Hermitian operators

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Section 8.4 Hermitian operators

A self-adjoint or Hermitian matrix is a complex \(n\times n\) matrix satisfying \(A = A\ad\text{.}\) Such matrices have a wealth of structural results available to describe them. For example, such matrices have real eigenvalues, can be unitarily diagonalized, and carry a partial order (called the Loewner order

¹

en.wikipedia.org/wiki/Loewner_order

) with a notion of positivity. We’re interested in developing analogous results in the infinite dimensional setting.

Definition 8.4.1.

Let \(\hilbert\) be a Hilbert space and \(A \in \L(\hilbert)\text{.}\) We say that \(A\) is Hermitian (or self-adjoint) if \(A = A\ad\text{.}\)

Hermitian operators have a nice expression for calculating norms.

Theorem 8.4.2.

If \(A\) is a Hermitian operator on a Hilbert space \(\hilbert\text{,}\) then

\begin{equation*} \norm{A} = \sup_{\norm{x} = 1} \abs{\ip{Ax}{x}}. \end{equation*}

Proof.

As an immediate consequence of Theorem 2.2.4, for any \(\norm{x} = 1\text{,}\) we have

\begin{equation*} \abs{\ip{Ax}{x}}\leq \norm{A}\norm{x}^2 = \norm{A}. \end{equation*}

In the other direction we have a bit more work to do. Let \(M = \sup_{\norm{x} = 1} \abs{\ip{Ax}{x}}\text{.}\) Note that for all \(u\text{,}\) we have \(\abs{\ip{Au}{u}} \leq M\norm{u}^2\) (which one should check!). Now assume that \(\norm{x} = 1\) and \(\norm{y} = 1\text{.}\) Using that \(A\ad = A\)),

\begin{equation*} \ip{A(x\pm y)}{x\pm y} = \ip{Ax}{x}\pm 2 \RE \ip{Ax}{y} + \ip{Ay}{y}. \end{equation*}

If we subtract one of these equations from the other, we get

\begin{align*} 4\RE\ip{Ax}{y} \amp= \ip{A(x+y)}{x+y} - \ip{A(x-y)}{x-y}\\ \amp \leq M(\norm{x + y}^2 + \norm{x-y}^2) \end{align*}

Now apply Theorem 2.2.8 to get

\begin{equation*} 4 \RE \ip{Ax}{y} \leq 2M(\norm{x}^2 + \norm{y}^2) = 4M. \end{equation*}

Suppose that \(\ip{Ax}{y} = e^{it} \abs{\ip{Ax}{y}}\text{.}\) Then replacing \(x\) with \(e^{-it} x\) in the inequality gives

\begin{equation*} \abs{\ip{Ax}{y}} \leq M \end{equation*}

when \(\norm{x} = \norm{y} = 1\text{.}\) Notice that when \(y = Ax/\norm{Ax}\text{,}\) we get

\begin{equation*} \norm{Ax}^2 \leq M \norm{Ax} \end{equation*}

and so

\begin{equation*} \norm{Ax} \leq M \end{equation*}

for all \(\norm{x} = 1\text{.}\) But then \(\norm{A} \leq M = \sup_{\norm{x} = 1} \abs{\ip{Ax}{x}}\text{.}\)