Let $A \in \mathbb C^{m \times n}$. There exists a singular value decomposition such that $A = U\Sigma V^H$, where $U = (u_1, \cdots, u_m)$ and $V = (v_1, \cdots, v_n)$. Suppose that $m \geq n$. According to here: https://www.cs.utexas.edu/users/flame/laff/alaff/chapter02-the-reduced-svd.html
$A = U\Sigma V^H = \sum_{i=1}^n \sigma^i u_i v^H_i$.
Algebraically, this makes sense since $A$ is just a matrix that maps each $v_i \mapsto \sigma^i u_i$, and every $x \in \mathbb C^m$ can be represented as $x = \sum \langle x, v_i\rangle v_i$. Hence, $Ax = A(\sum \langle x, v_i\rangle v_i) = \sum \sigma^i \langle x, v_i\rangle u_i$. Note that in $(\sum_{i=1}^n \sigma^i u_i v^H_i)$, the expression $v^H_i$ is just equivalent to $\langle \cdot, v^H_i\rangle$.
However, I just cannot figure out in matrix-multiplication sense that $U \Sigma V^H = \sum_{i=1}^n \sigma^i u_i v^H_i$. If $A,B$ are matrices in $\mathbb R^{k \times k}$ in general, is it generally true that $AB = \sum a_i b^i$, where $a_i$ is $i$-th col of $A$ and $b^i$ is $i$-th row of $B$? (sorry for bad notation, I don't mean $a_i b^i$ to be a tensor product with Einstein notation. It is merely a $\mathbb R^{n \times n}$ matrix)
===================edit====================
After some work, I think $AB = \sum a_i b^i$ is true. Let
\begin{equation*} \hat a_1:= \begin{pmatrix} \mid & \mid & \cdots & \mid\\ a_1 & 0 & \cdots & 0 \\ \mid & \mid & \cdots & \mid\end{pmatrix}, \end{equation*} and define similarly for $\hat a_i$ for other $i$'s.
I may define \begin{equation*} \hat b^1 = \begin{pmatrix} - & b^1 & - \\ - & 0 & - \\ &\vdots& \\ - & 0 & - \end{pmatrix} \end{equation*} and define similarly for all other $\hat b^j$'s.
Then, $AB = (\sum \hat a_i)(\sum \hat b^j)$. Furthermore,
\begin{equation*} \hat a_i \hat b^j = \delta_i^j a_ib^j \end{equation*}
so that $AB = \sum_{i,j} \hat a_i \hat b^j = \sum \delta_i^j a_ib^j = \sum_{i=1} a_i b^j$.
Is this correct algebra?
