# Prove compact SVD using normal SVD

Please read the EDIT first - it's the bit that will be helpful to anyone coming across this post.

Question is in the title (SVD means SVD decomposition). The answer from my textbook says that $$U$$ and $$V$$ both contain bases for the column and row space respectively (which is true - I can understand this myself with a little bit of effort). It then goes on to say that because $$C$$ and $$B$$ also contain the same respectively, then you can write $$U=CF$$ and $$V=BG$$ for two invertible $$r \times r$$ matrices $$F$$ and $$G$$ (where $$r=rank(A)$$). But here is where the issue arises for me: SVD decomposition splits an $$m \times n$$ matrix $$A$$ into the following matrices: an $$m \times m$$ $$U$$, an $$m \times n$$ $$\Sigma$$ and an $$n \times n$$ $$V^T$$. But both $$C$$ and $$B$$ have at max $$r$$ columns because they are limited by rank (they will be $$m \times r$$ and $$n \times r$$ respectively). This means that the products $$CF$$ and $$BG$$ will come out as being the same dimensions as $$C$$ and $$B$$ originally and respectively - not as square matrices. What am I missing here? (for this textbook proof) Or is there another method for proving the title question?

EDIT: After some searching $$A=CMB^T$$ is actually compact SVD ($$A=U_1\Sigma_1 V_1^T$$, with $$U \in \mathbb{R}^{m \times r}$$, $$\Sigma \in \mathbb{R}^{r \times r}$$ and $$V \in \mathbb{R}^{n \times r}$$) instead of "normal" SVD. A proof will be below for any weary travellers who stumble open this post.

Let $$A \in \mathbb{R}^{m \times n}$$, with $$m$$, $$n \mathbb{R}$$ Assume $$m < n$$. We will use the usual SVD decomposition $$A=U\Sigma V^T$$ with $$U \in \mathbb{R}^{m \times m}$$, $$\Sigma \in \mathbb{R}^{m \times n}$$ and $$V \in \mathbb{R}^{n \times n}$$. Let $$u_i$$ denote the $$i$$th column vector of $$U$$. $$A=U\Sigma V^T$$ $$\implies{A=\sum^m_{i=1}\sigma_iu_iv_i^T}$$ We know that $$\sigma_i=0$$ $$\forall i$$ s.t. $$r < i \leq m$$ (number of non-zero eigenvalues is the same as the rank for diagonalizable matrices - good proof here: Proving number of non zero eigen values.) $$\implies{A=\sum^r_{i=1}\sigma_iu_iv_i^T}$$ $$\therefore A=U_1\Sigma_1V_1^T \text{ with } U \in \mathbb{R}^{m \times r}\text{, } \Sigma \in \mathbb{R}^{r \times r}\text{, (with }E^{-1}\text{ existing) and }V \in \mathbb{R}^{n \times r}$$ To prove the compact SVD for $$m > n$$, it's trivial to prove it from the first proof given here - with some minor alterations to where the dimension variables are used.