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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.

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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.

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