SVD of a matrix from SVD of its columns Assume a matrix A, and I know the left singular vectors of SVD(A(:,i)), i=1,2,...,# of columns, is there a simple/fast transformation to obtain the left singular vectors of SVD(A) (the whole matrix)?
 A: Absolutely not. The singular vectors of a single vector are trivial to calculate (there is a direct method you can use to do it). The singular vectors of a matrix are very hard to calculate (there is no, and there can be no, direct method if $n\geq 5$ where $n$ is the size of the matrix)
A: Start with a matrix $\mathbf{A}\in\mathbb{R}^{m/times n}_{\rho}$ and the definition of the singular value decomposition as
$$
 \mathbf{A} = \mathbf{U} \, \Sigma \, \mathbf{V}^{*}.
$$
Given $\mathbf{A}$ and $\mathbf{U}$, can we quickly compute $\mathbf{V}$?
@5xum provides the most general answer.
What we have is the identity,
$$
 \mathbf{U}^{*} \, \mathbf{A} = \Sigma \, \mathbf{V}^{*}.
$$
To move beyond this, we need either the singular values $\sigma_{k}$, $k=1,\rho$ to compute $\mathbf{V}$:
$$ 
\left[ \begin{array}{cc}
  \mathbf{S}^{-1} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} 
\end{array} \right]
%
\left[ \begin{array}{c}
  \color{blue}{\mathbf{U}^{*}_{\mathcal{R}}} \\ 
  \color{red}{\mathbf{U}^{*}_{\mathcal{N}}}
\end{array} \right]
\mathbf{A}
=
\left[ \begin{array}{c}
  \color{blue}{\mathbf{V}^{*}_{\mathcal{R}}} \\ 
  \mathbf{0}
\end{array} \right]
%
%
$$
or we need $\mathbf{V}$ to compute the singular value spectrum
$$
 \mathbf{U}^{*} \, \mathbf{A} \, \mathbf{V} = \Sigma
$$
