Conjecture : Any complex or real matrix is the sum of two unitary matrices.

Proof (ideas) :

We know that every complex matrix A could be diagonalized using two unitary matrices U and V : $A = UDV^{*}$ . The matrix D has positive elements : D=diag(d1,...d2) with  $d_1\geq d_2 \geq ...\geq d_n \geq 0$.

A basic result is the following : every diagonal matrix could be diagonalized with n unitary matrix. Indeed,  you just have to choose the good coefficient and use the set of matrix {$E_i$} where $E_i$ is a diagonal with -1 everywhere, except at position i  where there is a +1:

$\begin{pmatrix} -1 & & & & & & & & \\ & & \ddots & & & & 0 & & \\ & & & -1 & & & & & \\ & & & & 1 & & & & \\ & & 0 & & & -1 & & & \\ & & & & & & \ddots & & \\ & & & & & & & 1 & \end{pmatrix}$

The matrices in this set are unitary, and they are a basis of diagonal matrices.

Result for a sum of only two terms :

For n=2 we see that the previous matrices (Ei)  works, and there are only two of them.

For any n, I think that we could prove the hypothesis : Every diagonal matrix could be a sum of two block matrix as the following :

$\begin{pmatrix}<br /> (U) & & 0\\<br /> & & \\<br /> 0& & 1<br /> \end{pmatrix}<br /> ,<br /> \begin{pmatrix}<br /> 1& & 0 \\<br /> & & \\<br /> 0 & & (V)<br /> \end{pmatrix}$

I have not the end of the proof, but I was thinking of writing D as the sum or product of matrix of the form diag(1,d2,...dn+1) and diag(d1,d2,...dm,1). And applying a recurrence hypothesis, but I did not suceed.

Be Sociable, Share!

•