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Any matrix $A \in Gl(n, \mathbb{C})$ can be written as a finite linear combination of elements $U_i\in U(n)$:

$$ A = \sum_{i} \lambda_i U_i$$

Is this true, how could I prove it?

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Are the $\lambda_i$ complex or real numbers ? – Dietrich Burde Mar 23 at 14:17
    
I guess complex, but it's not clear in the paper, where this is stated – john Mar 23 at 14:22
    
@CaptainLama how do you figure? – john Mar 23 at 14:32
    
Ah, no, nevermind, the argument I had in mind is actually false... – Captain Lama Mar 23 at 14:35
    
At least if the eigenvectors are lineally independent the sentence is true. You can use the own eigenvectors of A to build the unitary matrices $U_i$, playing with the eigenvalues and the $\lambda_i$ to obtain the proper eigenvalues of $A$. However, it is not clear for me what does it happens when some of the eigenvectors are linearly dependent. – seoanes Mar 23 at 14:47
up vote 8 down vote accepted

The reference to the MathOverflow question is a good one. If $A$ is a complex matrix, you can normalize so that $\|A\| \le 1$. Then $$ A = B + iC $$ where $B$, $C$ are selfadjoint and given by $$ B = \frac{1}{2}(A+A^{\star}),\;\;\; C=\frac{1}{2i}(A-A^{\star}). $$ These selfadjoint operators also satisfy $\|B\| \le 1$ and $\|C\|\le 1$, which means that their eigenvalues--which must be real--are in $[-1,1]$. Then you can decompose $B$ and $C$ as $$ B = \frac{1}{2}(U_{B}+V_{B}),\;\;\; C=\frac{1}{2}(U_{C}+V_{C}) $$ where $U_{B}, V_{B}, U_{C}, V_{C}$ are unitary and given by $$ U_{B} = B + i\sqrt{I-B^2},\;\; V_{B}=B-i\sqrt{I-B^2} \\ U_{C} = C + i\sqrt{I-C^2},\;\; V_{C}=C-i\sqrt{I-C^2} $$ This makes sense becaue $I-B^2$ and $I-C^2$ are selfadjoint with their eigenvalues in $[0,1]$; so the square roots are defined that also have eigenvalues in $[0,1]$. You can check that $$ U_{B}U_{B}^{\star}= U_{B}^{\star}U_{B} = (B-i\sqrt{I-B^2})(B+i\sqrt{I-B^2})=B^2+(I-B^2)=I. $$ Then $\frac{1}{2}(U_{B}+V_{B})=B$, $\frac{1}{2}(U_{C}+V_{C})=C$ and $A=B+iC$ is a linear combination of unitary matrices.

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looks good, thanks a lot for the effort. – john Mar 23 at 17:00
    
@john : You're welcome. I made a couple of minor corrections where I dropped a factor of $2$, but that does not affect the final result. – TrialAndError Mar 23 at 17:04

It is known that every complex square matrix $A$ can be written as a linear combination of at most two unitary matrices. First, by scaling, you may assume that $\|A\|\le1$. Then, by singular value decomposition, you may also assume that $$ A=\operatorname{diag}(s_1,\ldots,s_n) $$ where the singular values $s_j$s are real nonnegative and bounded above by $1$. Now, as $s_j=\frac12(z_j+\bar{z}_j)$, where $z_j=s_j+i\sqrt{1-s_j^2}$ has unit modulus, it follows that $A$ is the average of two unitary matrices.

There is also an open conjecture that every real square matrix is a linear combination of at most four real orthogonal matrices. See Chi-Kwong Li and Edward Poon, Additive Decomposition of Real Matrices, Linear and Multilinear Algebra, 50(4):321-326, 2002.

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I do not see an elementary linear algebra proof right now, but an answer is given at this MO-question, namely that in a $C^*$-algebra, any operator is the linear combination of four unitary operators. The algebra $M(n, \mathbb{C})$ of $n × n$ matrices over $\mathbb{C}$ becomes a $C^*$-algebra if we consider matrices as operators on the Euclidean space $\mathbb{C}^n$, and use the operator norm $||.||$ on matrices.

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