Singular Value Decomposition of a $2\times2$ diagonal matrix Suppose we have a matrix $A = \begin{bmatrix}3 & 0\\0 & -2\end{bmatrix}$.
According to theory, the eigenvectors of $AA^T$ form $U$ and the eigenvectors of $A^TA$ form $V$.
Then, solving $\det(AA^T - \lambda I)=0$ I find $\lambda_1=9$ and $\lambda_2=4$.
And solving $AA^Tu_1 = 9u_1$ and $AA^Tu_2 = 4u_2$ I find that the eigenvectors of $AA^T$ are $u_1=[1,0]$ and $u_2=[0,1]$, this way obtaining $U = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}=I_{2\times2}$.
So far so good. But now if I compute $A^TA = AA^T$ yielding the same equations for finding the eigenvectors and I end up with the erroneous solution that $V=I_{2\times2}$ when in fact it should be $\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}$.
I do not want a solution by inverting $A$ to find $V$. Please, point me where in the calculation above I missed a $-1$. Thank you.
 A: The matrices $A^TA$ and $AA^T$ have the same eigenvalues, so we need not perform the eigen-decomposition twice. For example, let $AA^T$ has eigenvalue $\lambda$ and the corresponding eigenvector $v$, then we have,
$AA^Tv=\lambda v \implies (A^TA)(A^Tv)=\lambda (A^Tv)$
i.e., $A^TA$ also has the same eigenvalue $\lambda$ and the eigenvector $A^Tv$ (needs to be normalized).
Since, you already found eigenvalues for $AA^T$ as $9,4$ and the corresponding (left singular) eigenvectors as $\begin{bmatrix}1 \\ 0\end{bmatrix}$ and $\begin{bmatrix}0 \\ 1\end{bmatrix}$ respectively,
The eigenvalues for $A^TA$ will be $9, 4$ and the corresponding unnormalized eigenvectors will be $A^T\begin{bmatrix}1 \\ 0\end{bmatrix}=\begin{bmatrix}3 & 0 \\ 0 & -2\end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix}=\begin{bmatrix}3 \\ 0\end{bmatrix}$ and $A^T\begin{bmatrix}1 \\ 0\end{bmatrix}=\begin{bmatrix}3 & 0 \\ 0 & -2\end{bmatrix}\begin{bmatrix}0 \\ 1\end{bmatrix}=\begin{bmatrix}0 \\ -2\end{bmatrix}$ $\implies$ the corresponding normalized (right singular) eigenvectors are $\begin{bmatrix}1 \\ 0\end{bmatrix}$ and $\begin{bmatrix}0 \\ -1\end{bmatrix}$.
Since we have $A^TA=V\Sigma^2V^T$ and $AA^T=U\Sigma^2U^T$
$\implies$ we have $\Sigma=\begin{bmatrix}\sqrt{\lambda_1} & 0 \\ 
0 & \sqrt{\lambda_2} \end{bmatrix}=\begin{bmatrix}3 & 0 \\ 
0 & 2 \end{bmatrix}$, $V=I_{2\times 2}=\begin{bmatrix}1 & 0 \\ 
0 & 1 \end{bmatrix}$ and $U=\begin{bmatrix}1 & 0 \\ 
0 & -1 \end{bmatrix}$,
so that $U\Sigma V^T=\begin{bmatrix}1 & 0 \\ 
0 & -1 \end{bmatrix}\begin{bmatrix}3 & 0 \\ 
0 & 2 \end{bmatrix}\begin{bmatrix}1 & 0 \\ 
0 & 1 \end{bmatrix}^T=\begin{bmatrix}3 & 0 \\ 
0 & -2 \end{bmatrix}=A$
