User alexander - MathOverflow

A series representation

2012-06-28T13:23:27Z

How to find the end of a series representation of the product $$ \prod_{\substack{i=1...\infty\\ j=0...i\\ k=0...j}}\frac{1}{1-x^{i-j}y^{j-k}z^{k}}? $$

For example for product $$ \prod_{\substack{i=1...\infty\\ j=0...i}}\frac{1}{1-x^{i-j}y^j} $$ the ends of series is $$ ...+7x^5 + 12x^4y + 16x^3y^2 + 16x^2y^3 + 12xy^4 + 7y^5 + 5x^4 +\\ +7x^3y + 9x^2y^2 + 7xy^3 + 5y^4 + 3x^3 + 4x^2y + 4xy^2+\\ + 3y^3 + 2x^2 + 2xy + 2y^2 + x + y + 1 $$

Number of Jordan canonical forms for an nxn matrix

2012-06-23T13:41:52Z

How many Jordan canonical forms may have an nxn matrix?

In the article https://oeis.org/A000219 states that the number of Jordan canonical forms for an nxn matrix is the Number of planar partitions of n. But calculating the normal form of $4\times4$ matrix I'm obtained 14 distinct Jordan forms, and for $5\times5$ I'm obtained 27 distinct Jordan forms. Author of the article (https://oeis.org/A000219) states that the normal form 2 11 can't be obtained. I do not understand this. For example $$ J=\begin{pmatrix} \lambda&1&0&0\\ 0&\lambda&0&0\\ 0&0&\lambda'&0\\ 0&0&0&\lambda'\ \end{pmatrix}. $$

I think that the number of Jordan canonical forms for an nxn matrix is the number of partitions of n. (https://oeis.org/A001970). But how to prove it, I do not know.

I would like to hear your opinion on this matter. Sorry for my english.

Number of perturbations of the Jordan form

2011-11-24T20:23:57Z

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 2x2 $$J=\begin{pmatrix} \lambda_0 &1\\ 0&\lambda_0\end{pmatrix},$$ 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 mxm, 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.

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

When we have a Jordan form $$J=\begin{pmatrix}\lambda_0&0\\0&\lambda_0\end{pmatrix}\ \ \ \mbox{denote by}\ \ \ 1(\lambda_0)+1(\lambda_0),$$ we may obtained $$2(\lambda_0) \ \mbox{ or} \ \ 1(\lambda_0)+1(\lambda_1).$$ There are three variants.

If we have a Jordan form $2(\lambda_0)+1(\lambda_0)$ we may obtained $$3(\lambda_0), 2(\lambda_0)+1(\lambda_1), 1(\lambda_0)+1(\lambda_1)+1(\lambda_1),$$ $$1(\lambda_0)+1(\lambda_1)+1(\lambda_2).$$ There are five variants.

If we have a Jordan form $3(\lambda_0)+1(\lambda_0)$ we may obtained

$$4(\lambda_0), \ \ 3(\lambda_0)+1(\lambda_1), \ \ 2(\lambda_0)+2(\lambda_1),$$ $$2(\lambda_0)+1(\lambda_1)+1(\lambda_1), \ \ 2(\lambda_0)+1(\lambda_1)+1(\lambda_2),$$ $$1(\lambda_0)+2(\lambda_1)+1(\lambda_1)=2(\lambda_1)+1(\lambda_1)+1(\lambda_0),$$ $$1(\lambda_0)+1(\lambda_1)+1(\lambda_2)+1(\lambda_3).$$ There are eight variants. And so on.

Jordan form of compact operator

2011-11-20T16:17:50Z

Let $X$ be Banach space over a field $\mathbb{C}$. Consider the Banach space $L_c$ of compact operators in $X$. Let $A^0\in L_c$ be fixed and $\lambda^0\neq 0$ his eigenvalue with algebraic multiplicity $m$ which is associated with Jordan form $J_{\overline k}(\lambda^0)$.

In the canonical basis of the subspace that corresponding to eigenvalue $\lambda^0$, operator $A^0$ has the form $$ A^0=\begin{pmatrix}J_{\overline k}(\lambda^0)&0 \\ 0 & A_{\textbf{2,2}}^0 \end{pmatrix}. $$

Questions.

Can I speak about canonical basis in Banach space or better use a Hilbert space?
Is it natural to consider the Jordan form of compact operator on Banach or Hilbert space?
Does any non-zero eigenvalue of a compact operator have finite multiplicity?

21.11. Operator $A^0$ such that it has non-zero eigenvalues.

Distance between lattices of invariant subspaces of matrices

2011-08-24T11:02:38Z

For a linear transformation $A: C^n \to C^n$ let $Inv(A)$ be the lattice of all $A$-invariant subspaces. In work I.~Gohberg, L.~Rodman "On the Distance between Lattices of Invariant Subspaces of Matrices" analysis the distanse between $Inv(A)$ and $Inv(B)$ defined as follows: $$ dist(Inv(A),Inv(B))=\max\Big(\sup\limits_{M\in Inv(A)}\ \ \inf\limits_{L\in Inv(B)}||P_L-P_M||, \sup\limits_{M\in Inv(B)}\ \ \inf\limits_{L\in Inv(A)}||P_L-P_M||\Big), $$ where $P_N$ is orthogonal projector on the subspace $N$ in $C^n$.

I can not quite understand this definition. Why is there symmetry? Maybe someone met geometric Interpretation of these definition.

Thanks for your help.

Comment by Alexander

2013-03-06T13:53:17Z

@Robert Bryant: Thank you for your help.

Comment by Alexander

2013-03-05T17:41:12Z

Does for any scalar differential operators with constant coefficients associated function do not exist?

Comment by Alexander

2013-03-05T16:57:27Z

This equation has two eigenfunctions. I need one eigenfunction and associated function that corresponds to it. $y(x)$ - eigenfunction, if $$y′′=\lambda y$$. $u(x)$ - associated function to eigenfunction $y(x)$, if $$u′′=\lambda u+y$$.

Comment by Alexander

2013-03-05T16:24:53Z

Sorry, i mean associated function. Fix the question.

Comment by Alexander

2012-12-02T07:05:41Z

I inverstigate the submanifolds of compact operators with fixed Jordan structure of chosen eigenvalue. For this purpose i need an example of such operators. They can be constructed by a differential operators of the special form.

Comment by Alexander

2012-12-01T17:53:54Z

We are my scientific adviser and me. Sorry for my english.

Comment by Alexander

2012-08-17T05:22:37Z

In the question alredy right down that $I(0)<0$.

Comment by Alexander

2012-08-16T17:31:01Z

Thank you for your answer. This case has already been studied, but further progress is not possible.

Comment by Alexander

2012-08-15T14:43:44Z

Solution of the equation is the gap between the eigenvalues of the periodic problem of second order, which correspond to the eigenfunctions of the same oscillation.

Comment by Alexander

2012-08-15T13:54:57Z

It is my mistake. The function only has the property $$ \theta(x+2\pi)=\theta(x)+2\pi $$

Comment by Alexander

2012-06-28T17:54:14Z

Thank you, I already figured out.

Comment by Alexander

2012-06-28T16:58:53Z

Can you write youre Maple's commands?

Comment by Alexander

2012-06-28T16:57:26Z

Thank you. I am edit the quastion.

Comment by Alexander

2012-06-28T16:47:41Z

Thank you. How to get this? Or where I can see more members of the series?

Comment by Alexander

2012-06-28T16:33:18Z

If y=0 this product $$ \prod_{n=1}^\infty\frac{1}{1-x^n} $$ will be generating function of number of partitions of $n$ $$ 1+x+2x^2+3x^3+5x^4+7x^5... $$ (oeis.org/A000041).