Proving or disproving this matrix V is invertible.

[kaienfr]

Let us take $n\in \mathbb{N}^*$ bins and $d\in \mathbb{N}^*$ balls. Denote the set $B = \{\alpha^1, \ldots, \alpha^m\}$ to be all possible choices for putting $d$ balls into $n$ bins, such as
$$\alpha^1 = (d,0,\ldots, 0), ~ \alpha^2 = (0,d,\ldots, 0), \ldots$$
Let us define the matrix $V$ as:
$$V = \begin{bmatrix}
(\alpha^1)^{\alpha^1} & \cdots & (\alpha^1)^{\alpha^m}\\
(\alpha^2)^{\alpha^1} & \cdots & (\alpha^2)^{\alpha^m}\\
\vdots & \vdots & \vdots\\
(\alpha^m)^{\alpha^1} & \cdots & (\alpha^m)^{\alpha^m}
\end{bmatrix}$$
where the notation $(\alpha^i)^{\alpha^j} = \displaystyle\prod_{k=1}^{n}(\alpha^i_k)^{\alpha^j_k}$, under assumption that $0^0=1$.

Now, I'd like to know: "is the matrix $V$ invertible?"

I have tested several examples, and it seems that $V$ is always invertible, but I have no idea how to prove it or find a counterexample.
Here are two facts I understood:

1. all diagonal elements are strictly positives, so the trace of $V$ is strictly positive.
2. the matrix $V$ is not symmetric.

Does anyone can help to prove or disprove the invertibility of $V$? Thanks a lot in advance for sharing any idea.

 

[jvicens]

I would like to offer some insights on this problem.

Firstly, let's consider the dimensions of the matrix $V$. Since there are $m$ possible choices for putting $d$ balls into $n$ bins, the matrix $V$ will be an $m\times m$ matrix.

Next, let's take a closer look at the diagonal elements of $V$. The diagonal elements are of the form $(\alpha^i)^{\alpha^i}$, which is equal to $\displaystyle\prod_{k=1}^{n}(\alpha^i_k)^{\alpha^i_k}$. Since $0^0=1$, we can rewrite this as $\displaystyle\prod_{k=1}^{n}(\alpha^i_k)^{\alpha^i_k} = \displaystyle\prod_{k=1}^{n}1^{\alpha^i_k} = 1$. Therefore, all diagonal elements of $V$ are equal to $1$.

Now, let's consider the off-diagonal elements of $V$. These elements are of the form $(\alpha^i)^{\alpha^j}$, where $i\neq j$. Recall that $(\alpha^i)^{\alpha^j} = \displaystyle\prod_{k=1}^{n}(\alpha^i_k)^{\alpha^j_k}$. Since $i\neq j$, there must exist at least one $k$ such that $\alpha^i_k \neq \alpha^j_k$. This means that at least one term in the product will be greater than $1$, since $\alpha^i_k > \alpha^j_k$ or $\alpha^i_k < \alpha^j_k$. Therefore, all off-diagonal elements of $V$ are strictly greater than $1$.

Based on these observations, we can conclude that all diagonal elements of $V$ are equal to $1$, and all off-diagonal elements are strictly greater than $1$. This means that the determinant of $V$ will be strictly greater than $1$. In general, a matrix with all diagonal elements equal to $1$ and all off-diagonal elements strictly greater than $1$ will have a non-zero determinant. Therefore, we can conclude that $V$ is invertible.

To further support this conclusion, let's consider the inverse of $V$. The