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 $P$?
 A: 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.
A: 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,
$$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & \frac{-1 + i\sqrt{3}}{2} & 0 \\\ 0 & 0 & 0 & \frac{-1 - i\sqrt{3}}{2} \end{pmatrix}$$
