# 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 \(A\).

## No comments:

## Post a Comment