How to transform matrix to this form by unitary transformation?

Question

Without loss of gernerality, we can only consider $n$-dimensional diagonal matrix $M$ whose elements are all nonnegative, i.e. $$M=\operatorname{diag}(m_1,m_2,\cdots,m_n)\ (m_i \geq 0).$$ Then is there any $U,V\in U(n)$ such that $$UMV=\sum_{k=0}^{n-1}c_kS^k,$$ where $$S=\left( \begin{array}{cccc} 0 & 1 & & \\ & 0 & \ddots & \\ & & \ddots & 1 \\ & & & 0 \\ \end{array} \right),\ c_i \in \mathbb{C}$$ It is true for $n=2$, but what about a general $n$?

Note: This problem is arised in quantum theory, where unitary transform can be ignored. So, we want to find some simple representation and we do not know whether the above question is true or false. Any help or suggestion will be appreciated!

Answer 1

Eden, your last post is false! Let the $(n_i=m_i^2)$ be given non-negative numbers and $T=\sum_k c_kS^k$. We search the $(c_i)$ (assumed to be real numbers) s.t. the eigenvalues of $TT^T$ are the $(n_i)$. If $n=2$, then $CharPoly(TT^T)=X^2-(2c_0^2+c_1^2)X+c_0^4$ that has $2$ non-negative roots. More generally, we must solve a system of $n$ algebraic equations with $n$ unknowns. The calculation of $c_0\geq 0$ is staightforward. Then it "remains" to solve $n-1$ equations in $n-1$ unknowns. We have a look at the case $n=4$: we use the theory of Grobner basis. $c_1$ is some root of a polynomial of degree $48$ (recall that $c_0$ is known). After, it is easy: $c_2$ and $c_3$ are solutions of polynomials of degree $1$ (recall that $c_0,c_1$ are known). Some numerical calculations seem to "show" that the required result is true.

Answer 2

Re-edited in view of comments made on (incorrect) first post- which means that this is reduced to a couple of remarks. I assume that the $m_{i}$ are intended to be real and positive. We may label so that $m_{1} \geq m_{2} \geq \ldots \geq m_{n},$ possibly after modifying $U$ and $V$ by permutation matrices, which are unitary in any case. Notice that $UMV$ still has operator norm (wrt Euclidean norm on vectors) $m_{1}.$ Now $\sum_{k=0}^{n-1} c_{k} S^{k}$ has spectrum $\{ c_{0} \}.$ Hence $|c_{0}| \leq m_{1}$ if $UMV$ has this form. Now there is an inner product on the space of $n \times n$ complex matrices given by $\langle A,B \rangle = {\rm tr}(A\overline{B}^{T} )$. Note that $\langle A,B \rangle = \langle XA,XB \rangle$ whenever $X$ is unitary.

By Cauchy-Schwarz, we have $|\langle U\sqrt{M}, V^{-1}\sqrt{M} \rangle| \leq {\rm tr}(M).$ Hence we must have $|{\rm tr}(UMV)| \leq {\rm tr}(M).$ But if $UMV$ has the stated form, then we should have ${\rm tr}(UMV) = nc_{0}.$ Hence we in fact must have $|c_{0}| \leq \frac{1}{n} \left( \sum_{i=1}^{n} m_{i} \right)$ if $UMV$ has such an expression.