A Motivation of Normal Operators on Inner Product Spaces

A Motivation of Normal Operators

Let $A$ be a $n\times n$ matrix. We've already known that

$$A\text{ is symmetric}\Leftrightarrow A\text{ is diagonally diagonalizable}$$

Proof: (Theorem 6.17)

"$\Leftarrow$" is easy.

There are four parts for proving "$\Rightarrow$".

Part I: We show that the eigenvalues of $A$ are real. Suppose that $A\mathbf{v}=\lambda v$. Then

$$\lambda \langle \mathbf{v}, \mathbf{v}\rangle =\langle \lambda \mathbf{v}, \mathbf{v}\rangle =\langle A\mathbf{v}, \mathbf{v}\rangle =\langle \mathbf{v}, A^t\mathbf{v}\rangle =\langle \mathbf{v}, A\mathbf{v}\rangle =\langle \mathbf{v}, \lambda\mathbf{v}\rangle =\overline{\lambda}\langle \mathbf{v}, \mathbf{v}\rangle$$ This follows that $\lambda=\overline{\lambda}$ since $\mathbf{v}\neq \mathbf{0}$.

Part II: We show that the characteristic polynomial of $A$ splits. Let $f(x)$ be the characteristic polynomial of $A$. By the Fundamental Theorem of Algebra, $f(x)$ splits over $\mathbb{C}$. Since all eigenvalues of $A$ are real, $f(x)$ splits over $\mathbb{R}$.

Part III: Schur's Theorem. See Theorem 6.4.3 in Leon's Linear Algebra.

Part IV: Finally, we show that $A$ is diagonally diagonalizable. By Schur's Theorem, there exists an orthonormal basis for $\mathbb{R}^n$ such that $A=PUP^t$, where $P$'s columns consists of those orthonormal eigenvectors of $A$ and $U$ is an upper triangular matrix. Then

$$PU^t P^t=A^t=A=PUP^t.$$ This follows that $U=U^t$ and $U$ is a diagonal matrix.

A natural conjecture is

$$A\text{ is self-adjoint}\Leftrightarrow A\text{ is unitarily diagonalizable}$$

The proof of "$\Rightarrow$" is similar to the previous one. However, it is easy to find an counterexample for "$\Leftarrow$".

Let's recall the proof of "$A$ is symmetric $\Leftarrow$ $A$ is orthogonally diagonalizable" and examine it why the same method can't apply here.

When we were proving "$A$ is symmetric $\Leftarrow$ $A$ is orthogonally diagonalizable", we had

$$A^t=(PDP^t)^t=PD^t P=PDP^t=A.$$ At here, we only have $$A^*=(PDP^*)^*=PD^* P^*.$$ The proof failed because $D\neq D^*$.

However, above discussion gives us a theorem. (Corollary 3 of Theorem 6.25.)

$$A\text{ is self-adjoint}\Leftrightarrow A\text{ is unitarily diagonalizable and all eigenvalues are real}$$

However, this theorem doesn't give us any equivalent property of being unitarily diagonalizable.

Suppose that $A=PDP^*$. The key observation is that

$$\begin{array}{lll}AA^* &=& (PDP^*)(PDP^*)^* \\ &=& PDP^* PD^* P \\ &=& PDD^* P^* \\ &=& PD^* DP^* \\ &=& PD^* P^* PDP^* \\ &=& A^* A. \end{array}$$

This motivates us to define a matrix $A$ is normal if $AA^*=A^* A$.

Now, we prove that

$$A\text{ is normal}\Leftrightarrow A\text{ is unitarily diagonalizable}$$

Proof: (Theorem 6.16)

We have already proved "$\Leftarrow$".

For "$\Rightarrow$", we need a lemma: If $A$ is normal and $A\mathbf{v}=\lambda \mathbf{v}$, then $A^*\mathbf{v}=\overline{\lambda}\mathbf{v}$. See Theorem 6.15(c).

Now, back to our main theorem. Since we are over $\mathbb{C}$, by Schur's Theorem, there exists an orthonormal basis $\{\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n\}$ for $\mathbb{C}^n$ form a unitary matrix $P$ such that $A=PUP^*$. That is, $P=[\mathbf{v}_1|\mathbf{v}_2|\cdots|\mathbf{v}_n]$.

Note that $A\mathbf{v}_1=U_{11}\mathbf{v}_1$. That is, $\mathbf{v}_1$ is an eigenvector of $A$. We use induction. Suppose that $\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_{i-1}$ are eigenvectors of $A$. Then since $\{\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n\}$ is an orthonormal basis, we have

$$\begin{array}{cccccccccccc} A\mathbf{v}_i &=& U_{1i} & \mathbf{v}_1 & + & U_{2i} & \mathbf{v}_2 & + & \cdots & + & U_{ii} & \mathbf{v}_i \\ &=& \langle A\mathbf{v}_i, \mathbf{v}_1\rangle & \mathbf{v}_1 & + & \langle A\mathbf{v}_i, \mathbf{v}_2\rangle & \mathbf{v}_2 & + & \cdots & + & \langle A\mathbf{v}_i, \mathbf{v}_i\rangle & \mathbf{v}_i \\ &=& \langle \mathbf{v}_i, A^*\mathbf{v}_1\rangle & \mathbf{v}_1 & + & \langle \mathbf{v}_i, A^*\mathbf{v}_2\rangle & \mathbf{v}_2 & + & \cdots & + & \langle \mathbf{v}_i, A^*\mathbf{v}_i\rangle & \mathbf{v}_i \\ &=& \langle \mathbf{v}_i, \overline{\lambda}_1\mathbf{v}_1\rangle & \mathbf{v}_1 & + & \langle \mathbf{v}_i, \overline{\lambda}_2\mathbf{v}_2\rangle & \mathbf{v}_2 & + & \cdots & + & \langle \mathbf{v}_i, A^*\mathbf{v}_i\rangle & \mathbf{v}_i \\ &=& \lambda_1\langle \mathbf{v}_i, \mathbf{v}_1\rangle & \mathbf{v}_1 & + & \lambda_2\langle \mathbf{v}_i, \mathbf{v}_2\rangle & \mathbf{v}_2 & + & \cdots & + & \langle \mathbf{v}_i, A^*\mathbf{v}_i\rangle & \mathbf{v}_i \\ &=& \mathbf{0} & & + & \mathbf{0} & & + & \cdots & + & \langle \mathbf{v}_i, A^*\mathbf{v}_i\rangle & \mathbf{v}_i \end{array}.$$ Thus, $\mathbf{v}_i$ is also an eigenvector of $A$.