SVD of a data matrix after an orthogonal projection to a subspace Let's say I can know the SVD of some matrix $X$: $$X = USV^T$$
If I have an orthogonal matrix $A$ (i.e., $A$ is square and has orthonormal columns), then the SVD of $XA$ is
$$XA = USW^T$$ where $W = A^TV$.
But can anything be said about the SVD of $XB$ if $B$ has orthonormal columns but is not necessarily square? In other words, if the SVD of $XB$ is $XB = DEF^T$, can the matrices $D$, $E$, or $F$ be written in terms of the SVD of $X$ and $B$?

Update: @whuber suggests that I can extend $B$ to be orthogonal by adding in orthonormal columns until $B$ is square. Call this orthogonal matrix $\tilde B$.
$$ \tilde B = [B; B_{\perp}]$$
I know the SVD of $X\tilde B$ is $US(\tilde B^TV)^T$ (see above). But now I'm struggling to see if there's a way that I can write the SVD of $XB$ in terms of the SVD of $X\tilde B$.
 A: In the SVD $X = USV^\prime$, where $X$ is an $n\times p$ matrix, $V$ is an orthogonal $p\times p$ matrix.
Suppose $B$ is an orthogonal $p\times q$ matrix: that is, $B^\prime B = 1_q$.  Let
$$S V^\prime B = TDW^\prime\tag{1}$$
be an SVD of $S V^\prime B$.  Thus, by definition, $T$ is a $p\times q$ matrix, $D$ is a diagonal matrix of dimension $q$, and $W$ is an orthogonal $q\times q$ matrix.
Compute
$$XB = (USV^\prime) B = U(SV^\prime B) = U(TDW^\prime) = (UT)D(W^\prime).\tag{2}$$
Because $(UT)^\prime (UT) = T^\prime (U^\prime U) T = T^\prime T = 1_q$, $UT$ has orthonormal columns.  Because $D$ and $W^\prime$ are part of an SVD, then by definition $D$ is diagonal with non-negative entries and $W$ is a $q\times q$ orthogonal matrix.  Consequently, equation $(2)$ gives an SVD of $XB$.  Equation $(1)$ shows how this SVD is related to that of $X$ and $B$.
A: For a matrix $B$ with orthonormal columns (but not square), I would like a way of finding an SVD of $XB$ in terms of the SVD of $X = USV^T$.
As suggested by @whuber, a first step towards finding the SVD of $XB$ is to add columns to $B$ to make it square (and thus orthogonal). Call this matrix $\tilde B = [B; B_{\perp}]$, and let $k$ be the number of columns of $B_{\perp}$. Then because $\tilde B$ is orthogonal, if $X = USV^T$ is an SVD of $X$, then $X\tilde B = US(\tilde B^TV)^T$ is an SVD of $X \tilde B$.
Because $XB$ can be gotten from $X\tilde B$ by dropping the last $k$ columns, my original problem now reduces to the following: Given the SVD of a matrix $Y = DEF^T$, is there a way of finding the SVD of $Y' = D'E'F'^T$, where $Y'$ is the matrix resulting from dropping the last $k$ columns of $Y$? (Here I have $Y = X\tilde B$ and $Y' = XB$.)
This problem is referred to as "downdating the SVD", and in general, there seem to be many approaches for doing this. One relevant approach is found here, and more discussion here.
But in general, given that algorithms for downdating the SVD appear to be an area of active research, I'm concluding that there isn't a simple way of finding the SVD of $XB$ given only the SVD of $X$.
