This is a question from an old exam: Perform the svd decomposition of the following matrix
$$A=\begin{bmatrix} 1 & i\\ i & 1\end{bmatrix}$$
What I have done: I planned to use the svd decomposition formula $A=U\Sigma V^\dagger$ so I calculate $$A ^\dagger=\begin{bmatrix} 1 & -i\\ -i & 1\end{bmatrix}$$ where $A^\dagger$ is the complex-conjugated transpose of the the matrix A. Then I calculate $A^\dagger A$ and $A A^\dagger$ which both equals
$$\begin{bmatrix} 2 & 0\\ 0 & 2\end{bmatrix}$$ with double eigenvalues of 2. Then I take $A A^\dagger-2I$ which gives me the zero matrix and a basis for that would be the standard basis. In other words U should be $$\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$$ Doing the same thing for the right singular vector give me the same thing $A^\dagger A-2I$ and the zero matrix again.
The problem: First, according to the solution, it should be the V that is $$\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$$
Second, U should be $$\begin{bmatrix} 1/\sqrt{2} & i/\sqrt{2}\\ i/\sqrt{2} & 1/\sqrt{2}\end{bmatrix}$$.
What am I doing wrong?
Thanks!
There is nothing wrong. SVD is not unique. In your case, $\frac{1}{\sqrt{2}}A$ is unitary. Therefore $U(\sqrt{2}I)V^\ast$ is a SVD of $A$ as long as $U$ and $V$ are unitary matrices such that $UV^\ast=\frac{1}{\sqrt{2}}A$. So, there are infinitely many choices: just pick a random unitary matrix $V$ and set $U=\frac{1}{\sqrt{2}}AV$. In particular, you may take $(U,V^\ast)=(I,\frac{1}{\sqrt{2}}A)$ or $(\frac{1}{\sqrt{2}}A,I)$.