# Eigenvectors of a certain big upper triangular matrix

I'm looking at this matrix:

$$\begin{pmatrix} 1 & 1/2 & 1/8 & 1/48 & 1/384 & \dots \\ 0 & 1/2 & 1/4 & 1/16 & 1/96 & \dots \\ 0 & 0 & 1/8 & 1/16 & 1/64 & \dots \\ 0 & 0 & 0 & 1/48 & 1/96 & \dots \\ 0 & 0 & 0 & 0 & 1/384 & \dots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}$$

The first row contains the reciprocals of the double factorials $$2, \qquad 2 \cdot 4, \qquad 2 \cdot 4 \cdot 6, \qquad 2 \cdot 4 \cdot 6 \cdot 8, \qquad \dots$$ Each row is a shift of a scalar multiple of the first row, and the scalar multiple is in each case itself a reciprocal of a double factorial, so that the main diagonal is the same as the first row. A consequence is that each column is proportional to the corresponding row of Pascal's triangle. E.g. the last column shown is proportional to $$1, 4, 6, 4, 1.$$ This matrix is the matrix of coefficients in the "inversion formulas" section of this rant that I wrote.

I found the first three eigenvectors: $$\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ \vdots \end{pmatrix}, \begin{pmatrix} 1 \\ -1 \\ 0 \\ 0 \\ 0 \\ \vdots \end{pmatrix}, \begin{pmatrix} 5 \\ -14 \\ 21 \\ 0 \\ 0 \\ \vdots \end{pmatrix}$$ Meni Rosenfeld pushed this through some software and found that up to the 40th eigenvalue, the signs of the components of the eigenvectors alternate.

Can anything of interest be said about the eigenvectors?

• I found your answer interesting. I don't recalling saying I found anything misleading, and I didn't know you'd given a "hint". – Michael Hardy May 31 '10 at 20:42
• Michael, I edited my post and added a comment below. – Wadim Zudilin Jun 1 '10 at 12:53
• I won't claim that it is related to your matrices, but there are appearances of similar looking ones in the old literature (e.g. ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/…). – Wadim Zudilin Jun 9 '10 at 1:51

As far as I understand your construction, your matrix is $$\operatorname{diag}\biggl(1,\frac12,\frac18,\dots,\frac1{2^nn!},\dots\biggr) \cdot\exp\begin{pmatrix} 0 & \frac12 & 0 & 0 & 0 & \dots \cr 0 & 0 & \frac12 & 0 & 0 & \dots \cr 0 & 0 & 0 & \frac12 & 0 & \dots \cr 0 & 0 & 0 & 0 & \frac12 & \dots \cr \dots & \dots & \dots & \dots & \dots & \dots \end{pmatrix},$$ a diagonal matrix times the exponential of a nilpotent matrix. In your question you discuss some properties of truncations of your infinite matrix, finite $n\times n$ matrices. This corresponds to the truncations of the above diagonal and nilpotent matrices. I've never seen such matrices "in work" but this one could be a nice example of understanding the alteration property of entries of its eigenvectors. In view of the other response, this could be a good point of generalising the previous results in this area.

• Whether the matrix is of interest would be a misunderstanding of my question. I asked whether anything of interest could be said about the matrix. Isn't that a different question? – Michael Hardy May 30 '10 at 23:41
• Michael, I edited my post. I have no doubts that your expectations for the eigenvectors are provable, but the only method I see working is "brute force", definitely not the one suitable for MO but also requiring a lot of time. As for the hint, the eigen problems of the exponential of a matrix are really easy. – Wadim Zudilin Jun 1 '10 at 12:57
• OK, I think I see what you mean by saying you were giving a hint. To be continued....... – Michael Hardy Jun 2 '10 at 2:42

Your matrix is totally nonnegative (i.e., all minors are nonnegative). This is because it can be factorized as the matrix of binomial coefficients (which is totally nonnegative by the Karlin–McGregor–Lindström–Gessel–Viennot lemma) times a diagonal matrix with positive entries $1/(2k)!!$ on the diagonal.

For an oscillatory matrix (i.e., a totally nonnegative matrix such that some power of it is totally positive), there is a theorem by Gantmacher & Krein which says that the eigenvalues are real and simple, and the eigenvector corresponding to the $k$th largest eigenvalue has $k-1$ sign changes. (Theorem 5.3 in Pinkus, Totally positive matrices.)

Unfortunately that doesn't apply here, since a power of an upper triangular matrix is upper triangular, so that some minors (below the diagonal) are always zero; hence your matrix is not oscillatory. But perhaps it is possible to use similar ideas to prove the sign changes in your case?

Might be a wild intuition, I'd say the eigenvalues are the entries of the first row, and that the eigenvector coresponding to the $n$th eigenvalue, $k$ is made by adjoining a column of zeroes to the eigenvector coresponding to the eigenvector coresponding to the same eigenvalue for the first $n\times n$ minor of the matrix.

Example:

for eigenvaue $1$ we take the matrix $\begin{pmatrix} 1 \end{pmatrix}$ ,the eigenvetor corresponding to $1$ is $\begin{pmatrix} 1 \end{pmatrix}$ , so we obtain $\begin{pmatrix} 1 \cr 0 \cr 0 \cr 0 \cr 0 \cr \vdots \end{pmatrix}$ as the first eigevector.

for eigenvaue $1/2$ we take the matrix $\begin{pmatrix} 1& 1/2 \cr 0 & 1/2\end{pmatrix}$ ,the eigenvetor corresponding to $1/2$ is $\begin{pmatrix} 1 \cr -1\end{pmatrix}$ , so we obtain $\begin{pmatrix} 1 \cr -1 \cr 0 \cr 0 \cr 0 \cr \vdots \end{pmatrix}$ as the second eigevector .

The eigenvector coresponding to $1/8$ for $\begin{pmatrix} 1& 1/2 & 1/8 \cr 0 & 1/2 & 1/4 \cr 0 & 0 & 1/8 \end{pmatrix}$ is $\begin{pmatrix} 5 \cr -14 \cr 21 \cr \end{pmatrix}$, you get the ideea .Also , the eigenvectors span the entire space , ie if a possibly infinite (but convergent) sum of eigenvectors is $\vec 0$ then the coefficients of those vectors are $0$.

Here is an explicit formula for the eigenvectors: first select $M_n$, the $n\times n$ truncation of the matrix and calculate $M_n - I v_n$, the $n$th eigenvalue. Example: for $n=3$, we obtain \begin{pmatrix} 7/8 & 1/2 & 1/8 \cr 0 & 3/8 & 1/4 \cr 0 & 0 & 0 \end{pmatrix}. Now let $S$ be the $(n-1)\times(n-1)$ truncation of that , ie \begin{pmatrix} 7/8 & 1/2 \cr 0 & 3/8 & \end{pmatrix} Calculate $S^{-1} = \begin{pmatrix} 8/7 & -32/21 \cr 0 & 8/3 & \end{pmatrix}$, now multiply $S^{-1}$ with the truncation of the last column of $M_n$, \begin{pmatrix} 1/8 \cr 1/4 \end{pmatrix} You obtain $$\begin{pmatrix} -5/21 \\ 2/3 \end{pmatrix}.$$ Concatenating $-1$ to that, you obtain $$\begin{pmatrix} -5/21 \\ 2/3 \\ -1 \end{pmatrix},$$ the third eigenvector, or the $n$th eigenvector in the general case.