In SVD, why is V in the row space In Strang's book, on the topic of Singular Value Decomposition, when we have: $AV=U\Sigma$, he says that the vectors $u$ are in the column space of $A$, which is clear enough. But also that $v$ are in the row space of $A$. Now, why would that be?
For each vector, we have:
$$Av_i=\sigma_iu_i$$
So clearly $u_i$ is in the column space of $A$. But why is $v_i$ in the row space of $A$? Does it actually stem from the above equation?
 A: The row space of $A$ is $\{ x^T A \}_x = \operatorname{sp} \{ b_1^T A, ..., b_n^T A \}$, for any basis $b_1,...,b_n$.
Since $U^T A = \Sigma V^T$, $e_k^T U^T A = u_k^T A = \sigma_k v_k^T$, we see that
the row space of $A$ is given by $\operatorname{sp} \{ \sigma_k v_k^T \} = \operatorname{sp} \{ v_k^T \}_{ \sigma_k \neq 0} $.
Note that it is not true in general that the $v_k^T$ are in the row space of $A$, what is true is that the row space of $A$ is contained in the span of the $v_k^T$. 
A: The answer is simple: We have
\begin{equation}
AV = U \Sigma \Rightarrow A = U \Sigma V^T \Rightarrow A^T = V \Sigma U^T.
\end{equation}
Thus, as you know, columns of $V$ are in the column space of $A^T$, equivalently columns of $V$ are in the row space of $A$.
A: This depends on which singular values the vectors in V corresponds to. 

If you have a reduced singular value decomposition then this is certainly true for every vector $v$, as each vector v has a non-zero singular value, this implies that:
$A^{T}Av = \sigma^2v$ 
And therefore this implies that $Av \neq 0$, which means that $v$ does not live in $A$'s null space. Now since $\mathbb{R}^m = col(A) \oplus null(A)$, we know that $v$ must live in A's row space.

This is not true for any $v$ in a non-reduced singular value decomposition as in that case there are such $v$ that $A^{T}Av = 0$ and for these since we know $v \neq 0$ it means these live in $null(A)$ exclusively.
(Additionally, from the above and the fact we know these vectors are linearly independent, the first $r$ vectors form a basis for $row(A)$ and the last $n-r$ form a basis for null(A))

I suspect that the question actually asks for the first $r$ vectors in V that correspond to non-zero singular values.
