As we know, a matrix multiplied by a diagonal matrix can be viewed as a scaling of itself. And multiplied by a unitary matrix can be viewed as a rotation from the original matrix. So now I'm wondering how to analytically find the largest eigenvalue of the following mentioned matrix.

Could anyone please give any inspirations? Many thanks!

$\mathbf{M} = \mathbf{\Lambda_G V \Lambda_p V}^H\mathbf{\Lambda_G^H}$, where $\mathbf{\Lambda_G}$ is a complex diagonal matrix, and $\mathbf{\Lambda_p}$ is real and its diagonal elements are all non-negative. $\mathbf{V}$ is a unitary matrix such that $\mathbf{VV}^H=\mathbf{I}$.

The problem discription

  • $\begingroup$ So $\Lambda_P$ is real? "Non-negative" does not make any sense for complex matrices. $\endgroup$ – NickD Jul 6 '18 at 13:27
  • $\begingroup$ Sorry, my mistake... already fixed it. Thanks for pointing out! $\endgroup$ – R.F. Jul 6 '18 at 13:31

It is possible when the matrices involved are 2 x 2. See here for closed form solutions to 2 x 2 SVDs, and for the problem posed here, the singular values and the eigenvalues are the same thing. Likely doable for 3 x 3, since that should reduce to solving a third order polynomial, and there are (very messy) closed form solutions for doing that. Beyond that, good luck - you have entered the mathematics wastelands...

  • $\begingroup$ Thanks for your quick respond! I went through your paper and that's amazing! Now I think I need some time to digest it... $\endgroup$ – R.F. Jul 6 '18 at 13:45
  • $\begingroup$ Even fourth order is solvable. Not so messy as one generally thinks. $\endgroup$ – Yves Daoust Jul 6 '18 at 13:47
  • $\begingroup$ I have always wondered whether it is possible in general, once one understands the underlying structure of a unitary matrix see here so that the problem can be attacked via a decimation approach. $\endgroup$ – John Polcari Jul 6 '18 at 14:12

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