From Confusion to Clarity: Understanding Normal Operators Through Gelfand Duality

When I was learning Linear Algebra, I felt really confused by the definition of normal operators.

Let $V$$V$ be a vector space, with an inner product $⟨\cdot ,\cdot ⟩$$\langle \cdot, \cdot \rangle$. We say $T:V\to V$$T: V \to V$ is normal if $T{T}^{\ast }=T^{\ast }T$$TT^* = T^*T$.

Why do we define it this way? What is the deep structure underlying it?

After I learned some abstract algebra, I know this tells us that the smallest algebra containing $T$$T$ is a commutative ring.

But I still do not get the point. Lots of proofs in linear algebra around normal operators are long and dirty. I do not like that kind of math.

Another story is, when I was a first year student, my tutor Thomas suggested me to attend Eva's talk. He told me that I would find what I wanted since I had observed a kind of Algebra-Geometry Duality in operator theory. That talk was about Gelfand Duality.

Gelfand Duality

We only need to use the really basic idea of Gelfand Duality in this essay.

Definition. Let $F$$F$ be a field and $A$$A$ be a finite dimensional $F$$F$ algebra. We say $A$$A$ is a diagonalizable algebra if $A\cong F^n$$A \cong F^n$.

Lemma. Let $T:V\to V$$T: V \to V$ where $V$$V$ is a finite dimensional vector space. Then $T$$T$ is diagonalizable iff $\mathbb{C}[T]$$\mathbb{C}[T]$ is diagonalizable.

Proof. The minimal polynomial of a $T$$T$ is separable. $◻$$\square$

Lemma. Let $T:V\to V$$T: V \to V$ where $V$$V$ is a finite dimensional vector space over $F$$F$. Then the minimal polynomial of $T$$T$ exists.

Proof. Define $e{v}_T:F[x]\to {\mathrm{End}}_F(V)$$ev_T:F[x]\to\mathrm{End}_F(V)$ to be the evaluation map at $T$$T$. If it is injection then ${\mathrm{End}}_F(V)$$\mathrm{End}_F(V)$ contain $F[x]$$F[x]$ as an infinite dimensional vector space, contradiction. The monic generator of the kernel of $e{v}_T$$ev_T$ is the minimal polynomial. $◻$$\square$

Proposition. Let $T:V\to V$$T: V \to V$ where $V$$V$ is a finite dimensional inner product space over $\mathbb{C}$$\mathbb C$. Then $T$$T$ is diagonalizable if $T$$T$ is normal.

Remark. Huh, the minimal polynomial of a normal operator is separable. Galois! lol

Proof. We only need to prove that $\mathbb{C}[T]$$\mathbb{C}[T]$ has no nilpotent elements. Since $T$$T$ is normal, we know that the smallest $C^{\ast }$$C^*$ algebra containing $T$$T$ is $\mathbb{C}[T,T^{\ast }]$$\mathbb{C}[T,T^*]$, which is a commutative algebra. By Gelfand duality, $\mathbb{C}[T,T^{\ast }]$$\mathbb{C}[T,T^*]$ is isomorphic to $C(X)$$C(X)$ for a compact Hausdorff space $X$$X$. Since $C(X)$$C(X)$ is a continuous function algebra, it has no nilpotent elements, hence $\mathbb{C}[T]$$\mathbb{C}[T]$ as a subring of $C(X)$$C(X)$ has no nilpotent elements.

Thus

$$\begin{array}{c}\mathbb{C}[T]\cong \mathbb{C}[x]/(m_T(x))\cong \prod _{i=1}^k\mathbb{C}[x]/(x-{\lambda }_i)\cong {\mathbb{C}}^n\end{array}$$

By the above lemma, $T$$T$ is diagonalizable. $◻$$\square$

Then using the fact that $T$$T$ is normal iff $\mathrm{\forall }v\in V,\parallel Tv\parallel =\parallel T^{\ast }v\parallel$$\forall v\in V,\|Tv\|=\|T^*v\|$ hence $\parallel (T-\lambda )v\parallel =\parallel (T^{\ast }-\bar{\lambda })v\parallel =0$$\|(T-\lambda)v\|=\|(T^*-\overline\lambda)v\| =0$ if $v$$v$ is a eigenvector of $T$$T$.

Then let $\lambda ,\mu$$\lambda,\mu$ be two distinct eigenvectors of $T$$T$,

$$(\lambda -\mu )⟨v,w⟩=⟨\lambda v,w⟩-⟨v,\bar{\mu }w⟩=⟨Tv,w⟩-⟨v,T^{\ast }w⟩=0⟹\lambda =\mu$$

We could choice orthogonal eigenvalue to diagnolize $T$$T$.

Then using the fact that $T$$T$ is normal iff $T{T}^{\ast }=T^{\ast }T$$TT^* = T^*T$, we can show that if $v$$v$ is an eigenvector of $T$$T$, then it has special orthogonality properties.

Let $v_1,v_2$$v_1, v_2$ be two eigenvectors of $T$$T$ corresponding to distinct eigenvalues ${\lambda }_1\ne {\lambda }_2$$\lambda_1 \neq \lambda_2$. Then:

$$⟨T{v}_1,v_2⟩=⟨v_1,T^{\ast }{v}_2⟩,{\lambda }_1⟨v_1,v_2⟩=\overline{{\lambda }_2}⟨v_1,v_2⟩$$

Since ${\lambda }_1\ne \overline{{\lambda }_2}$$\lambda_1 \neq \overline{\lambda_2}$, we have $⟨v_1,v_2⟩=0$$\langle v_1, v_2 \rangle = 0$.

Therefore, we can choose orthogonal eigenvectors to diagonalize $T$$T$ unitarily.

Remark. Let $T$$T$ be $\mathbb{C}$$\mathbb C$ inner product space, then $T$$T$ is normal iff the smallest $C^{\ast }$$C^*$ algebra containing $T$$T$ in ${\mathrm{End}}_{\mathbb{C}}(V)$$\mathrm{End_\mathbb C}(V)$ is a commutative $C^{\ast }$$C^*$ algebra.

Remark. Let $A$$A$ be any $C^{\ast }$$C^*$ algebra, and $T\in A$$T\in A$. If $T{T}^{\ast }=T^{\ast }T$$TT^*=T^*T$, then $T$$T$ is not nillpotent.

Remark. For the real case, $T$$T$ is normal implies $\mathbb{R}[T]$$\mathbb R[T]$ is etale, $T$$T$ is self-adjoint implies $\mathbb{R}[T]$$\mathbb R[T]$ is diagonalizable.