Convex decomposition of a vector

232 Views Asked by At

Let $(a_i)_{i=1}^n$ be a probability vector, that is, $a_i\geq 0$ and $\sum_i a_i=1$ and let $(U_{ij})_{i,j=1}^n$ be a unistochastic matrix, that is, the pointwise square of a unitary matrix. Now consider the vector $v=U \cdot a$. It is a vector that is more mixed than $a$ (or more precisely $a$ majorizes $v$). It seems intuitive that there should exist a probabilistic ensemble $(t_i,x_i)_{i=1}^m$ (with $t_i$ probabilities and $x_i$ vectors) such that $v=\sum_{i=1}^m t_i x_i$ and the components of every $x_i$ is a permutation of the components of $a$.

One example to make things concrete. Let $a=(0.6,0.2,0.2)$ and

$$ U=\left(\begin{array}{ccc} 1/2 & 1/2 & 0 \\ 1/2 & 1/2 & 0 \\ 0 & 0 & 1 \end{array}\right) $$

then $v=(0.4,0.4,0.2)$ and it can be written as:

$$ v= 0.5 \left(\begin{array}{c} 0.6 \\ 0.2 \\ 0.2 \end{array}\right) + 0.5 \left(\begin{array}{c} 0.2 \\ 0.6 \\ 0.2 \end{array}\right) $$ that is, we can write $v$ as a convex combination of two permutations of $a$.

Then my question is, does this convex decomposition exist in general? I guess that the answer is yes, and that it is based on some well known litterature, so any pointers to the relevant texts would also be welcome.

1

There are 1 best solutions below

0
On BEST ANSWER

This is indeed true. A unistochastic matrix $U$ is doubly stochastic (i.e. real, non-negative, and each row and column sums to $1$, see also here: http://en.wikipedia.org/wiki/Unistochastic_matrix). Moreover, the Birkhoff-von Neumann-Theorem states that every doubly stochastic matrix is a convex combination of permutation matrices (see here: http://en.wikipedia.org/wiki/Birkhoff-von_Neumann_Theorem). So we have a representation $$ U = \sum_{j=1}^m t_j P_j $$ with some $m \in \mathbb N$, permutation matrices $P_j$, and real coefficients $t_j \geq 0$ with $\sum_{j=1}^mt_j=1.$ If we put $x_j = P_ja$, then the coordinates of $x_j$ are a permutation of those of $a$, and we get $$ v=UA=\left(\sum_{j=1}^mt_jP_j\right)a = \sum_{j=1}^mt_j\left(P_ja\right) = \sum_{j=1}^mt_jx_j, $$ as desired.