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 $U$?