Given a matrix with its Singular Value Decomposition:
$$
X = \begin{bmatrix}
x_{1,1} & \dots & x_{1,m} \\[0.3em]
\dots & \dots & \dots \\[0.3em]
x_{n,1} & \dots & x_{n,m}
\end{bmatrix} = U\Sigma V^T$$
If I build a new matrix, $X_{perm}$, whose rows are a permutation of the $X$ ones, e.g.
$$X_{perm1} = \begin{bmatrix}
x_{n,1} & \dots & x_{n,m} \\[0.3em]
\dots & \dots & \dots \\[0.3em]
x_{1,1} & \dots & x_{1,m}
\end{bmatrix}$$
(where the first and the last rows are exchanged), I thought that the $V$ matrix of the SVD won't change. Actually, computing it with some tools it changes.
Can you please briefly explain me why?
Best Answer
If $E$ is the elementary matrix that effects exchange of the first and last rows, then for $X = U \Sigma V^\ast$ an SVD of $X$, $X_{\mathrm{perm}} = (EU)\Sigma V^\ast$ ought to be a perfectly valid SVD of $X_{\mathrm{perm}} = EX$, since $E$ is a perfectly respectable real orthogonal matrix. But then again, SVD's aren't unique (only $\Sigma$ is, up to permutation of the diagonal entries), so it might just be an idiosyncrasy (but not an error) of your software package that it ends up computing different $V$'s for $X$ and $X_{\mathrm{perm}}$?