41.3 On the Subject of 2×2 Matrices

Lemma 41.3.1.label 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 $M_{2}(A)$.

Proof, [Lemma 3.1, Po02]. Using the Gelfand-Naimark Theorem, 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 Cauchy-Schwarz inequality,

\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 $y \ge x^{*}x$.$\square$

Proposition 41.3.2 (Schur Complement Condition for Positivity, Invertible Case).label Let $H$ be a complex Hilbert space, $P, Q, A \in B(H)$. Suppose that $P$ is invertible, then

(P1)
If $P$ is also positive, then
\[\begin{bmatrix}P&A \\ A^{*}&Q\end{bmatrix} = \begin{bmatrix}P^{1/2}&0 \\ A^{*}P^{-1/2}&I\end{bmatrix}\begin{bmatrix}I&0 \\ 0&Q - A^{*}P^{-1}A\end{bmatrix}\begin{bmatrix}P^{1/2}&P^{-1/2}A \\ 0&I\end{bmatrix}\]
(P2)
$\left[\begin{smallmatrix}P&A \\ A^{*}&Q\end{smallmatrix}\right] \ge 0$ if and only if $P \ge 0$ and $Q - A^{*}P^{-1}A \ge 0$.

The operator $Q - A^{*}P^{-1}A$ is the Schur complement of $P$. Similarly, if $Q$ is invertible, then

(Q1)
If $Q$ is also positive, then
\[\begin{bmatrix}P&A \\ A^{*}&Q\end{bmatrix} = \begin{bmatrix}I&AQ^{-1/2}\\ 0&Q^{1/2}\end{bmatrix}\begin{bmatrix}P - AQ^{-1}A^{*}&0 \\ 0&I\end{bmatrix}\begin{bmatrix}I&0 \\ Q^{-1/2}A^{*}&Q^{1/2}\end{bmatrix}\]
(Q2)
$\left[\begin{smallmatrix}P&A \\ A^{*}&Q\end{smallmatrix}\right] \ge 0$ if and only if $Q \ge 0$ and $P - AQ^{-1}A^{*} \ge 0$.

Proof, [Exercise 3.2, Po02]. (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.$\square$

Proposition 41.3.3 (Schur Complement Condition for Positivity).label 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:

(1)
$\left[\begin{smallmatrix}P&A \\ A^{*}&Q\end{smallmatrix}\right] \ge 0$.
(2)
For each $\xi, \eta \in H$, $|\dpn{A\xi, \eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{P\eta, \eta}{H}^{1/2}$.
(3)
$P \ge 0$, $A(H) \subset P^{1/2}(H)$, and $Q - (P^{-1/2}A)^{*}(P^{-1/2}A) \ge 0$.
(4)
$Q \ge 0$, $A^{*}(H) \subset Q^{1/2}(H)$, and $P - (Q^{-1/2}A)^{*}(Q^{-1/2}A) \ge 0$.

Proof, [Exercise 3.2, Po02]. (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 Proposition 41.3.2, $P + \eps \ge 0$ and $Q - A^{*}(P + \eps)^{-1}A \ge 0$. Thus for any $\xi, \eta \in H$, by the Cauchy-Schwarz inequality,

\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 Riesz Representation Theorem, 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 Closed Graph Theorem, $P^{-1/2}A \in B(H)$, 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*}

\[\dpn{(Q - (P^{-1/2}A)^*(P^{-1/2}A))\xi, \xi}{H}\ge \dpn{Q\xi, \xi}{H}- \dpn{Q\xi, \xi}{H}= 0\]

(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 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*}

\[0 \le Q - (P^{-1/2}A)^{*}(P^{1/2}A) \le Q - A^{*}(P+\eps)^{-1}A\]

and $\left[\begin{smallmatrix}P + \eps&A \\ A^{*}&Q\end{smallmatrix}\right] \ge 0$ by (P2) of Proposition 41.3.2. As this holds for all $\eps > 0$, $\left[\begin{smallmatrix}P&A \\ A^{*}&Q\end{smallmatrix}\right] \ge 0$.$\square$

Proposition 41.3.4.label 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

(1)
For each $\xi \in H$,
\[0 \le \dpn{(P + A + A^* + Q)\xi, \xi}{H}\le (\dpn{P\xi, \xi}{H}^{1/2}+ \dpn{Q\xi, \xi}{H}^{1/2})^{2}\]
(2)
$\normn{P + A + A^* + Q}_{B(H)}\le (\norm{P}_{B(H)}^{1/2}+\norm{Q}_{B(H)}^{1/2})^{2}$.
(3)
If $P = Q$, then $A^{*}A \le \norm{P}_{B(H)}P$ and $\norm{A}_{B(H)}\le \norm{P}_{B(H)}$.

Proof, [Exercise 3.2, Po02]. (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 Proposition 41.2.10 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 Proposition 41.3.2, 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 $\norm{A}_{B(H)}\le \norm{P}_{B(H)}$.$\square$

Jerry's Digital Garden

41.3 On the Subject of 2×2 Matrices

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