Recovering the singular values of SVD from orthogonal (not orthonormal) basis Let $A:\mathbb C^n \to \mathbb C^m$ be a complex linear map. We have the singular value decomposition $A=X\Sigma Y^*$ where columns of $X$, denoted $\{x_1,x_2,\dots, x_m\}$ and $Y$ denoted $\{y_1, y_2,\dots, y_n \}$ are orthonormal basis in $\mathbb C^m$ and $\mathbb C^n$, resp. $X$ and $Y$ are both unitary. The matrix $\Sigma$ is diagonal with positive entries $\sigma_1,\dots, \sigma_r$.
By multiplying both sides of SVD by $y_i$. We get $Ay_i=\sigma_i x_i$.
Now assume we have a new set of orthogonal basis:
$$\{y'_1,\dots, y'_n \} \subset \mathbb C^n, \{x'_1,\dots, x'_m \}\subset \mathbb C^m$$
satisfying $Ay'_i=x'_i$ for $i=1,\dots,r$ and $Ay'_i=0$ for $i\ge r+1$. I read from a paper that we can recover $\sigma_i$ as 
$$\sigma_i=\|A y'_i\|/\|y'_i\|, i=1.\dots,r.$$
How is the last equation derived? I only know the special cases like when $y'_i$ is a multiple of $y_i$ and $x'_i$ is a multiple of $x_i$.
 A: Suppose there are orthogonal basis $ \left[\tilde{y_{i}'}\right] _{1 \leq i \leq n}$ and $ \left[\tilde{x_{i}'}\right] _{1 \leq i \leq m}$ of $\mathbb{C}^{n}$ and $\mathbb{C}^{m}$, respectively, such that $A \tilde{y_{i}'} = \tilde{x_{i}'}, \: 1 \leq i \leq r$ and $A \tilde{y_{i}'} = 0, i > r$. Let's define $\tilde{X'} := \left[\tilde{x_{1}'},...,\tilde{x_{m}'}\right] \in \mathbb{C}^{m \times m}$ and  $\tilde{Y'} := \left[\tilde{y_{1}'},...,\tilde{y_{m}'}\right] \in \mathbb{C}^{n \times n}$. Now, normalize every column of $\tilde{X'}$, $\tilde{Y'}$: $\rightsquigarrow$ $X'$, $Y'$. The new matrices are now unitary (since e.g. $\bar{X'}^{T} X' = 1_{m \times m}$).
$\Rightarrow A Y' = \left[\tilde{x_{1}'}/ ||\tilde{y_{1}'}||,..., \tilde{x_{r}'}/ ||\tilde{y_{r}'}||,0,...,0  \right]$.
Let's multiply this by the left with $\bar{X'}^{T}$:
$\Rightarrow \bar{X'}^{T} A Y' = \left[ \begin{matrix} c_{1} & 0 & 0 & 0 &  ... \\0 & c_{2} & 0 & 0 & ...\\ 0 & 0 & ... & 0 & ... \\ 0& 0&0& c_{r} & ... \\ 0&0&0&0&0 \end{matrix} \right] = : \Sigma$. 
Note that $c_{i} = \dfrac{|| \tilde{x_{i}'} ||} {||  \tilde{y_{i}'} ||} = \dfrac{|| A \tilde{y_{i}'} ||} { ||  \tilde{y_{i}'} || }, 1 \leq r \leq r $.
If we then multiply $\Sigma$ with $\bar{Y'}^{T}$ by the right and with $X'$ by the left we get: $A = X' \Sigma \bar{Y'}^{T}$, which corresponds to the singular value decomposition of $A$. This means that the singular values $\sigma_{i}$ correspond to the the constants $c_{i}$ found above.
