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Mathematics LibreTexts

11.2 Normal operators

 

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Normal operators are those that commute with their own adjoint. As we will see, this includes many important examples of operations.

Definition 11.2.1. We call \(T\in\mathcal{L}(V)\) normal if \(TT^*=T^*T\).

        Given an arbitrary operator \(T \in \mathcal{L}(V)\), we have that \(TT^*\neq T^*T\) in general. However, both \(TT^*\) and \(T^*T\) are self-adjoint, and any self-adjoint operator \(T\) is normal. We now give a different characterization for normal operators in terms of norms.

Proposition 11.2.2. Let \(V\) be a complex inner product space, and suppose that \(T\in\mathcal{L}(V)\) satisfies

\begin{equation*}
    \inner{Tv}{v} = 0, \quad \text{for all \(v\in V\).}
\end{equation*}
Then \(T=0\).

Proof. You should be able to verify that
\begin{equation*}
\begin{split}
    \inner{Tu}{w} = \frac{1}{4} & \left\{ \inner{T(u+w)}{u+w} - \inner{T(u-w)}{u-w}\right.\\
    & \left.+i  \inner{T(u+iw)}{u+iw} - i \inner{T(u-iw)}{u-iw} \right\} .
\end{split}
\end{equation*}
Since each term on the right-hand side is of the form \(\inner{Tv}{v}\), we obtain 0 for each \(u,w\in V\).
Hence \(T=0\).

Proposition 11.2.3.  Let \(T\in \mathcal{L}(V)\). Then \(T\) is normal if and only if

\begin{equation*}
    \norm{Tv} = \norm{T^* v}, \quad \text{for all \(v\in V\).}
\end{equation*}

Proof.  Note that

\begin{equation*}
\begin{split}
    \text{\(T\) is normal} & \Longleftrightarrow T^*T-TT^* =0\\
    & \Longleftrightarrow \inner{(T^*T-TT^*)v}{v} = 0, \quad \text{for all \(v\in V\)}\\
    & \Longleftrightarrow \inner{TT^* v}{v} = \inner{T^*T v}{v}, \quad \text{for all \(v\in V\)}\\
    & \Longleftrightarrow \norm{Tv}^2 = \norm{T^*v}^2, \quad \text{for all \(v\in V\).}
\end{split}
\end{equation*}

Corollary 11.2.4.  Let \(T \in \mathcal{L}(V)\) be a normal operator.

  1. \(\kernel(T) = \kernel(T^*)\).
  2. If \(\lambda\in\mathbb{C}\) is an eigenvalue of \(T\), then \(\overline{\lambda}\) is an eigenvalue of \(T^*\) with the same eigenvector.
  3. If \(\lambda,\mu\in\mathbb{C}\) are distinct eigenvalues of \(T\) with associated eigenvectors \(v,w\in V\), respectively, then \(\inner{v}{w}=0\).
 
Proof.  Note that Part~1 follows from Proposition 11.2.3 and the positive definiteness of the norm.

To prove Part~2, first verify that if \(T\) is normal, then \(T-\lambda I\) is also normal with \((T-\lambda I)^* = T^* - \overline{\lambda} I\). Therefore, by Proposition 11.2.3, we have
\begin{equation*}
     0 = \norm{(T-\lambda I) v} = \norm{(T-\lambda I)^* v} = \norm{(T^*-\overline{\lambda} I)v},
\end{equation*}
and so \(v\) is an eigenvector of \(T^*\) with eigenvalue \(\overline{\lambda}\).

Using Part~2, note that

\begin{equation*}
    (\lambda-\mu)\inner{v}{w} = \inner{\lambda v}{w} - \inner{v}{\overline{\mu} w}
    = \inner{Tv}{w} - \inner{v}{T^* w} = 0.
\end{equation*}

Since \(\lambda-\mu\neq 0\) it follows that \(\inner{v}{w}=0\), proving Part~3.