Using SVD, prove the following: Let $A \in \mathbb{R}^{m\times n}$ and $T(x)=Ax$, a linear transformation for which A is the representative matrix in the canonical coordinates. The existence of SVD implies that we can select a basis $\cal{B}$ of $\mathbb{R}^n$ and a basis $\cal C$ of $\mathbb{R}^m$ such that $[T]^\cal B_\cal C$ is diagonal, and the diagonal elements are the singular values.
I think I understand the intuition behind this, but not sure how to formulate it...
I assume $[T]^\cal B_\cal C$ means transformation from $\cal B$ to $\cal C$ ?
$Ax = U\Sigma V'x$. So we can choose $V'$ to be the basis of $\cal B$. And $U'$ as the basis of $\cal C$. Then $U$ is the transformation back to the canonical coordinates (as $UU' = I$). And then $[T]^\cal B_\cal C = \Sigma$ which is the diagonal matrix of singular values.