# Determine the Jordan normal form of a complex matrix

I want to determine the Jordan normal form of a complex matrix $A$ with characteristic polynomial $\chi_A(x)=(x+1)^4(x+2)^2$, minimal polynomial $m_A(x)=(x+1)^2(x+2)^2$ and that has the property that the matrix $A+I_6$ has order $3$.



I have done the following:

The eigenvalues of $A$ are the roots of $\chi_A$. So the eigenvalues of $A$ are $-1$ and $-2$, where the multiplicity of the eigenvalue $-1$ is $4$ and the multiplicity of the eigenvalue $-2$ is $2$.

From the minimal polynomial we get the information that there is a Jordan block for the eigenvalue $-1$ that is $2\times 2$ (since we have the power $2$). We also have a Jordan block for the eigenvalue $-2$ that is $2\times 2$ (since we have the power $2$).

Is everything correct so far?

What information do we get from the fact that the matrix $A+I_6$ has order $3$ ?

• Try to find all possible Jordan normal forms matching the given data. Now observe that we can assume without restriction of generality that $A$ is already in this form. (Since $I=I_6$ commutes with the base change matrix.) Which is the multiplicative order of $A+I$ for each possible form of $A$? – dan_fulea Sep 2 '18 at 13:53
• I haven't really understood what I have to do now. Could explain it further to me? @dan_fulea – Mary Star Sep 2 '18 at 14:33
• OK, first of all, we know that $A$ is of the shape$$\begin{bmatrix}-1&*&*&* &&\\&-1&*&*&&\\&&-1&*&&\\&&&-1&&\\&&&&-2&*\\&&&&&-2\end{bmatrix}\ .$$ Which are now all possibilities to fill in the stars (up to conjugation with permutation matrices)? Now assume $A$ is in the above list. What is $A+I$? (What is the order of a matrix? Its rank?) – dan_fulea Sep 2 '18 at 14:53
• Do we have that form because of the multiplicity of the eigenvalues? @dan_fulea – Mary Star Sep 2 '18 at 15:15
• It is ok so far, the two 2x2 Jordan blocks do not cover all six rows / columns, so there is a need of two more 1x1 blocks or one more 2x2 block... – dan_fulea Sep 2 '18 at 16:41

The information you have so far is correct. This narrows it down to two forms:

$$\begin{bmatrix} -1 & 1 & 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 & 0 & 0\\ 0 & 0 & -1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0 & -2 & 1\\ 0 & 0 & 0 & 0 & 0 & -2\\ \end{bmatrix} \quad\text{ and } \quad \begin{bmatrix} -1 & 1 & 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 & 0 & 0\\ 0 & 0 & -1 & 1 & 0 & 0\\ 0 & 0 & 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0 & -2 & 1\\ 0 & 0 & 0 & 0 & 0 & -2\\ \end{bmatrix}$$

Now the first matrix has $\operatorname{rank} (A + I) = 3$ and the second one has $\operatorname{rank} (A + I) = 4$.

Therefore the first one is the one you need.

• Why do we get these two forms? I got stuck right now. Could you explain that further to me? – Mary Star Sep 2 '18 at 15:32
• @MaryStar The characteristic polynomial gives that $-1$ appears $4$ times, and $-2$ appears $2$ times on the diagonal. The minimal polynomial gives that the size of the largest block for $-1$ is $2 \times 2$, and the size of the largest block for $-2$ is also $2 \times 2$. Hence it only remains to check whether there are two $2\times 2$ blocks for $-1$, or there is one $2\times 2$ block and two $1 \times 1$ blocks. – mechanodroid Sep 2 '18 at 15:37
• Ah ok!! But how do we place the $1$s at the second diagonal? Do we place a 1 if we have a 2x2 block? – Mary Star Sep 6 '18 at 12:37
• @MaryStar Yes, a $2\times 2$ block is $\begin{bmatrix} \lambda & 1 \\ 0 & \lambda\end{bmatrix}$, a $3 \times 3$ block is $\begin{bmatrix} \lambda & 1 & 0\\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{bmatrix}$ and so on. – mechanodroid Sep 6 '18 at 12:59
• Ah I see! Thanks a lot!! :-) – Mary Star Sep 6 '18 at 21:23