# How to transform matrix to this form by unitary transformation?

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!

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In what context does this problem arise? – David Roberts Aug 17 '13 at 4:47
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. – Eden Harder Aug 17 '13 at 5:44
In the special case where M has positive real entries, I think the answer is yes: pick $c_k$ such that $\sum_k c_kS^k$ has $m_i$ as its singular values. I believe that your desired form has enough flexibility to do this, though haven't thought it through. – Benjamin Young Aug 17 '13 at 16:58
If I was right about the above comment, then I guess you should be able to reduce to this case by multiplying a general M by a diagonal complex matrix with entries of norm 1? – Benjamin Young Aug 17 '13 at 17:02
@Benjamin You, probably, wanted to say "Pick $c_k$ such that ... has prescribed singular values". It is clear that it is all that is needed (see any linear algebra textbook). But why are you so sure it is always possible? – fedja Aug 17 '13 at 17:17

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.

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Thanks! I took 'singular value' as 'eigenvalue'. Could you give the details of computation for $n=4$ using the theory of Grobner basis? – Eden Harder Aug 19 '13 at 11:09
I use the library "Groebner" of Maple. I ask for the solutions in the unknowns $[c_2,c_3],[c_0,c_1]$, that is the software seeks, during the first step, a subsystem in the unknowns $c_0,c_1$ only. For $n=3$ we obtain a polynomial in $c_1^2$ of degree $4$ and for $n=4$ we obtain a polynomial in $c_1^2$ of degree $24$. The coefficients of these polynomials are in $\mathbb{Q}[c_0^2]$ where $c_0^{2n}=\Pi n_i$. – loup blanc Aug 19 '13 at 16:58
Thanks! It seems that we can not go any further in this way, since the polynomial may have no root. Right? – Eden Harder Aug 20 '13 at 1:11
Right Eden ! For instance, let $n=5$. If $(n_i)=(65,72,25,69,2)$, then I think (but I am not sure) that there are no real solutions in $(c_i)$. If $(n_i)=(95,69,68,44,27)$, then there are real solutions in $(c_i)$. – loup blanc Aug 21 '13 at 23:39
Thanks! Then we can consider the case $c_i$ may be not real. I will correct the question. – Eden Harder Aug 22 '13 at 1:44

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.

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"Notice that UMV still has spectral radius $m_1$" - why? The singular values are preserved, but the spectrum may change. – Federico Poloni Apr 19 '14 at 12:37
Thanks! Will unitary transformation keeps the spectral radius? – Eden Harder Apr 19 '14 at 12:37
@Frederik Poloni: I was aware that $UMV$ would not have the same spectrum as $M.$ However, it is true that $UMV$ has the same operator norm (with respect to Euclidean norm on vectors ) as $M,$ namely $m_{1}.$ I was still careless, because this certainly need not imply that $UMV$ has spectral radius $m_{1}.$ – Geoff Robinson Apr 19 '14 at 13:48