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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
\[
s_1(X)\dots+s_{k}(X)\geq s_1(Y)+\dots+s_{k}(Y).
\]

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