For every matrix $A$ of dimensions $m\times n$ we have some decomposition of the form:
$$
A=U\Sigma V^{\dagger}
$$
So that $U,V$ are orthogonal matrices of dimensions $m\times m$,$n\times n$ respectively and $\Sigma$ some diagonal matrix of dimensions $m\times n$ with non-negative eigenvalues.
We can find $U$ by looking at $AA^{\dagger}$:
$$
AA^\dagger=U\Sigma V^\dagger V\Sigma^\dagger U^\dagger=UD^2U^\dagger
$$
And looking for the diagonalizing matrix $U$. and so for V:
$$
A^\dagger A=V\Sigma^\dagger U^\dagger U\Sigma V^\dagger=V\tilde{D}^2V^\dagger
$$
But if I do so for some hermitian matrix such as:
$$
A=\frac{1}{2}\left(\begin{array}{cc}1 & 1 \\ 1 & -1\end{array}\right)
$$
I get that both $A^\dagger A$ and $AA^\dagger$ are $\frac{1}{2}\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)$ for which I cannot find a suitable $U$ which fits the SVD correctly.
What am I missing?
I know that I can just diagonalize A, but as far as I'm concerned, this "algorithm" should work just fine.
Best Answer
Note that $A$ has orthogonal columns.
From $A^* A=AA^*=\frac{1}{2} I$ you know $\Sigma = \frac{1}{\sqrt{2}} I_2$. Since $A^* A$ is a multiple of the identity, any orthogonal matrix $V$ will do; we can even take $V=I$. Then simply taking $U=\sqrt{2} A$ gives the SVD $A=U\Sigma V^*$. There are several other SVD decompositions as well (e.g., take $U=I$ and $V=\sqrt{2} A$).