Left eigenvectors = right eigenvectors with alternating sign

Question

I've come across a matrix with an interesting property. Let $M\in \mathbb{Z}^n$ be the lower triangular matrix given by $$ M_{ij}=\begin{cases} \dfrac{n!(i-1)!}{j!}, &i\ge j\\ 0, &\text{otherwise}. \end{cases}$$ $M$ has an eigendecomposition $PDP^{-1}$ where $D=\operatorname{diag}\left(\frac{n!}{1},\frac{n!}{2},\ldots,\frac{n!}{n}\right)$ and $P$ is the lower triangular matrix given by $$ P_{ij}=\begin{cases} \dfrac{i!}{j!}\large\binom{i-1}{j-1}, &i\ge j\\ 0, &\text{otherwise}. \end{cases} $$ A surprising fact (to me) is that $(P^{-1})_{ij}= (-1)^{i+j} P_{ij}$. I've never seen this happen before. Does this property have a name? Is there some deeper explanation for it?

Answer 1

That's not surprising: the Pascal matrix $B_{n,m}= \binom {n}{m}$ (indexed from $0$) has the same property.
More generally, ${B^{\,r}}_{n,m}=r^{n-m}\, B_{n,m} \; |\, r \in \mathbb C$

In your case $P = D B D^{-1}$, being $D$ the diagonal matrix with entries $n!$.

Answer 2

To complete the (very nice) answer by G Cab:

The $B$ matrix is the matrix of the linear map $P(X)\mapsto P(X+1)$ in the basis $\{1,X,X^2,X^3, \}$ : this is the binomial theorem.

The $B^{-1}$ matrix is then matrix of the linear map $P(X)\mapsto P(X-1)$ in the same basis.

I don't know if there is something deeper to it.