# How do I find the Jordan normal form of a matrix with complex eigenvalues?

I'm trying to obtain the Jordan normal form and the transformation matrix for the following matrix:

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

I've calculated its characteristic and minimum polynomials as $(λ - 1)^2(λ^2 + λ + 1)$, and thus the eigenvalues are $λ = 1$ (with an algebraic multiplicity of $2$) and $λ = \frac{-1 \pm i\sqrt{3}}{2}$.

An eigenvector for $λ = 1$ is $\begin{pmatrix} 0 \\\ 1 \\\ 1 \\\ 1 \end{pmatrix}$.

Since the minimum polynomial contains two identical factors, there must be at least a $2 x 2$ Jordan block associated with the eigenvalue $λ = 1$, and so the Jordan normal form must look something like the following:

$A = \begin{pmatrix} 1 & 1 & 0 & 0 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & \frac{-1 + i\sqrt{3}}{2} & 0 \\\ 0 & 0 & 0 & \frac{-1 - i\sqrt{3}}{2} \end{pmatrix}$

However, I don't know how to derive a transformation matrix $P$ such that $PJ = AP$. How would I go about solving for $P$?

• If your calculus is correct and you find two linearly independent eigenvectors for $\lambda=1$ then the matrix is diagonalizable – user63181 Dec 8 '13 at 9:01
• @modocache: Do you still need help with this? – Amzoti Dec 8 '13 at 13:07
• @Amzoti: I realize that in the question I posted, I listed 2 eigenvectors, but the second one isn't quite right. I've been reading up on Jordan normal form but still don't have much of a clue on how to find the transformation matrix. I'm trying to find a way to reword my question to pinpoint just what it is I'm not understanding. Ideally I'd like to see a step-by-step explanation of the mechanics behind finding the transformation matrix, but I can't find any good resources. – modocache Dec 10 '13 at 2:13
• Do you mean $J = P^{-1} A P$ and you are looking for $P$? – Amzoti Dec 10 '13 at 2:18
• Okay, I will add an answer. Stand by. – Amzoti Dec 10 '13 at 2:28

We are given:

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

We find that characteristic polynomial by solving $|A - \lambda I| = 0$, yielding:

$$(\lambda -1)^2 (\lambda^2 + \lambda +1) = 0$$

This yields a double and a complex conjugate pair of eigenvalues:

$$\lambda_{1,2} = 1, \lambda_{3,4} = -\dfrac{1}{2} \pm i\dfrac{\sqrt{3}}{2}$$

To find the eigenvectors, we solve $[A -\lambda_i I]v_i = 0$, so for $\lambda_1 = 1$, we get:

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

This only gives us a single linearly independent eigenvector, so to find a generalized one, we set up and solve $[A -\lambda_1 I]v_2 = v_1$, yielding (using RREF):

$$v_2 = (3,2,1,0)$$

Next, we have a complex eigenvalue and follow the same procedure and what is nice is that the eigenvector will give us both of them since they too will be complex conjugates. We set up and solve $[A - \lambda_3 I]v_3 = 0$, where $\lambda_3 = -\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}$, which yields:

$$v_3 = (0, \dfrac{1}{2}(-1 + i \sqrt{3}), \dfrac{1}{2}(-1 - i \sqrt{3}), 1)$$

We can now write the last eigenvector as the complex conjugate, yielding:

$$v_4 = (0, \dfrac{1}{2}(-1 - i \sqrt{3}), \dfrac{1}{2}(-1 + i \sqrt{3}), 1)$$

Now, we have $P$ as a linear combination of these column eigenvectors:

$$P = [~v_1 ~| ~v_2 ~| ~v_3 ~| ~v_4 ~]$$

$$J = P^{-1} A P$$

However, we can figure out the JNF as:

$$J = \begin{pmatrix} 1 & 1 & 0 & 0 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & \frac{-1 - i\sqrt{3}}{2} & 0 \\\ 0 & 0 & 0 & \frac{-1 + i\sqrt{3}}{2} \end{pmatrix}$$

You could have also calculated it from what I wrote above $J = P^{-1}AP$.

I would work this forward and backward to get your hands around it.

• Thank you very much! It took me a while, but I got it. I think what confused me most was "We set up and solve $[A−λ3I]v3=0$, where $λ3=−12−i3‾‾√2$, which yields: ...". I couldn't for the life of me solve for $v_3$. I think I need more practice dealing with complex numbers. Thank you so much for such a detailed explanation, though! – modocache Dec 10 '13 at 8:22
• You are very welcome and glad I could help! Regards – Amzoti Dec 10 '13 at 13:38

Well, the dimension of the eigenspace corresponding to $\lambda=1$ is two and therefore there are two Jordan blocks for eigenvalue 1. Further calculation is unnecessary as we know that distinct eigenvalues gives rise to distinct Jordan blocks. Therefore in this case the Jordan normal form is,

$$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & \frac{-1 + i\sqrt{3}}{2} & 0 \\\ 0 & 0 & 0 & \frac{-1 - i\sqrt{3}}{2} \end{pmatrix}$$

• I see, thanks. I guess what you are implying here is that the OP has posted the eigenvectors for the eigenvalue $\lambda =1$ incorrectly. There is only one eigenvector and hence the jordan normal form should be, $$\begin{pmatrix} 1 & 1 & 0 & 0 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & \frac{-1 + i\sqrt{3}}{2} & 0 \\\ 0 & 0 & 0 & \frac{-1 - i\sqrt{3}}{2} \end{pmatrix}$$ I am sorry I just didn't do the problem completely but thought that the OP's eigenvectors are correct. :) – user112535 Dec 8 '13 at 13:29