Let $J$ be the $n$ by $n$ matrix whose each entry is $1$. Also define $f(n)$ to be the least $m$ so that there is a $\lambda>0$ so that $\lambda J$ is the sum of at most $m$ unitary matrices. Note $f(2)=2$.
What is the value of $f(n)$ for $n>2$?
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6$\begingroup$ In a C$^*$-algebra any operator is the linear combination of 4 unitaries... Indeed, every self-adjoint $h$ with $\|h\|\leq 1$ can be written as $(u+u^*)/2$ where $u=h+i(1-h^2)^{1/2}$. As your $J$ is the scalar-multiple of a projection, we conclude that $f(n)=2$ for all $n$. $\endgroup$– Matthew DawsCommented Jan 18, 2012 at 14:41
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2$\begingroup$ A way to picture it geometrically: consider the line $x_1 = x_2 = \dots x_n$ (the $x_i$ being the coordinates of the vector space) and the rotation of 180 degrees around this line. The sum of a vector and his image rotation then lies on this line, from which one then can derive that the sum of the identity matrix and the matrix corresponding with the rotation is of the form $\lambda J$. $\endgroup$– Koen SCommented Jan 18, 2012 at 14:51
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1$\begingroup$ Am I wrong, or the 4-unitaries theorem does not apply here? The OP asks for a representation of the form $J=\alpha \sum Q_i$, not $J=\sum \alpha_i Q_i$. $\endgroup$– Federico PoloniCommented Jan 18, 2012 at 15:30
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1$\begingroup$ @Federico: Yes; but that's why I wrote "Indeed..." To be more explicit, there is $\lambda>0$ with $\lambda J=h$ a self-adjoint contraction. Then $2h=u+u^*$ for the $u$ I gave, and so $J = (1/2\lambda)(u+u^*)$. $\endgroup$– Matthew DawsCommented Jan 18, 2012 at 15:58
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2 Answers
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Edit: I correct the mistake pointed out by Matthew in his comment.
In fact, any matrix $X$ can be written as $X=\lambda(U+V)$ for unitary matrices $U,V$, and $\lambda$ can be taken as a half of the operator norm of $X$. For a proof, see this question. This is slightly stronger than Matthew's comment, and the proof works in any finite von Neumann algebra (i.e. when the partial isometry in the polar decomposition can be taken as a unitary). In a $C^*$-algebra this is not possible (consider $z \mapsto z$ in the $C^*$-algebra of continuous functions on the unit disc of the complex plane) but, as noted by Matthew, you can get the same decomposition with $4$ unitaries.
answered Jan 18, 2012 at 16:10
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$\begingroup$ Ah, nice! But in a general von Neumann algebra, wouldn't you get partial isometries in general? For example if $X$ is the unilateral shift on $\ell^2(\mathbb N)$. $\endgroup$ Commented Jan 18, 2012 at 16:17
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1$\begingroup$ You are right Matthew. I am too much working with finite von Neumann algebras. $\endgroup$ Commented Jan 18, 2012 at 16:23
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Just for fun ... This is more a comment than an answer. If $n=2$, the unit sphere $\Sigma$ of $M_2(\mathbb R)$ (the set of matrices whose operator norm equals $1$) is the joint of $SO_2$ and of $O_2^-$ (the set of orthogonal symmetries). The joint of two subsets $X$ and $Y$ is by definition the union of the segments $[x,y]$ with $x\in X$ and $y\in Y$. In particular, if $A\in\Sigma\setminus O_2$, then it is a unique convex combination of a rotation and a symmetry.
More interestingly, both $SO_2$ and $O_2^-$ are circles in the $3$-dimensional sphere $\Sigma$, and they are linked. In other word, every disk $D\in\Sigma$ with boundary $SO_2$ respectively $O_2^-$) must intersect $O_2^-$ (resp. $SO_2$).
answered Jan 18, 2012 at 17:18