Is it possible to decompose an n by n unitary matrix U, such that $U=O_1DO_2 $ with D being diagonal(obviously just has complex phase factors) and $O_1,O_2$ being real orthogonal matrices.


Hint: Let $M$ a complex $n\times n$ matrix. Is it always possible to decompose it as $$M = O_1 D O_2$$ ? Dimension considerations say no. Indeed, the space of complex matrices has $2n^2$ degrees of freedom, while on the RHS we get $\binom{n}{2} + 2n + \binom{n}{2} = n^2 + n$. There must be some extra conditions required for $n>1$.

Note that from $M= O_1 D O_2$ we get $$\bar M = O_1 \bar D O_2$$

So we see that $M$ and $\bar M$ (or, if you want, $\mathcal{Re}M$ and $\mathcal{Im}M$) have "real singular value decomposition with the same $O_1$, $O_2$. When is this possible. If we go through the proof of the singular value decomposition theorem we notice that the matrices $M^t M$ are used. So we get

$$M = O_1 D O_2\\ M^t = O_2^t D O_1^{t}\\ M^t M = O_2^t D^2 O_2$$


$$\bar M = O_1 \bar D O_2\\ \bar M^t = O_2^t \bar D O_1^{t}\\ \bar M^t \bar M = O_2^t \bar D^2 O_2$$

The last lines are the key

$$M^t M = O_2^{-1} D^2 O_2\\ \bar M^t \bar M = O_2^{-1} \bar D^2 O_2$$

The matrices $M^t M$ and $\bar M^t \bar M$ can be conjugated by the same orthogonal map to diagonal matrices. This implies that these matrices commute

$$M^t M \bar M^t \bar M= \bar M^t \bar M M^t M$$

It is not very hard (using the simultaneous reduction of commuting symmetric matrices to diagonal form by the same orthogonal) that this condition is also sufficient.

Now we ask: does this hold for every unitary matrix? I will leave this for you to check.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.