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# 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.

### Contributors

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