# Jordan form from the minimal polynomial $m_A$

Let the matrix

\begin{equation} A=\begin{bmatrix} 1 & 0 & -1 \\ 4 & 3 & 2 \\\ 2 & 1 & 1 \end{bmatrix}. \end{equation}

So far I found the characteristic polynomial $C_A(x)=(x-3)(x-1)^2$ and the minimal polynomial $m_A(x)=(x-3)(x-1)^2$. The solution sheet explains that the Jordan matrix has the form

\begin{equation} J_1=\begin{bmatrix} 3 & * & *\\ * & 1 & * \\\ * & * & 1 \end{bmatrix}. \end{equation} or

\begin{equation} J_2=\begin{bmatrix} 3 & * & *\\ * & 1 & 1 \\\ * & * & 1 \end{bmatrix}. \end{equation}

I know that since $m_a$ has a double root, we know that $A$ is not diagonalizable, so the Jordan form is the matrix $J_2$.

Does someone could explain where come from the matrix $J_1$ and $J_2$? What outcome can allow us to conclude the $J_1$ and $J_2$ form?

• @Peter Which blocks are you talking about? – user332681 Apr 19 '16 at 20:16
• I have three different blocks in head actually : \begin{equation} J_2=\begin{bmatrix} 3 & * & *\\ * & 1 & * \\\ * & * & 1 \end{bmatrix} \end{equation}, \begin{equation} J_2=\begin{bmatrix} 3 & * & *\\ * & 1 & 1 \\\ * & * & 1 \end{bmatrix} \end{equation} and \begin{equation} J_2=\begin{bmatrix} 3 & 1 & *\\ * & 1 & 1 \\\ * & * & 1 \end{bmatrix} \end{equation}... Why two? – user332681 Apr 19 '16 at 20:20
• @Dr.Dray Your last matrix is not in Jordan normal form. – Irregular User Apr 19 '16 at 20:22
• $\pmatrix {1&1\\0&1}$ is a Jordan-block of size $2$. – Peter Apr 19 '16 at 20:23
• So, in the given case, we have the block containing only one element (the $3$) and the block corresponding to $1$ – Peter Apr 19 '16 at 20:28

The solution sheet is wrong.

The characterstic equation and minimal polynomial are enough to tell us what the Jordan normal form of a $2 \times 2$ or a $3 \times 3$ matrix is.

Here $C_A(x)=(x-3)(x-1)^2$, so we know that the Jordan normal form will have exactly one $3$ and two $1$s on the diagonal.

Since the minimal polynomial is $m_A(x)=(x-3)(x-1)^2$, we know that the longest Jordan chain of $\lambda = 3$ is of length $1$, and the longest Jordan chain of $\lambda = 1$ is of length $2$, so the Jordan normal form is your $J_2$.

If on the other hand we had $m_A(x)=(x-3)(x-1)$, then we would have $J_1$.

• I think, that the exercise was to rule out the first case. The minimal polynomial probably was not given in advance. – Peter Apr 19 '16 at 20:26
• @Peter That would make more sense. Though, in this case, given the characteristic equation, it is easy enough to test that $(A - 3)(A-1) \neq 0_3$, so we must have $C_A(x) = m_A(x)$. – Irregular User Apr 19 '16 at 20:28

It easy to see that for $J=\left(\begin{array}{ccc}1&1&0\\0&1&0\\0&0&3\end{array}\right)$ we have $(J-3E)(J-E)^2=0$ where $E$ is the identity matrix, but $(J-3E)(J-E)\neq0$. Then the minimal polynomial and the characteristic polynomial are equal.

• so, $A$ and $J_2$ are in the same conjugacy class of $3\times 3$ matrices – janmarqz Apr 20 '16 at 15:05
• $J_2$ comes from the process of pseudo-diagonalization for square matrices – janmarqz Apr 20 '16 at 15:08
• use as in this example: m.wolframalpha.com/input/… to robotize some part of the calculations – janmarqz Apr 20 '16 at 17:25