Normal Operator
Definition
A Hilbert space $H$ and a bounded linear operator $T : H \to H$ are called a normal operator if they satisfy the following.
$$ T^{\ast}T = TT^{\ast} $$
$T^{\ast}$ is the adjoint operator of $T$.
Explanation
By the definition, if an operator is self-adjoint $(T^{\ast}=T)$ then it is normal, and if it is unitary $(T^{\ast}=T^{-1})$ then it is normal. The converses do not hold in general.
In finite dimensions, i.e., in the matrix picture, $T^{\ast}$ is the conjugate transpose matrix of $T$. In other words, a normal operator is a generalization of a normal matrix.
Properties
(a) Let two normal operators $S$ and $T$ satisfy $ST^{\ast} = T^{\ast}S$ and $TS^{\ast} = S^{\ast}T$. Then their sum and product are also normal operators. $$ \text{$S+T$ and $ST$ are normal} $$
(b) For a complex vector space $H$ and a bounded linear operator $T : H \to H$, the operator $T$ is normal if and only if $\| T^{\ast}x \| = \| Tx \|$ $\forall x \in H$.
(c) For a normal $T$, $\| T^{2} \| = \| T \|^{2}$
Proof
(a)
To show that $S + T$ is normal, it suffices to show that $(S + T)^{\ast}(S + T) = (S + T)^{\ast}(S + T)$.
$$ \begin{align*} (S + T)^{\ast}(S + T) &= (S^{\ast} + T^{\ast})(S + T) \\ &= S^{\ast}S + T^{\ast}S + S^{\ast}T + T^{\ast}T \\ &= SS^{\ast} + ST^{\ast} + TS^{\ast} + TT^{\ast} \\ &= S(S^{\ast} + T^{\ast}) + T(S^{\ast} + T^{\ast}) \\ &= (S + T)(S^{\ast} + T^{\ast}) \end{align*} $$
To show that $ST$ is normal, it suffices to show that $(ST)^{\ast}(ST) = (ST)^{\ast}(ST)$.
$$ \begin{align*} (ST)^{\ast}(ST) &= T^{\ast}S^{\ast} ST = T^{\ast}(S^{\ast}S)T \\ &= T^{\ast}S S^{\ast}T = (T^{\ast}S) (S^{\ast}T) \\ &= (ST^{\ast}) (TS^{\ast}) = S(T^{\ast}T)S^{\ast} \\ &= ST T^{\ast}S^{\ast} \\ &= ST(ST)^{\ast} \end{align*} $$
■
(b)
Assume $(\implies)$ $T$ is a normal operator. Then, by the definitions of the adjoint and a normal operator, we have
$$ \begin{align*} \| T^{\ast}x \| = \braket{T^{\ast}x, T^{\ast}x} &= \braket{x, TT^{\ast} x} \\ &= \braket{x, T^{\ast}T x} \\ &= \braket{Tx, Tx} = \| Tx \| \end{align*} \qquad \forall x \in H $$
■
Now assume $(\impliedby)$ $\| T^{\ast}x \| = \| Tx \|$ $\forall x \in H$. Then we obtain
$$ \begin{align*} && \braket{T^{\ast}x, T^{\ast}x} &= \braket{Tx, Tx} \\ \implies && \braket{TT^{\ast}x, x} &= \braket{T^{\ast}Tx, x} \\ \implies && \braket{(TT^{\ast} - T^{\ast}T)x, x} &= 0 \end{align*} \qquad \forall x \in H $$
For a complex vector space $X$, if $Q : X \to X$ satisfies $\braket{Qx, x} = 0$ for all $x \in X$, then $Q = 0_{\text{op}}$.
By the property of the zero operator, the following holds.
$$ TT^{\ast} - T^{\ast}T = 0 \implies TT^{\ast} = T^{\ast}T $$
■
(c)
By the properties of the adjoint operator we have $\| T^{\ast} T \| = \| T T^{\ast} \| = \| T \|^{2}$. Therefore it suffices to show $\| TT^{\ast} \| = \| TT \|$. By (b), $\forall (Tx) \in H$ $\| T^{\ast}(T x) \| = \| T(Tx) \|$ hold. Hence we obtain
$$ \begin{align*} \| T^{\ast}T x \| &= \| TTx \| \\ &= \sup\limits_{\| y \| = 1} | \braket{TTx, y} | \\ &\le \sup\limits_{\| y \| = 1} \| TTx \| \| y \| \\ &= \| TTx \| \end{align*} $$
The second equality follows from the properties of inner product and norm, and the third inequality follows from the Cauchy–Schwarz inequality. By the definition of the operator norm $(\| T T \|)$ we obtain
$$ \| T^{\ast}T x \| \le \| TTx \| \le \| TT \| \| x \| $$
Again by the definition of the operator norm $(\| T^{\ast} T \|)$ we obtain
$$ \| T^{\ast} T \| \le \| TT \| $$
In a similar way we obtain the reverse inequality as follows.
$$ \begin{align*} \| TTx \| &= \| T^{\ast} Tx \| \\ &= \sup\limits_{\| y \| = 1} | \braket{T^{\ast}Tx, y} | \\ &\le \sup\limits_{\| y \| = 1} \| T^{\ast}Tx \| \| y \| \\ &= \| T^{\ast}Tx \| \end{align*} $$
$$ \implies \| TTx \| \le \| T^{\ast}Tx \| \le \| T^{\ast}T \| \| x \| $$
$$ \| TT \| \le \| T^{\ast}T \| $$
Hence the following holds.
$$ \| T^{\ast}T \| \le \| TT \| \le \| T^{\ast}T \| \implies \| T^{\ast}T \| = \| TT \| $$
■