Eigenvectors of harmonic series matrix I have a finite upper triangular $n$ by $n$ matrix $A$ which has the following form:
$$ A = \begin{pmatrix}
 1 & 1\over 2 & 1\over 3 & \cdots & 1\over n \\
 0 & 1\over 2 & 1\over 3 & \cdots & 1\over n \\
 0 & 0 & 1\over 3 & \cdots & 1\over n \\
\vdots &\vdots &\vdots &\ddots &\vdots \\
0 & 0 & 0 & \cdots& 1\over n
 \end{pmatrix}$$
Or alternatively $A =TH$, where $T$ is the all-ones upper $n$ by $n$ triangular matrix and $H$ is the first $n$ terms of the harmonic series on the diagonal.
I'm interested in diagonalizing this matrix, as I'm interested in understanding and computing $A^k$ (specifically the final column). Since all elements on the diagonal are distinct the eigenvalues are simply the first $n$ terms of the harmonic series.
Doing some numerical analysis using numpy gave fairly opaque numbers, however Mathematica produced very interesting eigenvectors. E.g. for $n = 5$ the matrix with the eigenvectors as columns is:
$$P =
\begin{pmatrix}
 1 & -1 &  1 & -1 &  1 \\
 0 &  1 & -2 &  3 & -4 \\
 0 &  0 &  1 & -3 &  6 \\
 0 &  0 &  0 &  1 & -4 \\
 0 &  0 &  0 &  0 &  1 \\
\end{pmatrix},
P^{-1} = \begin{pmatrix}
1 & 1 & 1 & 1 & 1 \\
0 & 1 & 2 & 3 & 4 \\
0 & 0 & 1 & 3 & 6 \\
0 & 0 & 0 & 1 & 4 \\
0 & 0 & 0 & 0 & 1 \\
\end{pmatrix}
$$
This seems to be some alternating sign Pascal triangle. In fact, $P^{-1}$ seems the be the actual Pascal triangle! And indeed $PHP^{-1} = A$ (no PHP pun intended), for the cases I've tried.
I have two questions:


*

*Is there a good explanation for this seeming connection between the eigenvectors of the truncated harmonic series matrix and Pascal's triangle, assuming my numerical exploration is not a coincidence?

*Can we use this to say anything interesting about (the final column of) $A^k$? A closed form that does not require constructing the Pascal triangles perhaps?
 A: Since the given matrix is upper triangular, the eigenvalues are trivially $\frac{1}{1},\frac{1}{2},\frac{1}{3},\ldots$ and the shown decomposition is just an instance of the following identity:
$$\frac{1}{x(x+1)(x+2)\cdots(x+n)} = \frac{1}{n!}\sum_{k=0}^{n}\frac{(-1)^k}{x+k}\binom{n}{k}$$
that follows from the residue theorem.
A: Let's indicate your matrix as 
$$
{\bf A}_{\,h}  = \left\| {\;a_{\,n,\,m} \quad \left| {\;0 \le n \le m \le h} \right.\;} \right\| = \left\| {\;{{\left[ {n \le m} \right]} \over {m + 1}}\;} \right\|_{\,h} 
$$
so that it is a $(h+1) \times (h+1)$ matrix, indexed from $0$, and where the square brackets indicate the Iverson's bracket.
Then let's define
$$
\eqalign{
  & {\bf E}_{\,h}  = \left\| {\;\left[ {n + 1 = m} \right]\;} \right\|_{\,h}   \cr 
  & {\bf S}_{\,h}  = \left\| {\;\left[ {n \le m} \right]\;} \right\|_{\,h}  = {{{\bf I}_{\,h} } \over {{\bf I}_{\,h}  - {\bf E}_{\,h} }}\quad   \cr 
  & \left( {f(n) \circ {\bf I}_{\,h} } \right) = \left\| {\;f(n)\left[ {n = m} \right]\;} \right\|_{\,h}  \cr} 
$$
so that
${\bf E}_{\,h}$ is the Shifting matrix, with $1$ on the first upper sub-diagonal and null otherwise;
${\bf S}_{\,h}$ is the Summing matrix, upper triangular with all $1$ 's;
$\left( {f(n) \circ {\bf I}_{\,h} } \right)$ is the diagonal matrix, with diagonal elements $f(0),f(1), \cdots, f(h)$.
Then, as you correctly pointed out, we can write:
$$
{\bf A}_{\,h}  = {\bf S}_{\,h} \left( {{1 \over {n + 1}} \circ {\bf I}_{\,h} } \right)
$$
which means
$$
{\bf A}_{\,h} ^{\; - \,{\bf 1}}  = \left( {\left( {n + 1} \right) \circ {\bf I}_{\,h} } \right)\left( {{\bf I}_{\,h}  - {\bf E}_{\,h} } \right)
$$
It is not difficult to demonstrate that
$$
\eqalign{
  & \left( {f(n) \circ {\bf I}_{\,h} } \right)\;{\bf E}_{\,h}  = {\bf E}_{\,h} \;\left( {f(n - 1) \circ {\bf I}_{\,h} } \right) =   \cr 
  &  = \left( {\prod\limits_{0\, \le \,k\, \le \,n - 1} {f(k)}  \circ {\bf I}_{\,h} } \right)^{\; - \,{\bf 1}} \;{\bf E}_{\,h} \left( {\prod\limits_{0\, \le \,k\, \le \,n - 1} {f(k)}  \circ {\bf I}_{\,h} } \right) \cr} 
$$
and that, given in general
$$
{\bf A}_{\,h} (f) = {\bf S}_{\,h} \left( {f(n) \circ {\bf I}_{\,h} } \right)\quad \left| {\;f(m) \ne f(n)} \right.
$$
the relevant matrix of eigenvectors $\mathbf W$ is :
$$
{\bf W}_{\,h} (f) = \left\| {\;\left[ {n \le m} \right]{{f(m)^{\,m - n} } \over {\prod\limits_{n\, \le \,k\, \le \,m - 1} {\left( {f(m) - f(k)} \right)} }}\;} \right\|_{\,h}  = \left\| {\;\left[ {n \le m} \right]\prod\limits_{n\, \le \,k\, \le \,m - 1} {\left( {{{f(m)} \over {f(m) - f(k)}}} \right)} \;} \right\|_{\,h} 
$$
while the inverse is:
$$
\eqalign{
  & {\bf W}_{\,h} ^{\; - \,{\bf 1}} (f)\quad \left| {\;0 \ne f(0)} \right.\quad  =   \cr 
  &  = \left\| {\;\left[ {n \le m} \right]{{f(m)f(n)^{\,m - n - 1} } \over {\prod\limits_{n + 1\, \le \,k\, \le \,m} {\left( {f(n) - f(k)} \right)} }}\;} \right\|_{\,h}  = \left\| {\;\left[ {n \le m} \right]{{f(m)} \over {f(n)}}\prod\limits_{n + 1\, \le \,k\, \le \,m} {{{f(n)} \over {\left( {f(n) - f(k)} \right)}}} \;} \right\|_{\,h}  \cr} 
$$
Now, in the present case, we obtain
$$
\eqalign{
  & {\bf W}_{\,h}  = \left\| {\;\left[ {n \le m} \right]\prod\limits_{n\, \le \,k\, \le \,m - 1} {\left( {{{k + 1} \over {k - m}}} \right)} \;} \right\|_{\,h}  =   \cr 
  &  = \left\| {\;\left[ {n \le m} \right]{{m!} \over {n!}}\prod\limits_{1\, \le \,j\, \le \,m - n} {\left( { - {1 \over j}} \right)} \;} \right\|_{\,h}  = \left\| {\;\left( { - 1} \right)^{\,m - n} \left( \matrix{
  m \cr 
  n \cr}  \right)\;} \right\|_{\,h}  = \overline {\bf B} _{\,h} ^{\; - \,{\bf 1}}  \cr} 
$$
where $ \overline {\bf B} _{\,h} $ is the transpose of the Lower Triangular Pascal array.
Finally
$$
{\bf A}_{\,h}  = {\bf S}_{\,h} \left( {{1 \over {n + 1}} \circ {\bf I}_{\,h} } \right) = \overline {\bf B} _{\,h} ^{\; - \,{\bf 1}} \;\left( {{1 \over {n + 1}} \circ {\bf I}_{\,h} } \right)\;\overline {\bf B} _{\,h} 
$$
which demonstrates your assumption.  
Concerning your last question instead, we have of course that
$$
{\bf A}_{\,h} ^{\;{\bf q}}  = \overline {\bf B} _{\,h} ^{\; - \,{\bf 1}} \;\left( {{1 \over {\left( {n + 1} \right)^{\;q} }} \circ {\bf I}_{\,h} } \right)\;\overline {\bf B} _{\,h} 
$$
whose single components are:
$$
\eqalign{
  & a^{\left( q \right)} _{\,n,\,m}  = \sum\limits_{0\, \le \,\left( {n\, \le } \right)\,k\,\left( { \le \,m} \right)\, \le \,h} {\left( { - 1} \right)^{\,k - n} \left( \matrix{
  k \cr 
  n \cr}  \right){1 \over {\left( {k + 1} \right)^{\;q} }}\left( \matrix{
  m \cr 
  k \cr}  \right)}  =   \cr 
  &  = \sum\limits_{0\, \le \,\left( {n\, \le } \right)\,k\,\left( { \le \,m} \right)\, \le \,h} {\left( { - 1} \right)^{\,k - n} \left( \matrix{
  m \cr 
  n \cr}  \right)\left( \matrix{
  m - n \cr 
  k - n \cr}  \right)\left( {k + 1} \right)^{\; - \,q} }  =   \cr 
  &  = \left( \matrix{
  m \cr 
  n \cr}  \right)\sum\limits_{0\, \le \,\left( {n\, \le } \right)\,k\,\left( { \le \,m} \right)\, \le \,h} {\left( { - 1} \right)^{\,k - n} \left( \matrix{
  m - n \cr 
  k - n \cr}  \right)\left( {k - n + n + 1} \right)^{\; - \,q} }  =   \cr 
  &  = {{m!} \over {n!}}\sum\limits_{\,0\, \le \,j\, \le \,m - n} {{{\left( { - 1} \right)^{\,j} } \over {j!\left( {m - n - j} \right)!}}\left( {j + n + 1} \right)^{\; - \,q} }  \cr} 
$$
Cannot find at the moment a "more compact" expression.
