(A-2I)$.

The question is, where do I place them inside of $P$? in what order?

Answer 1

In order to find a matrix $Q$ such that $QAQ^{-1}=J$, let$$q_1=\begin{bmatrix}0\\1\\0\\0\end{bmatrix}\text{ and }q_2=\begin{bmatrix}0\\0\\1\\0\end{bmatrix}.$$You already know that $\operatorname{span}\{q_1,q_2\}=\ker(A-2I)$. So $q_1$ and $q_2$ are eigenvectors of $A$ with eigenvalue $2$ and therefore one of them will be the first column of $Q$ and the other one will the fourth one.

Now, let$$q_3=\begin{bmatrix}0\\0\\0\\1\end{bmatrix}.$$You know that $\operatorname{span}\{q_1,q_2,q_3\}=\ker(A-2I)^2$. Besides, $A.q_3=q_2+2q_3$, There is no $q_1$ here. Finally, let$$q_4=\begin{bmatrix}1\\0\\0\\0\end{bmatrix}.$$You know that $\operatorname{span}\{q_1,q_2,q_3,q_4\}=\ker(A-2I)^3$. Besides, $A.q_4=2q_4+q_3$. Again, there is no $q_1$ here. So, $Q$ will be such that:

• the first column of $Q$ is $q_2$;
• the second column of $Q$ is $q_3$;
• the third column of $Q$ is $q_4$;
• the fourth column of $Q$ is $q_1$.

In other words$$Q=\begin{bmatrix}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{bmatrix}.$$And therefore $P=Q^{-1}$, which happens to be equal to $Q$.

Answer 2

To find a Jordan basis, the standard algorithm consists in cascading in the following way (the placement of the 1s will be in reverse order). Well denote the coordinates of a vector as $(x,y,z,t)$ (in that order):

• Start with any vector in $\ker(A-2I)^3\smallsetminus\ker(A-2I)^2=K^4\smallsetminus\ker(A-2I)^2$. As $$(A-2I)^2=\begin{bmatrix}0&0&0&0\\0&0&0&0\\1&0&0&0\\0&0&0&0\end{bmatrix},$$ this simply means $x\ne 0$. So wset $p_4=(1,0,0,0)$.
• Next we set $p_3=(A-2I)(p_4)=(0,0,0,1)$. Note $p_3\in\ker(A-2I)^2\smallsetminus\ker(A-2I)$.
• Next we set $p_2=(A-2I)(p_3)=(0,0,1,0)$. This one is an eigenvector.
• We know the eigenspace has dimension, since it is defined by the equations $x=t=0$. So we end by completing the vector $p_2$ with another, linearly independent, eigenvector $p_1$, say $(0,1,0,0)$.

In the basis $\mathcal B=(p_1,p_2,p_3,p_4)$, the matrix $A$ takes the form $$J=\begin{bmatrix}2&0&0&0\\0&2&1&0\\0&0&2&1\\0&0&0&2\end{bmatrix}$$ since, by construction, $$Ap_4=2p_4+p_3,\quad Ap_3=2p_3+p_2,\quad Ap_2=2p_2, \quad Ap_1=2p_1,$$ and the change of basis matrix is $$P=\begin{bmatrix}p_1&p_2&p_3&p_4\strut\end{bmatrix}=\begin{bmatrix}0 &0&0&1\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{bmatrix}$$