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6.5 Self-adjoint

f is self-adjoint \implies \chi_f(x) has all its roots in \mathbb{R}\implies f is diagonalizable...

Ex 1: Let \(A \in C^{n \times n}\) be a hermitian matrix. Prove that for each \(m \in \mathbb{N}\), there exists \(B \in C^{n \times n}\) s.t. \(A=B^m\) (\(\sqrt[m]{A}=B\)).

Answer: We know that \(A\) is diagonalizable and that each eigenvalue is in \(\mathbb{R}\).

\(A=PDP^{-1}\) where \(D\) is diagonal and \(P\) unitary \((P^{-1}=P^*)\).

Thus, if \(D = \begin{bmatrix} \lambda_1 \\ & \lambda_2 \\ & & \ddots \\ & & & \lambda_n \end{bmatrix}\) then define \(B = P \begin{bmatrix} \sqrt[m]{\lambda_1} \\ & \sqrt[m]{\lambda_2} \\ & & \ddots \\ & & & \sqrt[m]{\lambda_n} \end{bmatrix} P^{-1}\)

And \(B^m = P \begin{bmatrix} \sqrt[m]{\lambda_1} \\ & \sqrt[m]{\lambda_2} \\ & & \ddots \\ & & & \sqrt[m]{\lambda_n} \end{bmatrix}^m P^{-1} = PDP^{-1} = A\)

Def: \(f\) is normal if \(f f^* = f^* f\). A matrix, \(A\) is normal \(A A^* = A^* A\).

Ex 2: Consider \(\mathbb{C}^n\) with the standard inner product and let \(f:\mathbb{C}^n \to \mathbb{C}^n\) given by \(f(x_1, \dots, x_n) = (x_2, x_3, \dots, x_n, x_1)\)
Prove that \(f\) is normal and if \(p \in \mathbb{R}[x]\), then \(p(f)\) is normal
Ans:

  1. \(\langle f(x_1, \dots, x_n), (y_1, \dots, y_n) \rangle = \langle (x_2, x_3, \dots, x_n, x_1), (y_1, y_2, \dots, y_n) \rangle = x_2 \bar{y}_1 + x_3 \bar{y}_2 + \dots + x_n \bar{y}_{n-1} + x_1 \bar{y}_n\)
    ​\(= x_1 \bar{y}_n + x_2 \bar{y}_1 + \dots + x_n \bar{y}_{n-1}\) \(= \langle (x_1, \dots, x_n), (y_n, y_1, \dots, y_{n-1}) \rangle\) \(= \langle (x_1, \dots, x_n), f^*(y_1, \dots, y_n) \rangle\)
    where \(f^*(y_1, \dots, y_n) = (y_n, y_1, \dots, y_{n-1})\). (\(f^*\) is unique)
    ​\(f \circ f^*(x_1, \dots, x_n) = f(f^*(x_1, \dots, x_n)) = f(x_n, x_1, \dots, x_{n-1}) = (x_1, x_2, \dots, x_n) = id(x_1, \dots, x_n)\)
    ​\(f^* \circ f(x_1, \dots, x_n) = f^*(x_2, \dots, x_n, x_1) = (x_1, x_2, \dots, x_n) = id(x_1, \dots, x_n)\)
    ​\(\Rightarrow f\) is unitary.

  2. Let \(p \in \mathbb{R}[x]\), \(p(x) = \sum_{i=0}^{m}a_{i} x^{i} \quad m \in \mathbb{N}\) and \(p(f) = \sum_{i=0}^{m} a_i f^i,\) \(\begin{cases}(\alpha f + \beta g)^* = \bar{\alpha} f^* + \bar{\beta} g^* \\(f \cdot g)^* = g^* f^*\end{cases}\)
    Thus \(p(f)^{*} = \left(\sum_{i=0}^{m} a_{i} f^{i}\right)^{*} = \sum_{i=0}^{m} \bar{a_i} (f^{i})^{*}\) \(= \sum_{i=0}^{m} \bar{a_i}(f^{*})^{i}=\sum_{i=0}^{m}a_i(f^*)^i=p(f^*)\)
    ​\(\therefore p(f)^{*}=p\left(f^{*}\right)\)

    \(p(f)\cdot p(f)^{*}=p(f)\cdot p(f^{*})\overset{\text{although we cannot commute it directly, we can expand it and use f is normal to change the order}} {=}p(f^{*})\cdot p(f)=p(f)^{*}\cdot p(f)\)

Theorem: \(F = \mathbb{C}\)

Let \(f: V \to V\) be a linear operator. \(f\) is a normal operator iff there exists a basis \(B\) for \(V\) such that each vector in \(B\) is an eigenvector for \(f\).

If \(v\) and \(w\) are eigenvectors of eigenvalues \(\lambda_1\) and \(\lambda_2\) (\(\lambda_1 \neq \lambda_2\)) then \(v \perp w\).

In particular, \(A \in \mathbb{C}^{n \times n}\) is normal iff there exists a unitary matrix \(U \in \mathbb{C}^{n \times n}\) such that \(UAU^*\) is diagonal.

Ex 3: Let \(A=\begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix}\). Find a unitary matrix \(U \in C^{2 \times 2}\) s.t. \(U A U^*\) is diagonal.
\(A = \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix}, \quad A^* = \begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix}.\)
\(A A^* = \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix} \begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}.\) \(A^* A = \begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix} \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}.\)
\(\therefore A\) is normal.

Thus \(A\) is diagonalizable:
\(\chi_{A}(x) = \begin{vmatrix} x-1 & -i \\ -i & x-1 \end{vmatrix} = (x-1)^{2} - i^{2} = (x-1-i)(x-1+i)\) \(= (x-(1+i))(x-(1-i))\)

\(E_{1+i}= \text{span}\left\{ (1,1) \right\}\) and \(E_{1-i}= \text{span}\left\{ (1,-1) \right\}\), then \(B = \left\{ \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right), \left( \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right) \right\}\)

\(U = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}\), \(U^* = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}\), note that \(U\) is not unique since we can change the eigenvector by \(\lambda\) or order

\(\implies A = U \begin{bmatrix} 1+i & \\ & 1-i \end{bmatrix} U^*\)

Ex 4: If \(f:V \to V\) is normal, then \(\text{Im} f = \text{Im} f^*\). (\(\dim V = n\))
Remark: \(f\) is a normal op. \(v\) is an eigenvector for \(f\) relative to the eigenvalue \(\lambda\) iff \(v\) is an eigenvector for \(f^*\) relative to the eigenvalue \(\bar{\lambda}\).

Since \(f\) is normal, \(\exists B\) o.n. basis for \(V\) s.t. each vector \(v_i\) in \(B\) is an eigenvector for \(f\).

Let us put \(B = \{v_1, ..., v_n\}\).

\(Im f = \text{span}\{f(v_i) \mid v_i \in B \text{ and } i \in [n]\}\) \(= \text{span}\{\lambda_i v_i \mid v_i \in B \text{ and } i \in [n]\}\) \(= \text{span}\{v_i \mid \lambda_i \neq 0, i \in [n]\}\).

\(Im f^* = \text{span}\{f^*(v_i) \mid v_i \in B \text{ and } i \in [n]\}\) \(= \text{span}\{\bar{\lambda_i} v_i \mid v_i \in B \text{ and } i \in [n]\}\) \(= \text{span}\{v_i \mid \lambda_i \neq 0, i \in [n]\}\).

\(\therefore Im f = Im f^*\).

Ex 5. Let \(A \in C^{4 \times 4}\) be a hermitian matrix s.t. \((A^2+I)(A^2-3A+2I)=0\).
Prove that for each \(x \in C^4 \setminus \{0\}\), \(\lang Ax, x\rang >0\).
Ans: If A is hermitian, \(A^*=A\) and \(A\)​ is diagonalizable with each eigenvalue in \(\R\).

On the other hand, Using Ex 6, List #7.
\(f\) is self-adjoint iff \(\lang f(v), v\rang\) is real for every \(v \in V\).
Since \(C^{4} = \bigoplus_{\lambda i}E_{\lambda i}\), \(v = \sum v_{i}\), \(v_{i} \in E_{\lambda i}\), then \(\lang Av, v\rang = \lang A(\sum v_{i}), \sum v_{i}\rang = \sum_{i,j}\lang Av_{i}, v_{j}\rang = \sum_{i,j}\lang \lambda _{i} v_{i}, v_{j}\rang\)

\(=\sum_{i}\lambda_{i}\|v_{i}\|_{\text{>0}}^{2}\quad\text{but}\quad m_{A}(x)|(x^{2} +1)(x^{2}-3x+2)\quad\text{and}\) \((x^2+1)(x^2-3x+2) = (x-i)(x+i)(x-2)(x-1)\)
So the roots of \(m_{A}\text{ are in }\{1,2\}.\) Thus \(\langle Av,v\rangle>0\)

Ex 6 = (Ex 6, List #7) \(V\) is a \(f.d.i.p.s\). \(f\) a \(lop\). Prove \(f\) is self-adjoint iff \(f\) \(\langle f(v), v \rangle\) is real \(\forall v \in V\).

\(\Rightarrow\)) \(f\) self-adjoint \(\Rightarrow f=f^*\) and \(\forall v \in V\), \(\langle f(v), v \rangle = \langle v, f^*(v) \rangle\) \(= \langle v, f(v) \rangle\) \(=\overline{\langle f(v), v \rangle}\)

\(\therefore \langle f(v), v \rangle\) is a real number.

\(\Leftarrow\)) \(\langle f(v),v\rangle\in\mathbb{R}\quad\forall v\in V.\) \(\text{w.p. } f^* = f \equiv \langle f(u), v \rangle = \langle u, f(v) \rangle \quad \forall u, v \in V.\)
​\(\text{Consider } \langle f(u+v), u+v \rangle = \underline{\langle f(u), u \rangle}_{\in \mathbb{R}} + \langle f(u), v \rangle + \langle f(v), u \rangle + \underline{\langle f(v), v \rangle}_{\in \mathbb{R}} \in \mathbb{R}\)
​\(\therefore \text{Im}(\langle f(u), v \rangle + \langle f(v), u \rangle) = 0\) \(\equiv\) \(\text{Im}(\langle f(u), v \rangle) = -\text{Im}(\langle f(v), u \rangle)\)
​\(\mathbb{R}\ni\langle f(u+iv),u+iv\rangle=\underline{\langle f(u),u\rangle}_{\in\mathbb{R}} -i\langle f(u),v\rangle+i\langle f(v),u\rangle+\underline{||i||^2}_{\in\mathbb{R}} \underline{\langle f(v),v\rangle}_{\in\mathbb{R}}.\in\mathbb{R}.\)