I often see this formula used in the rank 1 or rank 2 cases for Quasi-Newton methods, but I am wondering how this can be proven in the general rank $m$ case. As a linear algebra problem, I would like to show that for $v_1,v_2, \ldots,v_m \in \mathbb{R}^n$ linearly independent and $\lambda_1, \lambda_2, \ldots, \lambda_m \in \mathbb{R} \setminus \{0\}$ I have that $$ \sum_{i = 1}^{m}\lambda_i v_i v_i^T$$ has rank $m$
Any suggestions or proofs would be greatly appreciated.
A simple change of basis will do it. Let $M=\sum_{i=1}^m \lambda_m v_iv_i^T$ and let $B=\begin{bmatrix}v_1&v_2&\cdots&v_m&v_{m+1}&\cdots&v_n\end{bmatrix}$ be the matrix for a basis $\{v_i\}$ that extends the independent set $\{v_1,v_2,\dots,v_m\}$. This $B$ is invertible.
We claim that $M=BXB^T$, where $X$ is the diagonal matrix $$X=\begin{bmatrix}\lambda_1&0&\cdots&0&0&\cdots\\ 0&\lambda_2&\cdots&0&0&\cdots\\ \vdots&\vdots&\ddots&\vdots&\vdots&\ddots\\ 0&0&\cdots&\lambda_m&0&\cdots\\ 0&0&\cdots&0&0&\cdots\\ \vdots&\vdots&\ddots&\vdots&\vdots&\ddots\end{bmatrix}$$ This $X$ clearly has rank $m$. After multiplying on the left and right by invertible matrices $B$ and $B^T$, the rank will be unchanged, and $M$ will have rank $m$.
Why does this work? Just run the matrix multiplication: $$BX = \begin{bmatrix}v_1\lambda_1&v_2\lambda_2&\cdots&v_m\lambda_m&0&\cdots&0\end{bmatrix}$$ $$BXB^T = v_1\lambda_1v_1^T + v_2\lambda_2v_2^T+\cdots+v_m\lambda_mv_m^T+0+\cdots+0=\sum_{i=1}^m \lambda_iv_iv_i^T$$ There it is. Note that we didn't use anything about $\mathbb{R}$; this is true over any field.
A useful related concept: the tensor rank of a matrix. The rank of a matrix $A$ is equal to the smallest $k$ such that we can write $A=\sum_{i=1}^k u_iv_i^T$ as a sum of rank-1 tensors. This argument (well, a slight modification of it) shows that if each set $\{u_i\}$ and $\{v_i\}$ are linearly independent, then the tensor rank of the sum is the full $k$.