Making matrix upper triangular by finding Jordan Normal Form

Question

I must make the following matrix upper triangular:

\begin{bmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 1 \\ 1 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix}

To do this, I'm calculating the Jordan Normal form outlined here.

There is only one eigenvalue, 2. So I must find three generalized eigenvectors (ugh).

To find the first one, I try to solve $(A-λI)v_2=v_1$, where $v_1$ is the first eigenvector: \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}

So the equation is the following ($v_{21},v_{22},v_{23},v_{24}$ are the four components of $v_2$)

$$\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_{21} \\ v_{22} \\ v_{23} \\ v_{24} \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$$

My problem is that there is a column of all zeros in the first matrix, which means that there are infinite solutions for $v_{33}$? I'm not sure what I'm doing wrong here but any help would be appreciated!

Answer 1

HINT

Since there is only an eigenvalue the Jordan canonical form is:

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

Since $J$ is obtained by $A$ from a similarity transfomation: $$P^{-1}AP=J$$

to find the generalized eigenvectors you need to solve the following system:

$$AP=PJ$$

More in detail, let ($v_i$ are the generalized eigenvectors, columns of P):

$$P=[v_1,v_2,v_3,v_4]$$

then P has to satisfy the following systems:

$$\begin{cases}Av_1=2v_1 \\ Av_2=v_1+2v_2 \\ Av_3=v_2+2v_3 \\ Av_4=v_3+2v_4\end{cases}$$

You have already solved the first now you can find $v_2$ from the second system of equation and so on.

For example the second system $(A-2I)v_2=v_1$ is:

$$\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} v_2= \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$$

thus

$$ v_2= \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$

Finally you should obtain:

$$P=\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$