The singular value decomposition of a complex-values $n\times m$ matrix $A$ is defined as
$$A = U \Sigma V^*$$
where $U$ and $V$ are unitary $n \times n$ matrices and $m \times m$ matrices respectively, and $Σ$ is an $n \times m$ rectangular diagonal matrix, whose diagonal elements are the singular values $\sigma _I$ of $A$. $V^*$ denotes the conjugate transpose of $V$. The singular value decomposition s a generalization of the eigen decomposition to arbitrary $n\times m$ matrices
I have solved 4 other proofs about SVD but I am having trouble solving the following and I am looking for some help!
$A$ can be written as the sum of rank-one matrices, i.e $$A = \sum _{i=1}^r \sigma_i u_i v_i^*$$ where $r$ is the rank of $A$ and $u_i$ and $v_i$ are the ith columns of $U$ and $V$ respectively.
Best Answer
If your matrix is $A \in \mathbb{C}^{n \times m}$ then it can be written as $A = U \Sigma V^{*}$
A rank one matrix can be given by the outer-product of two vectors
$$ A = uv^{*} \tag{1}$$
now if multiply by a scalar $\sigma $ it is still a rank -1 matrix
$$ A = \sigma \cdot u v^{*} \tag{2}$$
you can generate a rank - $2$ matrix by the sum of two rank one matrices ..
$$ A = uv^{*} + wz^{*} \tag{3} $$
$$ A = \sigma_{1} uv^{*} + \sigma_{2} wz^{*} \tag{4} $$
then a rank $r$ matrix can be generated by taking $r$ outer products
$$ A = \sum_{i=1}^{r} \sigma_{i} u_{i} v_{i}^{*} \tag{5}$$