diff --git a/src/op/system/complete-upgrade.tex b/src/op/system/complete-upgrade.tex index 9ac84d3..fae2f83 100644 --- a/src/op/system/complete-upgrade.tex +++ b/src/op/system/complete-upgrade.tex @@ -73,7 +73,7 @@ The following is a slight generalisation of one form of the above example, but s \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1} &\le \phi_n(1_{M_n(A)})^{-1} \end{align*} - By the \hyperref[Schur complement condition]{lemma:schur-complement-positive}, + By the \hyperref[Schur complement condition]{proposition:schur-complement-positive}, \begin{align*} \phi_n(1_{M_n(A)}) &\ge \phi_n(T)\phi_n(1_{M_n(A)})^{-1}\phi_n(T)^* \\ &\ge \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1}\phi_n(T)\phi_n(T)^* \\ diff --git a/src/op/system/completely.tex b/src/op/system/completely.tex index dc816cd..93b323c 100644 --- a/src/op/system/completely.tex +++ b/src/op/system/completely.tex @@ -90,74 +90,6 @@ % Omitted -\begin{lemma} -\label{lemma:norm-matrix-test} - Let $A$ be a unital $C^*$-algebra, then for any $x \in A$ and $y \in A_{sa}$, $x^*x \le y$ if and only if - \[ - \begin{bmatrix} - 1 &x \\ - x^* &y - \end{bmatrix} \ge 0 - \] - - in $M_2(A)$. In particular, $\norm{x}_A \le 1$ if and only if - \[ - \begin{bmatrix} - 1 &x \\ - x^* &1 - \end{bmatrix} \ge 0 - \] - - in $M_2(A)$. -\end{lemma} -\begin{proof}[Proof, {{\cite[Lemma 3.1]{Paulsen}}}. ] - Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, assume without loss of generality that there exists a complex Hilbert space $H$ such that $A \subset B(H)$. For any $\xi, \eta \in H$, - \[ - \angles{ \begin{bmatrix} - 1 &x \\ - x^* &y - \end{bmatrix}\begin{bmatrix} - \xi \\ - \eta - \end{bmatrix}, \begin{bmatrix} - \xi \\ - \eta - \end{bmatrix}}_{H^2} = \norm{\xi}_H^2 + \dpn{x\eta, \xi}{H} + \dpn{\xi, x\eta}{H} + \dpn{y\eta, \eta}{H} - \] - - ($\Rightarrow$): If $y \ge x^*x$, then by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, - \begin{align*} - \angles{ \begin{bmatrix} - 1 &x \\ - x^* &y - \end{bmatrix}\begin{bmatrix} - \xi \\ - \eta - \end{bmatrix}, \begin{bmatrix} - \xi \\ - \eta - \end{bmatrix}}_{H^2} &\ge \norm{\xi}_H^2 + \dpn{x\eta, \xi}{H} + \dpn{\xi, x\eta}{H} + \dpn{x\eta, x\eta}{H} \\ - &\ge \norm{\xi}_H^2 - 2\norm{\xi}_H\norm{x\eta}_H + \norm{x\eta}_H^2 \ge 0 - \end{align*} - - ($\Leftarrow$): If the matrix is positive, then for each $\eta \in H$, - \begin{align*} - 0 &\le \angles{ \begin{bmatrix} - 1 &x \\ - x^* &y - \end{bmatrix}\begin{bmatrix} - -x\eta \\ - \eta - \end{bmatrix}, \begin{bmatrix} - -x\eta \\ - \eta - \end{bmatrix}}_{H^2} = \norm{x\eta}_H^2 -2\dpn{x\eta, x\eta}{H} + \dpn{y\eta, \eta}{H} \\ - &\le \dpn{y\eta, \eta}{H} - \dpn{x\eta, x\eta}{H} = \dpn{(y - x^*x)\eta, \eta}{H} - \end{align*} - - so $y \ge x^*x$. -\end{proof} - \begin{proposition} \label{proposition:unital-2-positive-contractive} Let $A, B$ be unital $C^*$-algebras, $S \subset A$ be an operator system, $\phi: S \to B$ be a unital $2$-positive map, then $\norm{\phi}_{L(S; B)} \le 1$. @@ -213,194 +145,6 @@ By \autoref{lemma:norm-matrix-test}, $\phi(x)^*\phi(x) \le \phi(x^*x)$. \end{proof} -\begin{lemma}[Schur Complement Condition for Positivity] -\label{lemma:schur-complement-positive} - Let $H$ be a complex Hilbert space, $P, Q, A \in B(H)$. Suppose that $P$ is invertible, then - \begin{enumerate}[label=(P\arabic*)] - \item If $P$ is also positive, then - \[ - \begin{bmatrix} - P &A \\ - A^* &Q - \end{bmatrix} = \begin{bmatrix} - P^{1/2} &0 \\ - A^*P^{-1/2} &I - \end{bmatrix}\begin{bmatrix} - I &0 \\ - 0 & Q - AP^{-1}A^* - \end{bmatrix}\begin{bmatrix} - P^{1/2} &P^{-1/2}A \\ - 0 &I - \end{bmatrix} - \] - - \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ if and only if $P \ge 0$ and $Q - A^*P^{-1}A \ge 0$. - \end{enumerate} - - The operator $Q - A^*P^{-1}A$ is the \textbf{Schur complement} of $P$. Similarly, if $Q$ is invertible, then - \begin{enumerate}[label=(Q\arabic*)] - \item If $Q$ is also positive, then - \[ - \begin{bmatrix} - P &A \\ - A^* &Q - \end{bmatrix} = \begin{bmatrix} - Q^{1/2} &Q^{-1/2}A \\ - 0 &I - \end{bmatrix}\begin{bmatrix} - I &0 \\ - 0 & P - A^*Q^{-1}A - \end{bmatrix}\begin{bmatrix} - Q^{1/2} &0 \\ - A^*Q^{-1/2} &I - \end{bmatrix} - \] - - \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ if and only if $Q \ge 0$ and $P - A^*Q^{-1}A \ge 0$. - \end{enumerate} - - Finally, let $P^{-1/2}: P^{1/2}(H) \to {P^{1/2}(H)}$ and $Q^{-1/2}: Q^{1/2}(H) \to {Q^{1/2}(H)}$ be inverses of $P^{1/2}$ and $Q^{1/2}$ on their images, respectively, $P^{-1} = (P^{-1/2})^2$, and $Q^{-1} = (Q^{-1/2})^2$, then the following are equivalent: - \begin{enumerate} - \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$. - \item For each $\xi, \eta \in H$, $|\dpn{A\xi, \eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{P\eta, \eta}{H}^{1/2}$. - \item $P \ge 0$, $A(H) \subset P^{1/2}(H)$, and $Q - A^*P^{-1}A \ge 0$. - \item $Q \ge 0$, $A(H) \subset Q^{1/2}(H)$, and $P - AQ^{-1}A^* \ge 0$. - \end{enumerate} - -\end{lemma} -\begin{proof}[Proof, {{\cite[Exercise 3.2]{Paulsen}}}. ] - (P1): Since $P$ is positive and invertible, so is $P^{1/2}$. Thus - \begin{align*} - \begin{bmatrix} - P &A \\ - A^* &Q - \end{bmatrix} &= \begin{bmatrix} - P^{1/2}P^{1/2} &P^{1/2}P^{-1/2}A \\ - A^*P^{-1/2}P^{1/2} &Q - \end{bmatrix} \\ - &= \begin{bmatrix} - P^{1/2} &0 \\ - A^*P^{-1/2} &I - \end{bmatrix}\begin{bmatrix} - P^{1/2} &P^{-1/2}A \\ - 0 &Q - A^*P^{-1}A - \end{bmatrix} \\ - \begin{bmatrix} - P^{1/2} &P^{-1/2}A \\ - 0 &Q - A^*P^{-1}A - \end{bmatrix} &= \begin{bmatrix} - I &0 \\ - 0 &Q - A^*P^{-1}A - \end{bmatrix} \begin{bmatrix} - P^{1/2} &P^{-1/2}A \\ - 0 &I - \end{bmatrix} - \end{align*} - - and - - \[ - \begin{bmatrix} - P &A \\ - A^* &Q - \end{bmatrix} = \begin{bmatrix} - P^{1/2} &0 \\ - A^*P^{-1/2} &I - \end{bmatrix}\begin{bmatrix} - I &0 \\ - 0 & Q - A^*P^{-1}A - \end{bmatrix}\begin{bmatrix} - P^{1/2} &P^{-1/2}A \\ - 0 &I - \end{bmatrix} - \] - - (P2): If $P$ is also positive and $Q - A^*P^{-1}A \ge 0$, then by (P1), $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right]$ is of the form $MDM^*$, where $D$ is positive. Thus the given matrix is also positive. - - On the other hand, if $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$, then as a diagonal entry, $P$ is also positive. In addition, - \[ - \begin{bmatrix} - I &0 \\ - 0 & Q - A^*P^{-1}A - \end{bmatrix} = \begin{bmatrix} - P^{-1/2} &0 \\ - -P^{-1}A^* &I - \end{bmatrix}\begin{bmatrix} - P &A \\ - A^* &Q - \end{bmatrix} \begin{bmatrix} - P^{-1/2} &-P^{-1}A \\ - 0 &I - \end{bmatrix} - \] - - is also of the form $MDM^*$, where $D$ is positive, so the diagonal matrix, and hence $Q - A^*P^{-1}A$ are also positive. - - (1) $\Rightarrow$ (2): If $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$, then $\left[\begin{smallmatrix} P + \eps & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ for all $\eps > 0$. By (P2), $P + \eps \ge 0$ and $Q - A^*(P + \eps)^{-1}A \ge 0$. Thus for any $\xi, \eta \in H$, by (P2) and the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, - \begin{align*} - |\dpn{A\xi, \eta}{H}| &= |\dpn{(P + \eps)^{-1/2}A\xi, (P + \eps)^{1/2}\eta}{H}| \\ - &\le \normn{(P + \eps)^{-1/2}A\xi}_{H} \normn{(P + \eps)^{1/2}\eta}_H \\ - &= \dpn{A^*(P + \eps)^{-1}A\xi, \xi}{H}^{1/2}\normn{(P + \eps)^{1/2}\eta}_H \\ - &= \dpn{Q\xi, \xi}{H}^{1/2}\dpn{(P + \eps)\eta, \eta}{H}^{1/2} - \end{align*} - - As the above holds for all $\eps > 0$, - \[ - |\dpn{A\xi, \eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{P\eta, \eta}{H}^{1/2} - \] - - (2) $\Rightarrow$ (3): Let $\xi \in H$, then for all $\eta \in P^{1/2}(H)$, - \[ - |\dpn{A\xi, P^{-1/2}\eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{\eta, \eta}{H}^{1/2} - \] - - By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists $\theta \in \ol{P^{1/2}(H)}$ such that - \[ - \dpn{A\xi, P^{-1/2}\eta}{H} = \dpn{\theta, \eta}{H} - \] - - for all $\eta \in P^{1/2}(H)$, so - \[ - \dpn{A\xi, P^{1/2}\eta}{H} = \dpn{\theta, P^{1/2}\eta}{H} = \dpn{P^{1/2}\theta, \eta}{H} - \] - - for all $\eta \in H$, and $A\xi \in P^{1/2}(H)$. In addition, for any $\xi \in H$, - \begin{align*} - \dpn{(Q - A^*P^{-1}A)\xi, \xi}{H} &= \dpn{Q\xi, \xi}{H} - \dpn{P^{-1}A\xi, A\xi}{H} \\ - &= \dpn{Q\xi, \xi}{H} - \dpn{A\xi, P^{-1}A\xi}{H} \\ - &= \dpn{Q\xi, \xi}{H} - \dpn{Q\xi, \xi}{H}^{1/2}\dpn{PP^{-1}A\xi, P^{-1}A\xi}{H}^{1/2} \\ - &\ge \dpn{Q\xi, \xi}{H} - \dpn{Q\xi, \xi}{H}^{1/2}\dpn{A^*P^{-1}A\xi, \xi}{H}^{1/2} \\ - &\ge \dpn{Q\xi, \xi}{H} - \frac{\dpn{Q\xi, \xi}{H}}{2} - \frac{\dpn{A^*P^{-1}A\xi, \xi}{H}}{2} \\ - &= \frac{1}{2}\dpn{(Q - A^*P^{-1}A)\xi, \xi}{H} - \end{align*} - - by \hyperref[Young's Inequality]{lemma:young-inequality}. Therefore $Q - A^*P^{-1}A \ge 0$. - - (3) $\Rightarrow$ (1): If $A(H) \subset P^{1/2}(H)$ and $Q - A^*P^{-1}A \ge 0$. For each $\xi \in H$, - \[ - \dpn{P^{-1}P^{1/2}\xi, P^{1/2}\xi}{H} = \dpn{P^{-1}P\xi, \xi}{H} = \dpn{\xi, \xi}{H} - \] - - and by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, - \[ - \dpn{(P+\eps)^{-1}P^{1/2}\xi, P^{1/2}\xi}{H} = \dpn{(P+\eps)^{-1}P\xi, \xi}{H} \le \dpn{\xi, \xi}{H} - \] - - Thus $\dpn{P^{-1}\xi, \xi}{H} \ge \dpn{(P+\eps)^{-1}\xi, \xi}{H}$ for all $\xi \in P^{1/2}(H)$. In particular, since $A(H) \subset P^{1/2}(H)$, - \begin{align*} - \dpn{P^{-1}A\xi, A\xi}{H} &\ge \dpn{(P+\eps)^{-1}A\xi, A\xi}{H} \\ - \dpn{A^*P^{-1}A\xi, \xi}{H} &\ge \dpn{A^*(P+\eps)^{-1}A\xi,\xi}{H} - \end{align*} - - so - - \[ - 0 \le Q - A^*P^{-1}A \le Q - A^*(P + \eps)^{-1}A - \] - - and $\left[\begin{smallmatrix} P + \eps & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$. As this holds for all $\eps > 0$, $\left[\begin{smallmatrix} P& A \\ A^* & Q \end{smallmatrix}\right] \ge 0$. - -\end{proof} \begin{example} diff --git a/src/op/system/index.tex b/src/op/system/index.tex index a9bc700..b2610e9 100644 --- a/src/op/system/index.tex +++ b/src/op/system/index.tex @@ -3,6 +3,7 @@ \input{./dilation.tex} \input{./os.tex} +\input{./two.tex} \input{./completely.tex} \input{./complete-upgrade.tex} \input{./module.tex} diff --git a/src/op/system/module.tex b/src/op/system/module.tex index 911007e..dfab3f2 100644 --- a/src/op/system/module.tex +++ b/src/op/system/module.tex @@ -104,3 +104,33 @@ By \autoref{lemma:matrix-positive-topleft-corner-zero}, $\phi(b)\phi(a) - \phi(ba) = 0$. \end{proof} + +\begin{definition}[Conditional Expectation] +\label{definition:cstar-conditional-expectation} + Let $A, B$ be unital $C^*$-algebras and $\phi: A \to B$ be a linear map, then $\phi$ is a \textbf{conditional expectation} if: + \begin{enumerate} + \item $\phi$ is unital and completely positive. + \item $\phi|_B$ is the identity. + \end{enumerate} +\end{definition} + + +\begin{example}[Conditional Expectation] + \label{example:matrix-conditional-expectation} + Let $\phi: M_n(\complex) \to \complex^n$ be defined by $\phi(A)_k = a_{kk}$, then: + \begin{enumerate} + \item $\phi$ is unital and completely positive. + \item The multiplicative domain of $\phi$ is the diagonal matrices on $M_n(\complex)$. + \end{enumerate} +\end{example} +\begin{proof} + (1): Identify $\complex^n$ as the diagonal matrices in $M_n(\complex)$, then for each $A \in M_n(\complex)$, + \[ + \phi(A) = \sum_{j = 1}^n E_{jj}AE_{jj} + \] + + By \autoref{proposition:multiply-completely-bounded}, $\phi$ is completely positive. + + (2): ??? +\end{proof} + diff --git a/src/op/system/os.tex b/src/op/system/os.tex index 5c55bea..771d55c 100644 --- a/src/op/system/os.tex +++ b/src/op/system/os.tex @@ -147,6 +147,7 @@ \begin{enumerate} \item The numerical radius $w: B(H) \to [0, \infty)$ is a norm on $B(H)$. \item For any $T \in B(H)$, $w(T) \le 1$ if and only if $2 + (\lambda T) + (\lambda T)^* \ge 0$ for all $\lambda \in \partial B_\complex(0, 1)$. + \item For each $T \in B(H)_{sa}$, $w(T) = \norm{T}_{B(H)}$. \item For each $T \in B(H)$, $w(T) \le \norm{T}_{B(H)} \le 2w(T)$. Both inequalities are sharp in general. \end{enumerate} \end{proposition} @@ -179,7 +180,21 @@ so $|\dpn{T\xi, \xi}{H}| \le 1$, and $w(T) \le 1$. - (3): For any $\xi \in H$ with $\norm{\xi}_H = 1$, $\dpn{T\xi, \xi}{H} \le \norm{T}_{B(H)}$, so $w(T) \le \norm{T}_H$. The sharp case is given by $I_H$. + (3): For any $\xi \in H$ with $\norm{\xi}_H = 1$, $\dpn{T\xi, \xi}{H} \le \norm{T}_{B(H)}$, so $w(T) \le \norm{T}_H$. On the other hand, if $w(T) \le 1$, then for each $\xi, \eta \in H$, + \begin{align*} + \dpn{T\xi, \eta}{H} &= \frac{1}{4}\sum_{k = 0}^3 i^k\dpn{T(\xi + i^k\eta), \xi + i^k\eta}{H} \\ + &= \frac{1}{4}(\dpn{T(\xi + \eta), \xi + \eta}{H} - \dpn{T(\xi - \eta), \xi - \eta}{H}) \\ + &\le \frac{w(T)}{4}(\norm{\xi + \eta}_H^2 + \norm{\xi - \eta}_H^2) \\ + &= \frac{1}{2}(\norm{\xi}_H^2 + \norm{\eta}_H^2) + \end{align*} + + by \hyperref[polarisation]{proposition:polarisation-complex} and the parallelogram law. In particular, + \begin{align*} + \norm{T\xi}_{H}^2 &\le \frac{1}{2}(\norm{\xi}_H^2 + \norm{T\xi}_H^2) \\ + \norm{T\xi}_H^2 &\le \norm{\xi}_H^2 + \end{align*} + + (4): For any $\xi \in H$ with $\norm{\xi}_H = 1$, $\dpn{T\xi, \xi}{H} \le \norm{T}_{B(H)}$, so $w(T) \le \norm{T}_H$. The sharp case is given by $I_H$. On the other hand, if $w(T) \le 1$, then $2 + T + T^* = 2 + 2\text{Re}(T) \ge 0$ and $2 - T - T^* = 2 - 2\text{Re}(T) \ge 0$ imply that $\norm{\text{Re}(T)}_{B(H)} \le 1$. Similarly, $\norm{\text{Im}(T)}_{B(H)} \le 1$ as well, so $\norm{T}_{B(H)} \le \norm{\text{Re}(T)}_{B(H)} + \norm{\text{Im}(T)}_{B(H)} \le 2$. Therefore $\norm{T}_{B(H)} \le 2w(T)$. diff --git a/src/op/system/two.tex b/src/op/system/two.tex new file mode 100644 index 0000000..03f5029 --- /dev/null +++ b/src/op/system/two.tex @@ -0,0 +1,349 @@ +\section{On the Subject of 2×2 Matrices} +\label{section:op-sys-2by2} + +\begin{lemma} +\label{lemma:norm-matrix-test} + Let $A$ be a unital $C^*$-algebra, then for any $x \in A$ and $y \in A_{sa}$, $x^*x \le y$ if and only if + \[ + \begin{bmatrix} + 1 &x \\ + x^* &y + \end{bmatrix} \ge 0 + \] + + in $M_2(A)$. In particular, $\norm{x}_A \le 1$ if and only if + \[ + \begin{bmatrix} + 1 &x \\ + x^* &1 + \end{bmatrix} \ge 0 + \] + + in $M_2(A)$. +\end{lemma} +\begin{proof}[Proof, {{\cite[Lemma 3.1]{Paulsen}}}. ] + Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, assume without loss of generality that there exists a complex Hilbert space $H$ such that $A \subset B(H)$. For any $\xi, \eta \in H$, + \[ + \angles{ \begin{bmatrix} + 1 &x \\ + x^* &y + \end{bmatrix}\begin{bmatrix} + \xi \\ + \eta + \end{bmatrix}, \begin{bmatrix} + \xi \\ + \eta + \end{bmatrix}}_{H^2} = \norm{\xi}_H^2 + \dpn{x\eta, \xi}{H} + \dpn{\xi, x\eta}{H} + \dpn{y\eta, \eta}{H} + \] + + ($\Rightarrow$): If $y \ge x^*x$, then by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, + \begin{align*} + \angles{ \begin{bmatrix} + 1 &x \\ + x^* &y + \end{bmatrix}\begin{bmatrix} + \xi \\ + \eta + \end{bmatrix}, \begin{bmatrix} + \xi \\ + \eta + \end{bmatrix}}_{H^2} &\ge \norm{\xi}_H^2 + \dpn{x\eta, \xi}{H} + \dpn{\xi, x\eta}{H} + \dpn{x\eta, x\eta}{H} \\ + &\ge \norm{\xi}_H^2 - 2\norm{\xi}_H\norm{x\eta}_H + \norm{x\eta}_H^2 \ge 0 + \end{align*} + + ($\Leftarrow$): If the matrix is positive, then for each $\eta \in H$, + \begin{align*} + 0 &\le \angles{ \begin{bmatrix} + 1 &x \\ + x^* &y + \end{bmatrix}\begin{bmatrix} + -x\eta \\ + \eta + \end{bmatrix}, \begin{bmatrix} + -x\eta \\ + \eta + \end{bmatrix}}_{H^2} = \norm{x\eta}_H^2 -2\dpn{x\eta, x\eta}{H} + \dpn{y\eta, \eta}{H} \\ + &\le \dpn{y\eta, \eta}{H} - \dpn{x\eta, x\eta}{H} = \dpn{(y - x^*x)\eta, \eta}{H} + \end{align*} + + so $y \ge x^*x$. +\end{proof} + + + +\begin{proposition}[Schur Complement Condition for Positivity, Invertible Case] +\label{proposition:schur-complement-positive} + Let $H$ be a complex Hilbert space, $P, Q, A \in B(H)$. Suppose that $P$ is invertible, then + \begin{enumerate}[label=(P\arabic*)] + \item If $P$ is also positive, then + \[ + \begin{bmatrix} + P &A \\ + A^* &Q + \end{bmatrix} = \begin{bmatrix} + P^{1/2} &0 \\ + A^*P^{-1/2} &I + \end{bmatrix}\begin{bmatrix} + I &0 \\ + 0 & Q - A^*P^{-1}A + \end{bmatrix}\begin{bmatrix} + P^{1/2} &P^{-1/2}A \\ + 0 &I + \end{bmatrix} + \] + + \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ if and only if $P \ge 0$ and $Q - A^*P^{-1}A \ge 0$. + \end{enumerate} + + The operator $Q - A^*P^{-1}A$ is the \textbf{Schur complement} of $P$. Similarly, if $Q$ is invertible, then + \begin{enumerate}[label=(Q\arabic*)] + \item If $Q$ is also positive, then + \[ + \begin{bmatrix} + P &A \\ + A^* &Q + \end{bmatrix} = \begin{bmatrix} + I &AQ^{-1/2} \\ + 0 &Q^{1/2} + \end{bmatrix}\begin{bmatrix} + P - AQ^{-1}A^* &0 \\ + 0 & I + \end{bmatrix}\begin{bmatrix} + I &0 \\ + Q^{-1/2}A^* &Q^{1/2} + \end{bmatrix} + \] + + \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ if and only if $Q \ge 0$ and $P - AQ^{-1}A^* \ge 0$. + \end{enumerate} +\end{proposition} +\begin{proof}[Proof, {{\cite[Exercise 3.2]{Paulsen}}}. ] + (P1): Since $P$ is positive and invertible, so is $P^{1/2}$. Thus + \begin{align*} + \begin{bmatrix} + P &A \\ + A^* &Q + \end{bmatrix} &= \begin{bmatrix} + P^{1/2}P^{1/2} &P^{1/2}P^{-1/2}A \\ + A^*P^{-1/2}P^{1/2} &Q + \end{bmatrix} \\ + &= \begin{bmatrix} + P^{1/2} &0 \\ + A^*P^{-1/2} &I + \end{bmatrix}\begin{bmatrix} + P^{1/2} &P^{-1/2}A \\ + 0 &Q - A^*P^{-1}A + \end{bmatrix} \\ + \begin{bmatrix} + P^{1/2} &P^{-1/2}A \\ + 0 &Q - A^*P^{-1}A + \end{bmatrix} &= \begin{bmatrix} + I &0 \\ + 0 &Q - A^*P^{-1}A + \end{bmatrix} \begin{bmatrix} + P^{1/2} &P^{-1/2}A \\ + 0 &I + \end{bmatrix} + \end{align*} + + and + + \[ + \begin{bmatrix} + P &A \\ + A^* &Q + \end{bmatrix} = \begin{bmatrix} + P^{1/2} &0 \\ + A^*P^{-1/2} &I + \end{bmatrix}\begin{bmatrix} + I &0 \\ + 0 & Q - A^*P^{-1}A + \end{bmatrix}\begin{bmatrix} + P^{1/2} &P^{-1/2}A \\ + 0 &I + \end{bmatrix} + \] + + (P2): If $P$ is also positive and $Q - A^*P^{-1}A \ge 0$, then by (P1), $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right]$ is of the form $MDM^*$, where $D$ is positive. Thus the given matrix is also positive. + + On the other hand, if $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$, then as a diagonal entry, $P$ is also positive. In addition, + \[ + \begin{bmatrix} + I &0 \\ + 0 & Q - A^*P^{-1}A + \end{bmatrix} = \begin{bmatrix} + P^{-1/2} &0 \\ + -A^*P^{-1} &I + \end{bmatrix}\begin{bmatrix} + P &A \\ + A^* &Q + \end{bmatrix} \begin{bmatrix} + P^{-1/2} &-P^{-1}A \\ + 0 &I + \end{bmatrix} + \] + + is also of the form $MDM^*$, where $D$ is positive, so the diagonal matrix, and hence $Q - A^*P^{-1}A$ are also positive. +\end{proof} + + +\begin{proposition}[Schur Complement Condition for Positivity] +\label{proposition:schur-complement-positive-general} + Let $H$ be a complex Hilbert space, $P, Q, A \in B(H)$, $P^{-1/2}: P^{1/2}(H) \to \ker(P)^\perp$ and $Q^{-1/2}: Q^{1/2}(H) \to \ker(Q)^\perp$ be partial inverses of $P^{1/2}$ and $Q^{1/2}$, respectively, then the following are equivalent: + \begin{enumerate} + \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$. + \item For each $\xi, \eta \in H$, $|\dpn{A\xi, \eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{P\eta, \eta}{H}^{1/2}$. + \item $P \ge 0$, $A(H) \subset P^{1/2}(H)$, and $Q - (P^{-1/2}A)^*(P^{-1/2}A) \ge 0$. + \item $Q \ge 0$, $A^*(H) \subset Q^{1/2}(H)$, and $P - (Q^{-1/2}A)^*(Q^{-1/2}A) \ge 0$. + \end{enumerate} +\end{proposition} +\begin{proof}[Proof, {{\cite[Exercise 3.2]{Paulsen}}}. ] + (1) $\Rightarrow$ (2): If $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$, then $\left[\begin{smallmatrix} P + \eps & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ for all $\eps > 0$. By (P2) of \autoref{proposition:schur-complement-positive}, $P + \eps \ge 0$ and $Q - A^*(P + \eps)^{-1}A \ge 0$. Thus for any $\xi, \eta \in H$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, + \begin{align*} + |\dpn{A\xi, \eta}{H}| &= |\dpn{(P + \eps)^{-1/2}A\xi, (P + \eps)^{1/2}\eta}{H}| \\ + &\le \normn{(P + \eps)^{-1/2}A\xi}_{H} \normn{(P + \eps)^{1/2}\eta}_H \\ + &= \dpn{A^*(P + \eps)^{-1}A\xi, \xi}{H}^{1/2}\normn{(P + \eps)^{1/2}\eta}_H \\ + &\le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{(P + \eps)\eta, \eta}{H}^{1/2} + \end{align*} + + As the above holds for all $\eps > 0$, + \[ + |\dpn{A\xi, \eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{P\eta, \eta}{H}^{1/2} + \] + + (2) $\Rightarrow$ (3): Let $\xi \in H$, then for all $\eta \in P^{1/2}(H)$, + \[ + |\dpn{A\xi, P^{-1/2}\eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{\eta, \eta}{H}^{1/2} + \] + + By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists $\theta \in \ol{P^{1/2}(H)}$ such that + \[ + \dpn{A\xi, P^{-1/2}\eta}{H} = \dpn{\theta, \eta}{H} + \] + + for all $\eta \in P^{1/2}(H)$, so + \[ + \dpn{A\xi, \eta}{H} = \dpn{\theta, P^{1/2}\eta}{H} = \dpn{P^{1/2}\theta, \eta}{H} + \] + + for all $\eta \in H$, and $A\xi \in P^{1/2}(H)$. + + Now, for each Cauchy sequence $\seq{\xi_n} \subset H$ such that $\bracsn{P^{-1/2}A\xi_n}_1^\infty \subset \ker(P)^\perp$ is Cauchy, then there exists $\eta \in \ker(P)^\perp$ such that $P^{-1/2}A\xi_n \to \eta$ as $n \to \infty$. By continuity of $P^{1/2}$, $P^{1/2}P^{-1/2}A\xi_n \to P^{1/2}\eta$ as $n \to \infty$. Since $\eta$ is the unique element of $\ker(P)^\perp$ with $P^{1/2}\eta = A\xi = A\limv{n}\xi_n$, $\eta = P^{-1/2}A\limv{n}\xi_n$. By the \hyperref[Closed Graph Theorem]{theorem:closed-graph}, $P^{-1/2}A \in B(H)$, and the second part of the condition is well-defined. + + + Now, for any $\xi \in H$, + \[ + \dpn{(Q - (P^{-1/2}A)^*(P^{-1/2}A))\xi, \xi}{H} = \dpn{Q\xi, \xi}{H} - \normn{P^{-1/2}A\xi}_H^2 + \] + + As $P^{-1/2}A\xi \in \ker(P)^\perp = \ol{P(H)}$, + \begin{align*} + \normn{P^{-1/2}A\xi}_H &= \sup_{\substack{\eta \in H \\ \normn{P^{1/2}\eta}_H \le 1}}|\dpn{P^{-1/2}A\xi, P^{1/2}\eta}{H}| = \sup_{\substack{\eta \in H \\ \normn{P^{1/2}\eta}_H \le 1}}|\dpn{A\xi, \eta}{H}| \\ + &\le \sup_{\substack{\eta \in H \\ \normn{P^{1/2}\eta}_H \le 1}}\dpn{Q\xi, \xi}{H}^{1/2}\dpn{P^{1/2}\eta, P^{1/2}\eta}{H}^{1/2} = \dpn{Q\xi, \xi}{H}^{1/2} + \end{align*} + + so + \[ + \dpn{(Q - (P^{-1/2}A)^*(P^{-1/2}A))\xi, \xi}{H} \ge \dpn{Q\xi, \xi}{H} - \dpn{Q\xi, \xi}{H} = 0 + \] + + + (3) $\Rightarrow$ (1): For each $\xi \in H$, + \[ + \dpn{P^{-1/2}P^{1/2}\xi, P^{-1/2}P^{1/2}\xi}{H} = \dpn{\xi, \xi}{H} + \] + + and by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, + \[ + \dpn{(P+\eps)^{-1}P^{1/2}\xi, P^{1/2}\xi}{H} = \dpn{(P+\eps)^{-1}P\xi, \xi}{H} \le \dpn{\xi, \xi}{H} + \] + + Thus $\dpn{P^{-1/2}\xi, P^{-1/2}\xi}{H} \ge \dpn{(P+\eps)^{-1/2}\xi, (P+\eps)^{-1/2}\xi}{H}$ for all $\xi \in P^{1/2}(H)$. In particular, since $A(H) \subset P^{1/2}(H)$, + \begin{align*} + \dpn{P^{-1/2}A\xi, P^{-1/2}A\xi}{H} &\ge \dpn{(P+\eps)^{-1/2}A\xi, (P+\eps)^{-1/2}A\xi}{H} \\ + \dpn{(P^{-1/2}A)^*(P^{1/2}A) \xi, \xi}{H} &\ge \dpn{A^*(P+\eps)^{-1}A\xi,\xi}{H} + \end{align*} + + so + + \[ + 0 \le Q - (P^{-1/2}A)^*(P^{1/2}A) \le Q - A^*(P+\eps)^{-1}A + \] + + and $\left[\begin{smallmatrix} P + \eps & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ by (P2) of \autoref{proposition:schur-complement-positive}. As this holds for all $\eps > 0$, $\left[\begin{smallmatrix} P& A \\ A^* & Q \end{smallmatrix}\right] \ge 0$. +\end{proof} + + +\begin{proposition} +\label{proposition:schur-complement-positive-estimate} + Let $H$ be a complex Hilbert space and $P, Q, A \in B(H)$ such that + \[ + \begin{bmatrix} + P & A \\ + A^* & Q + \end{bmatrix} \ge 0 + \] + + then + \begin{enumerate} + \item For each $\xi \in H$, + \[ + 0 \le \dpn{(P + A + A^* + Q)\xi, \xi}{H} \le (\dpn{P\xi, \xi}{H}^{1/2} + \dpn{Q\xi, \xi}{H}^{1/2})^2 + \] + + \item $\normn{P + A + A^* + Q}_{B(H)} \le (\norm{P}_{B(H)}^{1/2}+\norm{Q}_{B(H)}^{1/2})^2$. + \item If $P = Q$, then $A^*A \le \norm{P}_{B(H)}P$ and $\norm{A}_{B(H)} \le \norm{P}_{B(H)}$. + \end{enumerate} +\end{proposition} +\begin{proof}[Proof, {{\cite[Exercise 3.2]{Paulsen}}}. ] + (1): For each $\xi \in H$, + \[ + 0 \le \angles{\begin{bmatrix} + P & A \\ + A^* & Q + \end{bmatrix}\begin{bmatrix} + \xi \\ \xi + \end{bmatrix}, \begin{bmatrix} + \xi \\ \xi + \end{bmatrix}}_{H^2} = \dpn{(P + A + A^* + Q)\xi, \xi}{H} + \] + + and + \begin{align*} + \dpn{A + A^*\xi, \xi}{H} &= 2\text{Re}\dpn{A + A^*\xi, \xi}{H} \\ + |\dpn{A + A^*\xi, \xi}{H}| &\le 2 \dpn{P\xi, \xi}{H}^{1/2}\dpn{Q\xi, \xi}{H}^{1/2} \\ + \dpn{(P + A + A^* + Q)\xi, \xi}{H} &\le \dpn{P\xi, \xi}{H} + 2 \dpn{P\xi, \xi}{H}^{1/2}\dpn{Q\xi, \xi}{H}^{1/2} + \dpn{Q\xi, \xi}{H} \\ + &= (\dpn{P\xi, \xi}{H}^{1/2} + \dpn{Q\xi, \xi}{H}^{1/2})^2 \\ + \end{align*} + + (2): By (3) of \autoref{proposition:bh-numerical-radius-gymnastics} and (1), + \begin{align*} + \norm{P + A + A^* + Q}_{B(H)} &= \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\dpn{(P + A + A^* + Q)\xi, \xi}{H} \\ + &\le \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}(\dpn{P\xi, \xi}{H}^{1/2} + \dpn{Q\xi, \xi}{H}^{1/2})^2 \\ + &\le (\norm{P}_{B(H)}^{1/2}+\norm{Q}_{B(H)}^{1/2})^2 + \end{align*} + + (3): Using (2) of \autoref{proposition:schur-complement-positive}, assume without loss of generality that $P \ne 0$. For each $\xi, \eta \in H$, + \begin{align*} + 0 &\le \angles{\begin{bmatrix} + P & A \\ + A^* & P + \end{bmatrix}\begin{bmatrix} + \xi \\ \eta + \end{bmatrix}, \begin{bmatrix} + \xi \\ \eta + \end{bmatrix}}_{H^2} \\ + &= \dpn{P\xi, \xi}{H} + \dpn{P\eta, \eta}{H} + \dpn{A\eta, \xi}{H} + \dpn{A^*\xi, \eta}{H} + \end{align*} + + In particular, for $\xi = -A\eta/\norm{P}_{B(H)}$, + \begin{align*} + 0 &\le \frac{\dpn{PA\eta, A\eta}{H}}{\norm{P}_{B(H)}^2} + \dpn{P\eta, \eta}{H} - \frac{\dpn{A\eta, A\eta}{H}}{\norm{P}_{B(H)}} - \frac{\dpn{A^*A\eta, \eta}{H}}{\norm{P}_{B(H)}} \\ + &= \frac{\dpn{PA\eta, A\eta}{H}}{\norm{P}_{B(H)}} + \norm{P}_{B(H)}\dpn{P\eta, \eta}{H} - 2\norm{A\eta}_H^2 \\ + &\le\norm{A\eta}_H^2 + \norm{P}_{B(H)}\dpn{P\eta, \eta}{H} - 2\norm{A\eta}_H^2 \\ + &= \norm{P}_{B(H)}\dpn{P\eta, \eta}{H} - \dpn{A^*A\eta, \eta}{H} + \end{align*} + + so $A^*A \le \norm{P}_{B(H)}P$, and $\norm{A}_{B(H)} \le \norm{P}_{B(H)}$. + +\end{proof}