I have problem with the proof of the following statement.
If matrix $M\in M_{m \times n}(\mathbb{R})$ has rank $r \geq 1$, then it can be written as a sum of exactly $r$ rank-$1$ matrices
I try in this case:
The SVD of $M$ is given by $M = U\sigma V^T$, where $U$ is an $m \times r$ matrix whose columns are the left singular vectors of $M$, $V$ is an $n \times r$ matrix whose columns are the right singular vectors of $M$, and $\sigma$ is an $r \times r$ diagonal matrix whose diagonal entries are the singular values of $M$.
Using the SVD of $M$, we can express $M$ as a sum of rank 1 matrices as follows: \begin{align*} M &= U\sigma V^T \\ &= \sum_{i=1}^r \sigma_i u_i v_i^T \\ &= \sigma_1 u_1 v_1^T + \sigma_2 u_2 v_2^T + \cdots + \sigma_r u_r v_r^T, \end{align*} where $\sigma_i$ is the $i$th singular value of $M$, $u_i$ is the $i$th left singular vector of $M$, and $v_i$ is the $i$th right singular vector of $M$.
Each of the matrices $\sigma_i u_i v_i^T$ is a rank 1 matrix, since it is the outer product of two vectors of rank 1. Moreover, these matrices are linearly independent, since the left and right singular vectors are orthogonal to each other and have norm 1, and the singular values are positive and distinct. Therefore, we have expressed $M$ as a sum of $r$ rank 1 matrices, and we have shown that this is the minimum number of rank 1 matrices required to express $M$ in this way.
But i don't know that it is correct. Could you check it? And help
The matrix $M$ defines a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m.$ By assumption $r=\dim({\rm Im}M).$ Let $\{v_1,v_2,\ldots v_r\}$ denote a linear basis in ${\rm Im}M.$ Then for any vector $x\in \mathbb{R}^n$ there exist unique coefficients $\alpha_1(x),\alpha_2(x),\ldots, \alpha_r(x)$ such that $$Mx=\alpha_1(x)v_1+\alpha_2(x)v_2+\ldots +\alpha_r(x)v_r\quad (*)$$ The mappings $x\mapsto \alpha_k(x)$ are linear due to the uniqueness in $(*).$ The linear transformations $M_kx= \alpha_k(x)v_k$ have rank $1$ and $$M=M_1+M_2+\ldots +M_k$$ The formula gives a decomposition of the matrix $M$ into $r$ rank one matrices, associated with $M_k.$ Clearly if $M$ decomposed into $s$ rank one matrices then $\dim({\rm Im}M)\le s,$ hence $r$ is the smallest number of possible rank one matrices in the decomposition.