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I am looking for information about the number of Jordan forms that can be obtained from a given Jordan form of a small perturbation.

For example, if a Jordan form consists of a single cell $2 \times 2$ $$J=\begin{pmatrix} \lambda_0 &1\\\ 0&\lambda_0\end{pmatrix},$$ by small perturbation $\begin{pmatrix} 0&0\\\ 0&\varepsilon\end{pmatrix}$, we can only get the matrix $$\begin{pmatrix} \lambda_0 &0\\\ 0&\lambda_1\end{pmatrix}$$ i.e. only two variants are possible.

If the Jordan form consists of a single cell 3x3, there may be such cases: $$J=\begin{pmatrix} \lambda_0 &1&0\\\ 0&\lambda_0&1\\\ 0&0&\lambda_0\end{pmatrix},$$ $$J+ \begin{pmatrix} 0 &0&0\\\ 0&0&0\\\ 0&0&\varepsilon\end{pmatrix}\sim\begin{pmatrix} \lambda_0 &1&0\\\ 0&\lambda_0&0\\\ 0&0&\lambda_1\end{pmatrix},$$ $$J+ \begin{pmatrix} 0 &0&0\\\ 0&\varepsilon_1&0\\\ 0&0&\varepsilon_2\end{pmatrix}\sim\begin{pmatrix} \lambda_0 &0&0\\\ 0&\lambda_1&0\\\ 0&0&\lambda_2\end{pmatrix}.$$

i.e. only three variants are possible.

I think I proved that if the Jordan form consists of a single cell $m \times m$, then the number of variants equal to $p(m)$ (see http://en.wikipedia.org/wiki/Partition_%28number_theory%29).

It seems to me that these results have been obtained by someone, but I can not find them.

We are working over $\mathbb{C}$.

Able to show that the number of possible Jordan forms of the matrix $n\times n$ gives the number $a(n)$ (see http://oeis.org/A001970). Remains to determine how many Jordan forms can not be receive from this.

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1 Answer 1

I suppose:

  • You want to perturb a single $m$-by-$m$ Jordan block (instead of a Jordan form, which may contain one or more Jordan blocks).
  • Two Jordan forms are considered to have the same structure if they have the same multiset of Jordan block sizes. That is, only the sizes of the Jordan blocks matter and the eigenvalues for different Jordan blocks are just indistinguishable symbols. E.g. $$ \begin{pmatrix}a&1&0\\0&a&0\\0&0&b\end{pmatrix} \textrm{ and } \begin{pmatrix}c&0&0\\0&d&1\\0&0&d\end{pmatrix} $$ are viewed as equivalent structures regardless of whether $a,b,c,d$ are distinct or not.

Then the answer is obviously equal to the number of partitions of $m$. As you have already seen, it suffices to perturb only the diagonal entries to obtain all possible structures of the Jordan form. More precisely, for every partition $m=m_1+m_2+\ldots+m_k$, you can obtain a matrix whose Jordan form has Jordan block sizes $m_1,m_2,\ldots,m_k$ by perturbing the diagonal entries by $$ \underbrace{0,\ldots,0}_{m_1 \textrm{ entries}}, \underbrace{\epsilon,\ldots,\epsilon}_{m_2 \textrm{ entries}}, \underbrace{2\epsilon,\ldots,2\epsilon}_{m_3 \textrm{ entries}}, \ldots, \underbrace{(k-1)\epsilon,\ldots,(k-1)\epsilon}_{m_k \textrm{ entries}}. $$

Note, however:

  1. If you want to perturb a Jordan form instead of a Jordan block, the answer will be different. It may depend on both the geometric and algebraic multiplicities of the eigenvalues.
  2. If the eigenvalue symbols are distinguishable, the answer will be different, too. For example, if you perturb $X=\begin{pmatrix}c&1\\0&c\end{pmatrix}$ slightly to obtain a matrix $Y$, the Jordan form of $Y$ can take the form of $\begin{pmatrix}a&0\\0&b\end{pmatrix}$ for some $a\not=b$, but never the form $\begin{pmatrix}a&0\\0&a\end{pmatrix}=aI$. ($Y$ is similar to $aI$ iff $Y=aI$, but you cannot perturb $X$ slightly to obtain $aI$ because $X$ has an off-diagonal entry $1\gg\epsilon$.)
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1. I want to perturb a single m-by-m Jordan block. 2. Only the sizes of the Jordan blocks is matter. –  Alexander Dec 2 '11 at 5:46

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