Let $A \in \mathbb{R}^{n \times n}$ and $w \in \mathbb{R}^n$. Suppose that, $w_i>0$ and $a_{i,j} = w_i / w_j$ for all $i,j=1,\dots,n$.
Note that from the construction comes that $\operatorname{rank} A=1$.
Prove that the eigenvectors of $A$ are a basis in $\mathbb{R}^n$, or equivalently prove that $A$ is diagonalizable.
On
Here is a start -- perhaps you can fill in the details:
We have $w = (w_1,\ldots,w_n)$ with every entry positive. Let $v = (\frac{1}{w_1},\ldots,\frac{1}{w_n})$, inverting each entry. Observe that, as a linear transformation, $A$ is given by $A(x) = (x \cdot v) w$, where "$\cdot$" is the Euclidean dot product.
Note that your matrix can be written as product of a column matrix and a row matrix: $$ A=\pmatrix{w_1\\w_2\\\vdots\\w_n}\pmatrix{1/w_1&1/w_2&\ldots&1/w_n}, $$ which explains why it has rank$~1$ (provided that $n>0$; the case $n=0$ is trivial and I will henceforth not consider it). Now as you can for instance read in this answer, such a matrix is diagonalisable if and only if it has nonzero trace (the part that the condition is sufficient is easy: due to the rank, eigenvalue $0$ already has geometric multiplicity $n-1$, and there must be a nonzero eigenvalue because of, and equal to, the trace). Here the diagonal entries are all$~1$, so the trace is $n\neq0$ and $A$ is diagonalisable.