**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.