Do singular values of a linear transformation depend on inner products? Let $(V,\langle -,- \rangle _1), (W,\langle -,- \rangle _2)$ be finite-dimensional inner product spaces over a field $\mathbb{F}$ ($\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$)
Let $T:V\rightarrow W$ be a linear transformation with rank $r$.
Then, the singular values $\sigma_1≧...≧\sigma_r>0$ are uniquely determined.
Now, suppose another inner product is given to each vector spaces. Namely $\langle -,- \rangle_3$ and $\langle-,- \rangle_4$ for $V,W$ respectively.
Since inner product does not affect rank, singular values of $T$ can be determined as $\mu_1≧...≧\mu_r>0$.
My questions is that, are these two sequences the same?
That is $\mu_i = \sigma_i$?
 A: I recommend looking at the mathematical preliminaries of this text, which go into a lot of detail about coordinate-free transformations with arbitrary inner products, and define singular values $\gamma_i$ for real transformations $A$ with respect to orthonormal bases $u_i, w_i$ by the values that, for all $x$, satisfy
$$
Ax = \sum_{i=1}^r \gamma_i \langle u_i, x \rangle w_i
$$
A: Here's one way to see that the singular values depend on the inner product. First, note that the largest singular value of $T$ is equal to the operator norm of $T$. (Since $T^* T$ is self-adjoint, its largest eigenvalue is given by $\max_{\|v\|=1} \langle T^* T v, v \rangle
= \max_{\|v\|=1} \|Tv\|^2 =: \|T\|_{op}^2$.) 
But we can easily see that the operator norm depends on one's choice of inner product; suppose that $e_1,e_2$ is a basis for $V$ and $Te_1 = e_2, Te_2 = e_1$. If we choose an inner product such that $e_1,e_2$ is orthonormal, then $\|T\|_{op}=1$ (in fact $T$ is unitary). But if we choose an inner product such that $\|e_1\|=1$ and $\|e_2\| = 2$, then evidently $\|T\|_{op} \ge 2$.
