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Decomposition in unitary matrices

July 7th, 2009 | 2 Comments | Posted in Maths

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 :

$\begin{pmatrix} (U) & & 0\\ & & \\ 0& & 1 \end{pmatrix} , \begin{pmatrix} 1& & 0 \\ & & \\ 0 & & (V) \end{pmatrix}$

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.

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2 Responses to “Decomposition in unitary matrices”

Jon Says:
July 8th, 2009 at 11:27
Hi J.C. Certainly the matrix 10I (I being the identity matrix) is not a sum of two unitaries.
Jice Says:
July 8th, 2009 at 18:19
Do you mean the following matrix :

$\begin{pmatrix} 1 & & \\ & 0 & \\ & & I \end{pmatrix} = \begin{pmatrix} e^{i\pi/3} & & \\ & e^{2i\pi/3} & \\ & & e^{i\pi/3}I \end{pmatrix}+\begin{pmatrix} e^{-i\pi/3} & & \\ & e^{-2i\pi/3} & \\ & & e^{-i\pi/3}I \end{pmatrix}$

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