Suppose I have a matrix $M \in \mathbb{R}^{A \times B}$ with SVD $M = U S V^T$.
For a given vector $p \in \mathbb{R}^A$, can I compute the SVD of $p^T S V^T$ as a function of the original singular values $S$ and original right singular vectors $V$?
How do singular values of a matrix change when left multiplying by a row vector
linear algebramatricessingular valuessvd
Best Answer
Given the svd of the $A\times B$ matrix $M$, $M=USV^\top$, and the $1\times B$ matrix $L:=p^\top SV^\top$, then the svd of $L$ can be worked out from that of $M$.
The singular values of $L$ are the square roots of the eigenvalues of $LL^\top$ where $$\sigma^2v=LL^\top v=p^\top SV^\top VS^\top p=p^\top S^\top Sp=[r^2]$$ where $$r^2=p_1^2\sigma_1^2+\cdots+p_n^2\sigma_n^2$$ Thus there is only one right singular vector with singular value $r$.
The related eigenvector is that of $L^\top L$, so $$\sigma^2v=L^\top Lv=VS^\top pp^\top SV^\top pv=(p^\top SV^\top v)(VS^\top p)$$ so $v$ is the unit vector in the direction of $VS^\top p=p_1\sigma_1u_1+\cdots+p_n\sigma_nu_n$, where $u_n$ are the singular vectors of $M$.
Conclusion: The SVD of $L$ is $$\underbrace{[1]}_{1\times1}\underbrace{\begin{bmatrix}r,0,\ldots,0\end{bmatrix}}_{1\times B}\underbrace{\begin{bmatrix}v_1&v_2&\ldots&v_n\\&\cdots\end{bmatrix}}_{B\times B}$$ The first row of the new $V^\top$ matrix is the $v$ vector and the other rows are any set of orthonormal vectors (so it is not unique).