Normal Linear Operators

We are now going to look at another important type of linear operator known as Normal linear operators which we define below.

Definition: Let $V$ be a finite-dimensional inner product space. Then a linear operator $T \in \mathcal L(V)$ is said to be Normal if $TT^* = T^*T$.

Linear operators that are normal are not difficult to come by. In fact, if $T$ is self-adjoint, then $T$ is a normal linear operator since $TT^* = T^*T$. Of course, a normal operator need not be self-adjoint.

The following proposition will give us an important characteristic of normal linear operators.

Proposition 1: Let $V$ be a finite-dimensional nonzero inner product space. Then a linear operator $T \in \mathcal L(V)$ is normal if and only if for all $v \in V$ we have that $\| T(v) \| = \| T^*(v) \|$.

• Proof: $\Rightarrow$ Suppose that $T$ is normal. Then $TT^* = T^*T$, and so $TT^* - T^*T = 0$. From the Self-Adjoint Linear Operators over Complex Vector Spaces page we then have that $<(TT^* - T^*T)(v), v> = 0$ for all $v \in V$, and so for all $v \in V$ we have that:

(1)

\begin{align} <TT^*(v), v> - <T^*T(v), v> = 0 \\ <T^*T(v), v> = <TT^*(v), v> \\ \| T(v) \|^2 = \| T^*(v) \|^2 \end{align}

• $\Leftarrow$ Suppose that $\| T(v) \|^2 = \| T^*(v) \|^2$ for all $v \in V$. Then we have that for all $v \in V$:

(2)

\begin{align} \quad <TT^*(v), v> = <T^*T(v), v> \\ \quad <TT^*(v), v> - <T^*T(v), v> = 0 \\ \quad <(TT^* - T^*T)(v), v> = 0 \end{align}

• Therefore $TT^* - T^*T = 0$ and $TT^* = T^*T$, so $T$ is normal. $\blacksquare$

Proposition 2: Let $V$ be a finite-dimensional nonzero inner product space and let $T \in \mathcal L(V)$. If $T$ is normal then $T - \lambda I$ is normal.

• Proof: We begin by computing $(T - \lambda I)(T - \lambda I)^*(v)$:

(3)

\begin{align} \quad (T - \lambda I)(T - \lambda I)^*(v) = (T - \lambda I)(T^* - \overline{\lambda}I)(v) = (TT^* - \overline{\lambda} T - \lambda T^* + \lambda \overline{\lambda} I)(v) \end{align}

• Now we compute $(T - \lambda I)^*(T - \lambda I )(V)$:

(4)

\begin{align} \quad (T - \lambda I)^*(T - \lambda I)(v) = (T^* - \overline{\lambda} I)(T - \lambda I)(v) = (T^*T - \lambda T^* - \overline{\lambda} T + \overline{\lambda} \lambda I)(v) \end{align}

• Since $TT^* = T^*T$ (since $T$ is normal) then we see that indeed $(T - \lambda I)(T - \lambda I)^* = (T - \lambda I)^*(T - \lambda I)$ so $T - \lambda I$ is normal. $\blacksquare$

Proposition 3: Let $V$ be a finite-dimensional nonzero inner product space and let $T \in \mathcal L(V)$ be normal. Then if $v$ is an eigenvector corresponding to the eigenvalue $\lambda$ of $T$, then $v$ is also an eigenvector of corresponding to the eigenvalue $\overline{\lambda}$ of $T^*$.

• Proof: Let $v \in V$ be an eigenvector of $T$ corresponding to the eigenvalue $\lambda$. Then we have that $T(v) = \lambda v$. Now since $T$ is normal, we have that $T - \lambda I$ is normal from Proposition 2 above. Therefore, we apply Proposition 1 to get that:

(5)

\begin{align} \quad 0 = \| (T - \lambda I)(v) \| = \| (T - \lambda I)^*(v) \| = \| (T^* - \overline{\lambda}I)(v) \| \end{align}

• Therefore $\| (T^* - \overline{\lambda}I)(v) \| = 0$ implies that $(T^* - \overline{\lambda}I)(v) = 0$ so $T^*(v) = \overline{\lambda} v$. Thus $v$ is an eigenvector of corresponding to the eigenvalue of $\overline{\lambda}$ of $T^*$. $\blacksquare$

Corollary 1: Let $V$ be a finite-dimensional nonzero inner product space and let $T \in \mathcal L(V)$ be normal. Then the eigenvectors of $T$ that correspond to distinct eigenvalues are orthogonal to one another.

• Proof: Let $T$ be a normal linear operator and let $\lambda$ and $\mu$ be distinct eigenvalues. Then $\lambda \neq \mu$. Let $u$ be an eigenvalue corresponding to $\lambda$ and let $v$ be an eigenvalue corresponding to $\mu$. Therefore $T(u) = \lambda u$ and $T(v) = \mu b$. Now consider the following product:

(6)

\begin{align} \quad (\lambda - \mu)<u, v> = <\lambda u, v> - <u, \overline{\mu}v> = <T(u), v> - <u, T^*(v)> = 0 \end{align}

• Since $(\lambda - \mu) \neq 0$ the product above implies that $<u, v> = 0$ and so the corresponding eigenvectors to the distinct eigenvalues $\lambda$ and $\mu$ are orthogonal. $\blacksquare$