Change in singular values of matrix after left-multiply with a diagonal matrix Say that we have an SVD for a matrix $X = U \Sigma V^T$, giving trace norm $||X||_{tr} = ||\Sigma||_{tr} = \sum \Sigma_{ii}$. I am wondering what happens to the SVD and/or trace norm if we left multiply with a diagonal matrix that simply scales the row of $X$? Is it possible to write the SVD of $DX$ or the trace norm of $DX$ in terms of the diagonal of $D$ and the SVD of $X$?
 A: In a trivial way, you can write SVD of $DX$ in terms of $U$, $\Sigma$, $V$ by multiplying $DX=DU\Sigma V^T$  and taking the SVD of that. But of course this is not the question. I understand the question is whether there is a  transparent relation between the SVDs of $X$ and $DX$. To which the answer is: no. 
Consider the two-dimensional case. The image of the unit disk $\mathbb D$ under $X$ is an ellipse $E$. The SVD of $X$ tells us how long the semi-axes of $E$ are and how they are oriented: for example, the semi-axes are $3$ and $2$, with the major semi-axes making angle $\pi/3$ with the $x$-axis. Fine. 
Now $D=\begin{pmatrix}2&0\\0&1\end{pmatrix}$  shows up and stretches $E$ by the factor of $2$ in the horizontal direction. What are the new semiaxes? how are they oriented? There is no easy way to answer these questions by looking at the geometry; it's not even geometrically obvious that the stretched thing is an ellipse. What we do is return to the linear transformation $DX$ and compute from there.
A: Actually there is an answer, but somewhat complex. Denote the SVD decomposition of $X$ by $USV'$, denote $qr(DU) = QR$, and denote SVD decomposition of $RS$ by $U_1S_1V_1'$ then the SVD decomposition of $DX$ is $(QU_1)S_1(VV_1)'$. 
Proof: $DX= DUSV'=QRSV'=(QU_1)S_1(V_1'V') = (QU_1)S_1(VV_1)'$. $QU_1$ is unitary and $VV_1$ is unitary and $S_1$ is diagonal. 
Interpretation: the singular values of $DX$ are the singular values of $RS$. So the original singular values are multiplied by the upper triangular $R$ of the $qr$ decomposition of $DU$, i.e., $D$ after unitary transformation by right bases $U$ of $X$. 
I have no idea how to interpret this result geometrically or physically, but perhaps someone can.  
