# Jordan block in a matrix with a complex eigenvalue

Could you tell me, in general, how many Jordan blocks there are in the Jordan form of a matrix whose eigenvalue is complex, for example the matrix is $6 \times 6$ and the eigenvalue is $3i+2$?

What can we say about Jordan blocks of a matrix with two eigenvalues?

I know that if we have $f: X \rightarrow X$ an endomorphism of a real vector space, then its complexification $f^{\mathbb{C}} : X^{\mathbb{C}} \rightarrow X^{\mathbb{C}}$ has the following Jordan decomposition:

$X^{\mathbb{C}} = V_1 \bigoplus ... \bigoplus V_p \bigoplus W_1 ... \bigoplus W_p \bigoplus \overline{W_1} \bigoplus ... \overline{W_p}$,

where $V_i$ is are Jordan subspace with real eigenvalues and $W_i$ are subspaces with complex eigenvalues with positive imaginary part.

I also know that the numbers of Jordan blocks corresponding to $\lambda$ and $\overline{\lambda}$ are equal.

For example, if $M \in \mathcal{M}(6 \times 6, \mathbb{C})$ and $\lambda = a + bi$, then

$\left[\begin{array}{ccc}a+bi&0&0&0&0&0\\1&a+bi&0&0&0&0\\0&1&a+bi&0&0&0\\0&0&1&a-bi&0&0\\0&0&0&1&a-bi&0\\0&0&0&0&1&a-bi\end{array}\right]$ (one block $3 \times 3$ with $\lambda$ and one block with $\overline{\lambda}$).

We can also have $1$ block $1 \times 1$ with $\lambda$, $1$ block $2 \times 2$ with $\lambda$ and the same with $\overline{\lambda}$, and so on, until we exhaust all possible partitions of $3$. Is that reasoning corect?

My problem is how can we express this in terms of a matrix with real entries, I mean, with blocks $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$ and $\left[\begin{array}{ccc}a&b\\-b&a\end{array}\right]$?

Could you help me?

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If the $6 \times 6$ matrix has one eigenvalue, it can have anywhere from 1 to 6 Jordan blocks. If there are two eigenvalues, then there can be anything from 2 to 6 Jordan blocks. –  copper.hat Jun 6 '13 at 6:40

There is nothing special about complex eigenvalues. On the generalised eigenspace for $\lambda=-2+3i$, which is by definition the space of vectors that are ultimately annihilated by repeated applications of $A-\lambda I$, that matrix acts as a nilpotent matrix, and the problem of finding the Jordan normal form of $A$ on its generalised eigenspace for $\lambda$ is exactly the same as finding Jordan normal form of the nilpotent restriction of $A-\lambda I$ to that space. For a nilpotent matrix the only eigenvalue is$~0$, which simplifies thinking about it.

In case $A$ has multiple eigenvalues, the (complexified) vector space canonically decomposes into a direct sum of generalised eigenspaces for those eigenvalues; one performs this decomposition and then focusses on the restrictions to those generalised eigenspaces, which are completely independent. All in all the whole problem of finding Jordan normal forms boils down to the special case of nilpotent matrices.

Of course, if you start with a real matrix with complex (conjugate) eigenvalues, then the decomposition into generalised eigenspaces only exists in the complexicfied vector space: there is no decomposition at all of the initial real vector space that corresponds to it. This is no different from what happens in the case where the matrix is diagonalisable but with complex eigenvalues: there is no such thing as a real subspace that corresponds to each separate complex eigenvalue (although you can associate one real subspace space to each pair of complex conjugate eigenvalues; this is however not an eigenspace).

For the final question you added: one cannot do that expression in terms of real matrices. Did anybody tell you that one could? There may be some standard form you can give to the matrix over the real numbers, but describing it would be quite messy. If you really really want to know, you can try to read up on the Jordan-Chevalley decomposition.

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Could you tell me what the matrix with blocks $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$ and $\left[\begin{array}{ccc}a&b\\-b&a\end{array}\right]$ can look like if the matrix is $4 \times 4$ or $6 \times 6$? –  Hagrid Jun 6 '13 at 6:53
@Hagrid I just told you this cannot be done. Why are you insisting? –  Marc van Leeuwen Jun 6 '13 at 6:55
Could you take a look at this: ajm.asj-oa.am/36/1/AG_f6_Dalalyan.pdf? Also wikipedia says en.wikipedia.org/wiki/Jordan_normal_form#Real_matrices. Could you tell me how to arrange Jordan blocks given an eigenvalue and dimension of the space in $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$? –  Hagrid Jun 6 '13 at 7:04
I'm sorry if I got you confused. I just wanted to write everything I know on the subject. –  Hagrid Jun 6 '13 at 7:11
The WP entry appears to contradict what I said (that finding a standard for over the reals isn't easy) and I'll have to think about that. You can just try to perform what they say in the example: take a complex Jordan basis for one of the two complex conjugate eigenvalues, and take their real and imaginary parts giving twice as many real vectors, which joined together give a basis for the sum of the two generalised eigenspaces. So far things are clear; what am not sure about is whether the matrix expressed on this basis really gives the form displayed in the WP article. –  Marc van Leeuwen Jun 6 '13 at 7:53