# Complex eigenvalues Jordan real matrix

As I posted here and here I'm studying Jordan forms and similar concepts.

I've got a problem with complex eigenvalues in jordan real matrices. I know (at least I think so) how to compute the Jordan blocks for real eigenvalues. However, I don't know how to do it with complex ones. Let's illustrate this with an example:

Let $M$ be
$$\begin{bmatrix} 2 & 2 & 4 & 6 & 4\\ 0 & 0& -4 & -6 &-4\\ 0 & 0 & 2 & 4 & 4\\ 0 & 2 & 0 & -4 & -4\\ -2 & -4 & -2 & 2 & 2\\ \end{bmatrix}$$

The charasteristic polynomial is $(x-2)(x^2+4)^2$. I know the Jordan real form is: $$\begin{bmatrix} 2 & 0 & 0 & 0 & 0\\ 0 & 0& -2 & 0 &0\\ 0 & 2 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & -2\\ 0 & 0& 0 & 2 & 0\\ \end{bmatrix}$$

but I don't understand why. The block corresponding to the eigenvalue $2$ is clear, but I'm completely lost about the complex eigenvalues and how to arrive to that matrix.

• What's the characteristic polynomial and the eigenvalues of $\begin{pmatrix} 0& -2\\2 & 0 \end{pmatrix}$? Does this matrix have eigenvalues as an endomorphism in $\mathbb R^2$? – Roland Jan 3 '16 at 17:59
• We are not ignoring conjugates. The block I quoted represents precisely (one instance of) $2i$ and its conjugate $-2i$. Note that the CP of the large matrix is $(\lambda - 2)(\lambda^2 +4)(\lambda^2 +4)$ and that we have two identical factors each leading to $2i$ and its conjugate $-2i$. Each of the factors in brackets represents one block. The shape of the block, of course, depends on the eigenspace structure and the (possible) existence of Jordan chains. You can think of this matrix as a real representation of $diag(2i,-2i)$ – Roland Jan 3 '16 at 18:11
• @V.Gonzalez you have the right idea. In this context, the "1"s would be the $2 \times 2$ identity matrix. So, the Jordan real form of another matrix with this characteristic polynomial might be $$\pmatrix{ 2&0&0&0&0\\ 0&0&-2&1&0\\ 0&2&0&0&1\\ 0&0&0&0&-2\\ 0&0&0&2&0}$$ – Ben Grossmann Jan 3 '16 at 20:26
• @V.Gonzalez I said that you could use $M^2+4I$, not that the rules would be exactly the same. Because we're considering $\pm 2i$ at the same time, every number here is double what you would get with $M-2iI$. So, $4/2=2$ means we have $2$ blocks of size at least $1$. – Ben Grossmann Jan 4 '16 at 13:08