Null space for $\mathcal{N}(A)$ given SVD of $A$ Let $A \in \mathbb{R}^{m \times n}$ be a matrix with singular value decomposition
$$
A = U \tilde{\Sigma} V^{T}.
$$
Let $\text{rank}(A) = r \leq \min(m, n)$. My textbook notes that the first $r$ columns of $U$ then constitute a basis for $\mathcal{R}(A)$ (the range of $A$). That makes sense, since $Ax = U \tilde{\Sigma} V^{T} x = U \tilde{x}$ where $\tilde{x} = \tilde{\Sigma}V^{T} x$.
However, my book also states that the first $n-r$ columns of $V$ constitute a basis for $\mathcal{N}(A)$ (the null space of $A$). Shouldn't it instead be that the last $n-r$ rows of $V$ are a basis for $\mathcal{N}(A)$? If we want $V^{T} x = 0$, then multiplication by $x$ must produce a linear combination of the rows of $V$ that sum to the zero vector.
 A: Every matrix 
$$
\mathbf{A} \in \mathbb{C}^{m\times n}_{\rho}
$$
has a singular value decomposition
$$
\begin{align}
  \mathbf{A} &=
  \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\
%
 &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}_{\rho\times \rho} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V* 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
%
%
 &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cccc|cc}
     \sigma_{1} & 0 & \dots &  &   & \dots &  0 \\
     0 & \sigma_{2}  \\
     \vdots && \ddots \\
       & & & \sigma_{\rho} \\\hline
       & & & & 0 & \\
     \vdots &&&&&\ddots \\
     0 & & &   &   &  & 0 \\
  \end{array} \right]
% V 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
%
  & =
% U
   \left[ \begin{array}{cccccccc}
    \color{blue}{u_{1}} & \dots & \color{blue}{u_{\rho}} & \color{red}{u_{\rho+1}} & \dots & \color{red}{u_{n}}
  \end{array} \right]
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V
   \left[ \begin{array}{c}
    \color{blue}{v_{1}^{*}} \\ 
    \vdots \\
    \color{blue}{v_{\rho}^{*}} \\
    \color{red}{v_{\rho+1}^{*}} \\
    \vdots \\ 
    \color{red}{v_{n}^{*}}
  \end{array} \right]
%
\end{align}
$$
The $\rho$ singular values are ordered and satisfy 
$$
  \sigma_{1} \ge \sigma_{2} \ge \dots \sigma_{1} > 0
$$
The column vectors are orthonormal basis vectors:
$$
\begin{align} 
% R A
\color{blue}{\mathcal{R} \left( \mathbf{A} \right)} &=
\text{span} \left\{
 \color{blue}{u_{1}}, \dots , \color{blue}{u_{\rho}}
\right\} \\
% R A*
\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} &=
\text{span} \left\{
 \color{blue}{v_{1}}, \dots , \color{blue}{v_{\rho}}
\right\} \\
% N A*
\color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} &=
\text{span} \left\{
\color{red}{u_{\rho+1}}, \dots , \color{red}{u_{m}}
\right\} \\
% N A
\color{red}{\mathcal{N} \left( \mathbf{A} \right)} &=
\text{span} \left\{
\color{red}{v_{\rho+1}}, \dots , \color{red}{v_{n}}
\right\} \\
%
\end{align}
$$
The dimensions of the subspace matrices are
$$
\begin{align}
%
 \color{blue}{\mathbf{U}_{\mathcal{R}}}  &\in \mathbb{C}^{m\times\rho} \\
%
 \color{blue}{\mathbf{V}_{\mathcal{R}}}  &\in \mathbb{C}^{n\times\rho} \\
%
 \color{red}{\mathbf{U}_{\mathcal{N}}}  &\in \mathbb{C}^{m\times (m-\rho)} \\
%
 \color{red}{\mathbf{V}_{\mathcal{N}}}  &\in \mathbb{C}^{n\times (n-\rho)} \\
%
\end{align}
$$
To answer your question: Yes, the final $n-\rho$ columns of $\mathbf{V}$ are a span for $\color{red}{\mathbf{V}_{\mathcal{N}}}$.
A: The assertion is true if the singular values are ordered in a way that the first $n-r$ diagonal entries of $\tilde \Sigma$ are zero. 
Let $v_1, v_2, \dots $ be the columns of $V$. Let $$v= \sum_{i=1}^{n-r}x_i v_i$$ be a linear combination of the first $n-r$ columns. This means that in the first row of $v$, there's a linear combination of the first row of the $v_i$s, in the second row the second and so on. 
Thus we can write $v=Vx$, where $x= \begin{pmatrix}x_1\\ x_2\\ \vdots \\ x_{n-r}\\ 0 \\ \vdots \\0 \end{pmatrix}.$
Then we have that $V^T V = I$, so that 
$$Av = U \tilde \Sigma V^T V x = U \tilde  \Sigma x.$$
Now, since we required $\tilde \Sigma$ to be of a shape that the left-upper $(n-r)\times (n-r)$ block consists of zeros and this is precisely where $x$ doesn't have any zeros, so we get $$\tilde \Sigma x = 0, \mbox{ hence } Ax = 0.$$
If the zeros wouldn't be on the top of the diagonal of $\tilde \Sigma$, but on the bottom, then indeed, we'd need to look at the last $n-r$ columns of $V$.
Addendum: I'm pretty sure that you should be able to proof this with some kind of nice duality argument (since by $A= U\tilde \Sigma V^T$ we get $A^T = V \tilde \Sigma ^T U,$ which shows that the columns of $V$ span $\mathcal R(A^T) = (\mathcal N(A))^\perp$.)
