Singular value decomposition for matrices that are not square? I understand that the Singular Value Decomposition is defined as SVD = $U\Sigma V^T$ , but I am slightly confused about the calculations when the matrix is not square. 
For example, I have the matrix:
$$
\begin{bmatrix}
1 & -1 \\
-2 & 2 \\
2 & -2
\end{bmatrix}
$$
When I am solving for $V$, however, I am missing the last component. Have I done something wrong when calculating for matrices that are not square matrices?
$$\det(A^T A - \lambda I) = 
        \begin{bmatrix}
        2 - \lambda & -4 & 4 \\
        -4 & 8 - \lambda & -8\\
        4 & -8 & 8 - \lambda
        \end{bmatrix}
$$
$\lambda = 0, 2, 16$
Eigenvectors respectively are: 
\begin{bmatrix}             
        1  \\           
        1/2 \\
        0
 \end{bmatrix}
 \begin{bmatrix}             
        1  \\           
        2/7 \\
       -2/7
 \end{bmatrix}
\begin{bmatrix}             
        0  \\           
        1 \\
       -1
 \end{bmatrix}
Therefore $$\Sigma = 
\begin{bmatrix}             
        \sqrt 2  & 0 & 0  \\           
        0 & \sqrt 16 & 0 \\
        0 & 0 & 0
 \end{bmatrix}$$
Also, $$V = 
\begin{bmatrix}
       7/\sqrt 57 & 0 & 2/\sqrt 5 \\
       2/\sqrt 57 & 1/\sqrt 2 & 1/\sqrt 5 \\
       2/\sqrt 57 & -1/\sqrt 2 & 0
\end{bmatrix}$$
This is the portion I am confused about.
Is $U = AV / \sqrt\lambda $ ? What if I have am missing a vector so that I can only get the first two columns of $U$?
 A: First, correct a simple error
$$
\mathbf{A}^{*}\mathbf{A} = 
\left[
\begin{array}{rr}
 9 & -9 \\
 -9 & 9 \\
\end{array}
\right].
$$
This system matrix $\mathbf{A}$ is a rank one matrix (column 2 = $-$column 1) with singular value decomposition
$$
\begin{align}
  \mathbf{A} 
&= 
\mathbf{U}\, \Sigma \, \mathbf{V}^{*} \\
%
\left[
\begin{array}{rr}
 1 & -1 \\
 -2 & 2 \\
 2 & -2 \\
\end{array}
\right]
&= 
% U
\left[
\begin{array}{ccc}
\frac{1}{3}
\color{blue}{\left[
\begin{array}{r}
 -1 \\
  2 \\
 -2
\end{array}
\right]}
&
\frac{1}{\sqrt{5}}
\color{red}{\left[
\begin{array}{r}
 -2 \\
  0 \\
  1
\end{array}
\right]}
&
\frac{1}{3\sqrt{5}}
\color{red}{\left[
\begin{array}{r}
  2 \\
  5 \\
  4
\end{array}
\right]}
%
\end{array}
\right]
% sigma
\left[
\begin{array}{cc}
 3 \sqrt{2} & 0 \\
 0 & 0 \\
 0 & 0 \\
\end{array}
\right]
% V
\frac{1}{\sqrt{2}}
\left[
\begin{array}{rc}
 \color{blue}{-1} & \color{blue}{1} \\
 \color{red}{1}   & \color{red}{1} \\
\end{array}
\right].
\end{align}
$$
Blue vectors are in range spaces, red vectors in null spaces. 
The thin SVD uses the range space components only:
$$
  \mathbf{A} = 
% U
\frac{1}{3}
\color{blue}{\left[
\begin{array}{r}
 -1 \\
  2 \\
 -2
\end{array}
\right]}
% S
\left( 3\sqrt{2} \right)
% V
\frac{1}{\sqrt{2}}
\left[
\begin{array}{rc}
 \color{blue}{-1} & \color{blue}{1}
\end{array}
\right].
$$

You may benefit from this example: 
SVD and the columns — I did this wrong but it seems that it still works, why?
