- #1

kaienfr

- 3

- 0

$$\alpha^1 = (d,0,\ldots, 0), ~ \alpha^2 = (0,d,\ldots, 0), \ldots$$

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:

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