Dumb question, but just to be 1000% sure is the solution the column in V (aka the row in V^T), or the column in V^T ?
mpotoole
It is the column of $V$ (and equivalently, the row of $V^T$).
qryy
I don't get the purpose of the ||x||^2 = 1. constraint. Is this supposed to make sure we don't have the trivial zero vector solution for Ax=0? If so, how does this actually prevent that, and why do we choose it to equal 1?
mpotoole
Yes---this constraint is to make sure that we do not get the trivial solution. The reason that it prevents the trivial solution of $x = 0$ is because $|0| = 0$. So by making that the norm is strictly non-zero, we guarantee to avoid the trivial solution. And also, we could choose another value for this equality constraint other than 1; if will only change the magnitude of the corresponding solution.
Dumb question, but just to be 1000% sure is the solution the column in V (aka the row in V^T), or the column in V^T ?
It is the column of $V$ (and equivalently, the row of $V^T$).
I don't get the purpose of the ||x||^2 = 1. constraint. Is this supposed to make sure we don't have the trivial zero vector solution for Ax=0? If so, how does this actually prevent that, and why do we choose it to equal 1?
Yes---this constraint is to make sure that we do not get the trivial solution. The reason that it prevents the trivial solution of $x = 0$ is because $|0| = 0$. So by making that the norm is strictly non-zero, we guarantee to avoid the trivial solution. And also, we could choose another value for this equality constraint other than 1; if will only change the magnitude of the corresponding solution.
For anyone interested, I found this interesting derivation online for the SVD part: http://cmp.felk.cvut.cz/cmp/courses/XE33PVR/WS20072008/Lectures/Supporting/constrained_lsq.pdf (slide 13)
The most interesting takeaway was that the orthonormality of the $V^T$ matrix very naturally links with the constraint that $||x|| = 1$