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 $-e_i$.
 A: You need to match the left singular vectors to the right ones, or vice versa. E.g. after you have computed $e_1'$ and $e_2'$, you could get the two corresponding left singular vectors as $e_1=Ae_1'/\|Ae_1'\|=\frac1{\sqrt{5}}(1,0,2)^T$ and $e_2=Ae_2'/\|Ae_2'\|=(0,\color{red}{-1},0)^T$ (note: the sign of $e_2$ here is different from yours). The remaining left singular vector can be any unit vector orthogonal to the previous two singular vectors (in this case, it must be $\pm e_1\times e_2$, where the sign is unimportant). Then
\begin{align*}
A&=
\pmatrix{e_1&e_2&\pm e_1\times e_2}
\pmatrix{\frac{\|Ae_1'\|}{\|e_1'\|}&0\\ 0&\frac{\|Ae_2'\|}{\|e_2'\|}\\ 0&0}
\pmatrix{\frac{(e_1')^T}{\|e_1'\|}\\ \frac{(e_2')^T}{\|e_2'\|}}\\
&=\pmatrix{\frac{1}{\sqrt{5}}&0&\frac{\pm2}{\sqrt{5}}\\
0&-1&0\\ \frac{2}{\sqrt{5}}&0&\frac{\mp1}{\sqrt{5}}}
\pmatrix{\sqrt{10}&0\\ 0&\sqrt{8}\\ 0&0}
\pmatrix{\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}&\frac{1}{\sqrt{2}}}.
\end{align*}
A: 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 $U\Sigma V^H = -A$.
A: I'm posting an answer just to complete the previous answers.
Let $A=UDV'$. 
Then, if we have dim of $U$ greater than $V$ we can use the formula $v_i=\frac{1}{\sigma_i} A'u_i$, for given $U$.
However, if we have dim of $U$ smaller than $V$, we can use the formula $u_i=\frac{1}{\sigma_i}Av_i$, for given $V$.
