## Saturday, 25 January 2014

### Singular values before and after the action of a unital channel

Let $\X$ be a complex Euclidean space having dimension $n$, let $\Phi\in\Channel(\X)$ be a unital channel, let $X\in\Lin(\X)$ be an operator, and let $Y = \Phi(X)$. Following our usual conventions, let $s_1(X) \geq \cdots \geq s_n(X)$ and $s_1(Y) \geq \cdots \geq s_n(Y)$ denote the singular values of $X$ and $Y$, respectively, ordered from largest to smallest, and where we take $s_k(X) = 0$ when $k > \rank(X)$ and $s_k(Y) = 0$ when $k > \rank(Y)$.

Theorem:

$s_1(X) + \cdots + s_m(X) \geq s_1(Y) + \cdots + s_m(Y)$ for every $m \in \{1,\ldots,n\}$.

Proof:

Consider the space $\X\oplus\X$ and let
$\overline{\X}:= \begin{pmatrix} 0 & X\\ X^{\ast} & 0 \end{pmatrix}.$
Then it holds that $\overline{X}=\overline{X}^\ast$ so that $\overline{X}\in\Herm(\X\oplus\X)$. In addition, consider the channel $\overline{\Phi}\in\Channel(\X\oplus\X)$ defined as
$\overline{\Phi} \begin{pmatrix} A & B\\ C & D \end{pmatrix}= \begin{pmatrix} \Phi(A) & \Phi(B)\\ \Phi(C) & \Phi(D) \end{pmatrix}.$
Then it follows that
$\overline{\Phi}(\I_{\X\oplus\X}) \begin{pmatrix} \Phi(\I_{\X}) & 0\\ 0 & \Phi(\I_{\X}) \end{pmatrix}= \begin{pmatrix} \I_{\X} & 0\\ 0 & \I_{\X} \end{pmatrix}= \I_{\X\oplus\X},$
which implies that $\overline{\Phi}$ is unital. Moreover, letting $\Phi(X)=Y$ with $\Phi(X^\ast)=Y^\ast$ yields
$\overline{\Phi}(\overline{X})= \begin{pmatrix} 0&\Phi(X)\\ \Phi(X^\ast) & 0) \end{pmatrix}= \begin{pmatrix} 0 & Y\\ Y^\ast & 0 \end{pmatrix}=:\overline{Y}.$
It has now been shown that there exists a unital channel $\overline{\Phi}\in \Trans(\X\oplus\X)$ such that $\overline{\Phi}(\overline{X})=\overline{Y}$, where $\overline{X},\overline{Y}\in\Herm(\X\oplus\X)$. Therefore, by Uhlmann's theorem, this is equivalent to the statement that $\lambda(\overline{Y}) \prec \lambda(\overline{X})$, where $\lambda(\overline{Y})$ and  $\lambda(\overline{X})$ are the vector of eigenvalues of $\overline{Y}$ and $\overline{X}$, respectively.

In order to determine the singular values of  $\overline{Y}$ and $\overline{X}$, consider the singular value decompositions of $Y$ and $X$:
$X=\sum_{k=1}^{r_X}s_k(X)x'_kx_k^\ast \ \ \ \text{and} \ \ \ Y=\sum_{k=1}^{r_Y}s_k(Y)y'_ky_k^\ast,$
where $r_X=\rank(X)$, $r_Y=\rank(Y)$, $s(X)=(s_1(X),\dots,s_{r_X}(X))$ and $s(Y)=(s_1(Y),\dots,s_{r_Y}(Y))$ are the vectors of the non-zero singular values of $X$ and $Y$ (assumed to be written in decreasing order as the index increases), and
$\{x_1,\dots,x_{r_X}\}, \{x'_1,\dots,x'_{r_X}\}\subseteq\X \ \ \ \text{and} \ \ \ \{y_1\dots y_{r_Y}\}, \{y'_1\dots y'_{r_Y}\}\subseteq\Y$
are orthonormal sets of vectors in their respective spaces.

Then since the block matrix $\overline{X}$ can be diagonalized as
$\overline{\X}:= U \begin{pmatrix} 0 & X\\ X^{\ast} & 0 \end{pmatrix}U^\dagger= \begin{pmatrix} X & 0\\ 0 & -X^{\ast}, \end{pmatrix}$
with the unitary
$U=\frac{1}{\sqrt{2}}\begin{pmatrix} \I_\X & \I_\X\\ \I_\X & -\I_\X \end{pmatrix},$
the eigenvalues of $\overline{X}$ are given by
$\lambda(\X)=\{s_1(X),\dots,s_{r_X}(X),-s_{r_X}(X),\dots,-s_1(X)),$
where here they have been arranged in decreasing order. An equivalent argument shows that the the eigenvalues of $\overline{Y}$ are similarly given by
$\lambda(\Y)=\{s_1(Y),\dots,s_{r_Y}(Y),-s_{r_Y}(Y),\dots,-s_1(Y)).$

However the singular values of $\overline{X}$ and $\overline{Y}$ are related to the eigenvalues via the absolute value. Therefore the singular values $\overline{s}(X)$ of $\overline{X}$ and $\overline{s}(Y)$ of $\overline{Y}$ are positive and there are at least two equal values for each $s_k$. That is,
\begin{align*} \overline{s}(X)&=(s_1(X),s_1(X),\dots,s_{r_X}(X),s_{r_X}(X),\dots) \\ \overline{s}(Y)&=(s_1(Y),s_1(Y),\dots,s_{r_Y}(Y),s_{r_Y}(Y),\dots) , \end{align*}
where all values $s_{j}(X)$ and $s_{k}(Y)$for $j\geq r_X$ and $k \geq r_Y$ are assumed to be zero by the convention described in the problem statement.

Then it follows that for all $k\in\{1,\dots, n\}$,
$s_1(X)+s_1(X)+\dots+s_{k}(X)+s_{k}(X)\geq s_1(Y)+s_1(Y)+\dots+s_{k}(Y)+s_{k}(Y),$
or equivalently that
$s_1(X)\dots+s_{k}(X)\geq s_1(Y)+\dots+s_{k}(Y).$