Tensor rank of matrix $M = \sum_{i = 1}^k w_i (v_i \otimes v_i)$

Question

I encounter this situation while doing the whitening process of a tensor:

If $v_1, \ldots, v_k$ are linearly independent and $w_i \in \mathbb{R}^+$ then the matrix $M = \sum_{i = 1}^k w_i (v_i \otimes v_i)$ has rank $k$.

I can see rank of $M$ should be $\le k$ but not that it has to be exactly $k$. I could even use Linear independence of tensor product basis $\{ v_i \otimes w_j\}$ for $\{v_i\}$ and $\{w_j\}$ linearly independent. to show that the $v_i \otimes v_i$ forms a basis on the space of matrices. But I don't see the relation with the rank of the matrix.

Answer 1

If $\{ v_i\otimes v_i \}$ is a dependent set then there is a non trivial set of coefficient $\{a_i\}$ such that $\sum_i a_i v_i\otimes v_i=0$, in particular for any $j$ such that $a_j \neq 0$ if you right multiply this by $v_j$ you get \begin{align*} 0&=0\cdot v_j\\ &=\left(\sum_i a_i v_i\otimes v_i\right)\cdot v_j\\ &=\sum_i a_i (v_i\cdot v_j) v_i \end{align*} Now it is clear that $a_j(v_j\cdot v_j)\neq 0$ and therefore the coefficient $a_i(v_i\cdot v_j)$ are non trivial (not all equal to $0$), this means that $v_i$ are dependent.

Answer 2

Hint The rank of a $2$-tensor $T \in \Bbb V \otimes \Bbb V$ coincides with the rank of its matrix representation with respect to any basis of $\Bbb V$, and since $(v_1, \ldots, v_k)$ is linearly independent we can extend it to a basis $\mathcal B$. What is the matrix representation of $\sum w_i (v_i \otimes v_i)$ with respect to $\mathcal B$?