Polar decomposition of a linear combination of unitary matrices

linear algebramatricespositive-semidefinitesvdunitary-matrices

Consider a complex-valued square matrix $M$ of the form

$$M = \frac{1}{2}\left(U_1 + e^{-i\phi}U_2\right),$$

where $U_1$ and $U_2$ are unitary matrices and $\phi$ is a real number. Moreover, consider the polar decomposition of $M$:

$$M = U P,$$

where $U$ is a unitary matrix and where $P$ is a positive semi-definite Hermitian matrix.

I want to know if the matrices $U$ and $P$ can be directly expressed in closed form as a function of $U_1$, $U_2$, and $\phi$. Is that possible? I somehow doubt it, but I thought I'd ask.

We can assume that $M$ is nonsingular for simplicity.

EDIT: Shortly after posting this question, I discovered that there are at least expressions for the polar decomposition of 2 by 2 real matrices and of 2 by 2 complex matrices. Though I'm still wondering if something more can be said for the structure above in any dimension.

Best Answer

There exists a unitary matrix such that $V^2 = U_1^\dagger e^{-i\phi} U_2$ and $V+V^\dagger$ is positive semi-definite (It can be computed by diagonalizing $U_1^\dagger e^{-i\phi} U_2$ in an orthonormal basis and taking the square root of the eigenvalues, choosing a square root with non negative real part).

Then, we have : $$M = \frac 12 U_1V(V+V^\dagger)$$ so the polar decomposition is : $$U = U_1 V \text{ and } P= \frac 12 (V+V^\dagger)$$

This does not give a closed form expression for $U$ and $P$, but I don't think there is : I believe there is no closed form expression for $V$ and, if $P$ is non degenerate, then unicity of the polar decomposition means that finding $V$ is equivalent to finding $(U,P)$.

Related Question