Singular value decomposition works only for certain orthonormal eigenvectors, not all? I'm trying to find the SVD of the following matrix:
$$A=
\begin{pmatrix}
 1 & 1 \\
 2 & -2 \\
 2 & 2 \\
\end{pmatrix}
$$
I found the eigenvalues and vectors for $A'A$: 
$$
\begin{array}{cc}
 \lambda_1=10 & \lambda_2=8 \\
 e_1'=(1,1) & e_2'=(-1,1). \\
\end{array}
$$
I find the eigenvalues and vectors for $AA'$:
$$
\begin{array}{ccc}
 \lambda_1=10 & \lambda_2=8 & \lambda_3=0 \\
 e_1=(1,0,2) & e_2=(0,1,0) & e_3=(-2,0,1), \\
\end{array}$$
and so my SVD should be: 
$$\left(
\begin{array}{ccc}
 \frac{1}{\sqrt{5}} & 0 & -\frac{2}{\sqrt{5}} \\
 0 & 1 & 0 \\
 \frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\
\end{array}
\right).\left(
\begin{array}{cc}
 \sqrt{10} & 0 \\
 0 & \sqrt{8} \\
 0 & 0 \\
\end{array}
\right).\left(
\begin{array}{cc}
 \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
 -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array}
\right)$$
However, this gives 
$$\left(
\begin{array}{cc}
 1 & 1 \\
 -2 & 2 \\
 2 & 2 \\
\end{array}
\right)$$ instead of $A$. 
To get $A$ I need to decompose in the following way 
$$\left(
\begin{array}{ccc}
 \frac{1}{\sqrt{5}} & 0 & -\frac{2}{\sqrt{5}} \\
 0 & 1 & 0 \\
 \frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\
\end{array}
\right).\left(
\begin{array}{cc}
 \sqrt{10} & 0 \\
 0 & \sqrt{8} \\
 0 & 0 \\
\end{array}
\right).\left(
\begin{array}{cc}
 \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
 \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
\end{array}
\right)$$
This last decomposition is as if I had multiplied the first $e_2$ by $-1$. 
These eigenvalues and vectors were derived from Mathematica, just to be sure I was using the correct elements. 
Any help in explaining why my first decomposition doesn't work would be appreciated.
Edit: The book I'm using doesn't tell which of the orthonormal eigenvectors I have to use. For each eigenvalue, I have two orthonormal eigenvectors, $e_i$ and $-e_i$.

The answer is yes. Since your $m\times n$ matrix $A$ is such that $m\geq n$, you first compute $\sigma_i$s and $v_i$s and than get the $u_i$s via relation
$$u_i = \frac{1}{\sigma_i}Av_i\text{.}$$
This follows from the proof of "SVD theorem": There exist unitary matrices $U$ and $V$ with the columns $u_i$ and $v_i$ (and matrix $\Sigma$ ...) for which $A = U\Sigma V^H$.
If you choose, for example $-u_i$s from the upper relation (which are still eigenvectors of $AA'$), you would get $U\Sigma V^H = -A$.
