# $AB^H = \sum a_i b_i^H$?

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)

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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?

Yes, the result is correct and your answer is correct. A more succinct method is to use block matrix multiplication. In particular, if we treat the columns of $$A$$ and rows of $$B$$ as if they were numerical entries, we have (sticking to your notation) $$AB = \pmatrix{a_1 & \cdots & a_n} \pmatrix{b^1\\ \vdots \\ b^n} = a_1b^1 + \cdots + a_nb^n.$$