Finding Jordan canonical form of a $4\times 4$ matrix

Question

Let $$ A = \begin{pmatrix} 1&-2&1&0 \\ 1&-2&1&0 \\ 1&-2&1&0 \\ 1&-1&0&0 \end{pmatrix}$$

I need to find the Jordan canonical form and the minimal polynomial. Now I sense that their is a shortcut (due to the triple repeated rows) but I cannot quite work out what it is?

I know that it has an eigenvalue 0 of multiplicity 4, with the minimal polynomial of $x^2$

Also, I got an eigenvecctor to be $(1,1,1,0)$

Answer 1

Hints (fill in the details):

The characteristic polynomial is given by:

$$|A-\lambda I| = 0 \rightarrow \lambda^4 = 0 \rightarrow \lambda_{1,2,3,4} = 0$$

Since we have an eigenvalue of algebraic multiplicity $4$ and it is zero, we can find the null space of the matrix $A$ which has rank $2$. Thus, two eigenvectors are:

• $v_1 = (0,0,0,1)$
• $v_2 = (1,1,1,0)$

Now, we need two generalized eigenvectors and can find those by solving $[A - \lambda I]v_3 = v_1$, which has a RREF of:

$$\left[\begin{array}{cccc|c} 1& 0& -1& 0& 2\\ 0& 1& -1& 0 & 1\\ 0& 0& 0& 0& 0 \\ 0& 0& 0& 0& 0 \end{array}\right]$$

This leads to an eigenvector of $v_3 = (2,1,0,0)$.

We can repeat this for a fourth eigenvector as $[A - \lambda I]v_4 = v_2$, which has a RREF of:

$$\left[\begin{array}{cccc|c} 1 & 0 & -1 & 0 & -1\\ 0& 1& -1& 0 & 1\\ 0& 0& 0& 0& 0 \\ 0& 0& 0& 0& 0 \end{array}\right]$$

This leads to an eigenvector of $v_4 = (-1, -1,0,0)$.

Now, we know $A$ and $P$ and can find the Jordan block as $J = P^{-1} \cdot A \cdot P$.

Of course we could just figure out the Jordan block from the information above and there are obviously many ways to approach these problems.

Spoiler

The Jordan Normal Form is $J = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{bmatrix}$