I have the following matrix : $$ A= \begin{bmatrix} \sum_i w_i^0 & \sum_i w_i & \sum_i w_i^2 \\ \sum_i w_i & \sum_i w_i^2 & \sum_i w_i^3 \\ \sum_i w_i^2 & \sum_i w_i^3 & \sum_i w_i^4 \\ \end{bmatrix} $$ where the sum is finite, say it is $i=1,2,...,5$ and $w_i>0$. I want to prove that every linear system $Ax=b$ has a unique solution.
I have tried to prove that : 1) $ x^t Ax>0$, 2) $det(A)>0$ , 3) The pivots appeared to the Gaussian Elimination are positive. I failed to all of these ways because the expressions are not easy to manipulate. I don't remember why I tried to prove that the matrix A is Positive Definite maybe there is a simpler way to prove that the matrix is nonsingular.
So I appreciate if any of you would give me some advice. Thanks in advance.
I'm going to assume that the $(1,1)$ term can indeed be written as $\sum_i w_i^0$, not fixed at 5. If that's the case, then your matrix becomes
$$\begin{aligned} A &= \sum_i \begin{bmatrix} 1 \\ w_i \\ ... \\ w_i^{n-1} \end{bmatrix} \begin{bmatrix} 1 & w_i & ... & w_i^{n-1} \end{bmatrix}\\& = \begin{bmatrix} 1 & 1 & \dots & 1 \\ w_0 & w_1 & \dots & w_m \\ \vdots & \ \vdots & \ddots & \vdots \\ w_0^{n-1} & w_1^{n-1} & \dots & w_m^{n-1} \end{bmatrix} \begin{bmatrix} 1 & 1 & \dots & 1 \\ w_0 & w_1 & \dots & w_m \\ \vdots & \ \vdots & \ddots & \vdots \\ w_0^{n-1} & w_1^{n-1} & \dots & w_m^{n-1} \end{bmatrix}^T\triangleq B^TB\end{aligned}$$ So $A$ is definitely positively semidefinite, since it has a square root. The rank of $A$ is equal to the rank of $B$, which is a Vandermonde matrix. If $m\leq n$, then the rank is $m$ if and only if if the $w_i$ values are distinct. So it is not positive definite for all $w_i>0$, which means that the convex function involved is not strictly convex, just convex.