<s>[INST] 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?
Can anything of interest be said about this matrix?
 [/INST] 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.
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