# What exactly determines the block-sizes for Jordan forms?

For instance, with $T \in \mathcal{L}$(Mat($2,2,\mathbb{C}$)) we are given that the minimal polynomial of $T$ is $p(z) = (z - 2i)(z + 7)^2$. I want to find the possible Jordan Forms pertaining to this $T$.

We know that the characteristic polynomial of $T$ is a polynomial multiple of the minimal polynomial, thus it is either $(z - 2i)^2 (z+7)^2$ or $(z - 2i) (z+7)^3$. So the eigenvalue $2i$ will have multiplicity = $1$ or $2$, and the eigenvalue $7$ will have multiplicity = $2$ or $3$.

However, I am left wondering how we can determine the block sizes for each eigenvalue given only the information that the minimal polynomial is $p(z) = (z - 2i)(z + 7)^2 \,$?

• Does it have something to do with the fact that the minimal polynomial is, by definition, the unique monic polynomial of smallest degree s.t: $p(T) = 0$?? – Javier Dec 11 '16 at 2:51

## 2 Answers

Given the minimal polynomial of $T$, in order to give the possible Jordan normal forms, the following facts are useful:

• The minimal polynomial tells you what are the eigenvalues of $T$. In your case, there are two eigenvalues $\lambda_1=2i$ and $\lambda_2=-7$.

• The sum of the sizes of all Jordan blocks corresponding to an eigenvalue $\lambda_i$ is its algebraic multiplicity, which is given by the characteristic polynomial of $T$.

• The minimal polynomial divides the characteristic polynomial.

• Given an eigenvalue $\lambda_i$, its multiplicity in the minimal polynomial is the size of its largest Jordan block.

So what can you tell about the Jordan form of $T$ according to what we have above?

You have exploited the third bullet point to give the possible characteristic polynomials:

$$p_1(z)=(z-2i)^2(z+7)^2,\quad p_2(z)=(z-2i)(z+7)^3.$$

Now suppose it is $p_1$. ($p_2$ for your exercise.)

• The diagonal must consist of $2i$ and $-7$.
• Each of $2i$ and $-7$ appears twice in the diagonal.
• The biggest block for $-2i$ is $(-2i)$ and the one for $-7$ is $\begin{pmatrix}-7&1\\0&-7\end{pmatrix}$.

So you can conclude that the Jordan form for $T$ should be $$\begin{pmatrix} -2i&&&\\ &-2i&&\\ &&-7&1\\ &&0&-7 \end{pmatrix}.$$

Hint:

For a given eigenvalue $\lambda$, let $d_i=\dim \ker (T-\lambda I)^i$. A general result is the following:

The subspaces $\ker (T-\lambda I)^i$ constitute an eventually constant, non-decreasing sequence, and $\;d_i-d_{i-1}\;(i\ge 1)$ is equal to the number of Jordan blocks of size $\ge i$.