# 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 $ = 0$ for all $v \in V$, and so for all $v \in V$ we have that:

(1)

\begin{align}

`-` = 0 \\ = `\\ \| 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

`=` \\ \quad `-` = 0 \\ \quad = 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)

__= –____=__ – __= 0 \end{align}__

- Since $(\lambda – \mu) \neq 0$ the product above implies that $
__= 0$__and so the corresponding eigenvectors to the distinct eigenvalues $\lambda$ and $\mu$ are orthogonal. $\blacksquare$