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Mathematics LibreTexts

11.2 Normal operators

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.