# eigenvectors of a lower-diagonal matrix with a special form

I'm curious about the matrix mentioned in a now-closed Stackoverflow question. The matrix has the form:

1, 0, 0, 0, ... 0
2, 2, 0, 0, ... 0
3, 3, 3, 0, ... 0
...
n - 1, n - 1, n - 1, ... 0
n, n, n, ... n


for any number of rows n.

Does this matrix have a name? I wasn't able to find one.

The eigenvalues of this matrix are 1, 2, 3, ..., n. Is there a simple formula for the eigenvectors? Some experiments seem to show that the first one has elements (-1)^(k + 1)*k/(k - 1)! for k from 1 to n. But I wasn't able to puzzle out a formula for the others.

It would be easy enough to compute the eigenvectors for any specific value of n, but I am guessing there's a relatively simple general formula, which I've been too lazy to work out for myself. Thanks for any light you can shed on this problem.

Let $A$ be your matrix You want to find a vector in the nullspace of $A - kI$, for each $k\in \{1,\ldots,n\}$.
If you do row operations to $A - k I$, you can make it look like $$\left[\begin{array}{ccc|ccccc} 1 \\ &\ddots \\ && 1 \\ \hline &&& 0 \\ &&& k+1 & 1 \\ &&& k+2 & k+2 & 2 \\ &&& \vdots & \vdots & \vdots & \ddots \\ &&& n & n & n & \cdots & n-k \end{array}\right]$$ where the upper left block has size $(k-1) \times (k-1)$.
Thus, a particular $k$-eigenvector is $$\left(\underbrace{0,\ldots,0}_{k-1},-1,k+1, -k \frac{k+2}{2!}, \frac{k^2(k+3)}{3!} , -\frac{k^3(k+4)}{4!} , \ldots, \frac{(-k)^{n-k-1} n}{(n-k)!}\right),$$ or more compactly, the eigenvector is $(v_1,\ldots, v_n)$ where $$v_j := \begin{cases}0 & j < k \\ -1 & j=k \\ \frac{(-k)^{j-k-1}j}{(j-k)!}& j > k.\end{cases}$$
In the case $k=1$, we have $$\left(-1, 2, -\frac{3}{2!}, \frac{4}{3!}, -\frac{5}{4!}, \ldots, (-1)^{n} \frac{n}{(n-1)!}\right)$$ which is what you have written.