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# Lecture 18: Oct. 27, 2023 ## Proposition (Properties of Adjoints) Let $A, B \in \mathcal{L}(V,V)$ * $(A+B)^* = A^* + B^*$ * $(\lambda A)^* = \overline{\lambda} A^*$ * $(AB)^* = B^*A^*$ * $(A^*)^* = A$ * $I^* = I$ ### Proof Let $u, v \in V$. Recall that $\langle A^*v, u \rangle = \langle v, Au \rangle$ 1. $\langle (A + B)^* v, u \rangle = \langle v, (A+B)u \rangle = \langle v, Au + Bu \rangle = \langle v, Au \rangle + \langle v, Bu \rangle = \langle A^*v, u \rangle + \langle B^*v, u \rangle = \langle A^*v + B^*v, u \rangle = \langle (A^* + B^*)v, u \rangle$ $\therefore (A+B)^* = A^* + B^*$ 2. $\langle (\lambda A)^* v, u \rangle = \langle v, (\lambda A)u \rangle = \langle v, \lambda Au \rangle = \overline{\lambda} \langle v, Au \rangle = \overline{\lambda} \langle A^* v, u \rangle = \langle \lambda A^* v, u \rangle$ $\therefore (\lambda A)^* = \overline{\lambda} A^*$ 3. $\langle (AB)^* v, u \rangle = \langle v, (AB)u \rangle = \langle v, A(Bu) \rangle = \langle A^*v, Bu \rangle = \langle B^*A^*v, u \rangle$ $\therefore (AB)^* = B^*A^*$ 4. $\langle (A^*)^* v, u \rangle = \langle v, A^* u \rangle = \overline{\langle A^* u, v \rangle} = \overline{\langle u, Av \rangle} = \langle Av, u \rangle$ $\therefore (A^*)^* = A$ 5. $\langle I^* v, u \rangle = \langle v, Iu \rangle = \langle v, u \rangle = \langle Iv, u \rangle$ $\therefore I^* = I$ ## Definition $A \in \mathcal{L}(V,V)$ is **self-adjoint** if $A^* = A$. ## Theorem Let $A \in \mathcal{L}(V,V)$. Then $A$ is self-adjoint if and only if $\langle Av, v \rangle \in \mathbb{R}$ for all $v \in V$. ### Proof ($\Rightarrow$) Suppose $A$ is self-adjoint. Then $A^* = A$. Consider $\langle Av, v \rangle = \langle v, A^* v \rangle = \langle v, Av \rangle = \overline{\langle Av, v \rangle}$. Thus, $\langle Av, v \rangle = \overline{\langle Av, v \rangle}$, which implies $\langle Av, v \rangle \in \mathbb{R}$. ($\Leftarrow$) Suppose $\langle Av, v \rangle \in \mathbb{R}$ for all $v \in V$. Then $\langle Av, v \rangle = \overline{\langle Av, v \rangle} = \langle v, Av \rangle = \langle A^*v, v \rangle$ So $\langle Av, v \rangle = \langle A^*v, v \rangle$ $\langle Av, v \rangle - \langle A^*v, v \rangle = 0$ $\langle (A - A^*)v, v \rangle = 0$ We want to show that $A - A^* = 0$, i.e., $A = A^*$. Note that for any $u, w \in V$, $\langle (A - A^*)u, w \rangle = 0$ $\langle Au, w \rangle = \langle A^*u, w \rangle = \langle u, Aw \rangle$ Consider $u = x + y$ where $x, y \in V$. $\langle A(x+y), x+y \rangle = \langle Ax, x \rangle + \langle Ax, y \rangle + \langle Ay, x \rangle + \langle Ay, y \rangle$ Since $\langle Ax, x \rangle, \langle Ay, y \rangle \in \mathbb{R}$ $\langle A(x+y), x+y \rangle \in \mathbb{R}$ Thus $\langle A(x+y), x+y \rangle = \langle A(x+y), x+y \rangle$ $\langle Ax, x \rangle + \langle Ax, y \rangle + \langle Ay, x \rangle + \langle Ay, y \rangle = \langle Ax, x \rangle + \langle Ay, x \rangle + \langle Ax, y \rangle + \langle Ay, y \rangle$ $\langle Ax, y \rangle + \langle Ay, x \rangle = \overline{\langle Ax, y \rangle} + \overline{\langle Ay, x \rangle}$ $\langle Ax, y \rangle + \langle Ay, x \rangle = \langle y, Ax \rangle + \langle x, Ay \rangle$ $\langle Ax, y \rangle + \langle Ay, x \rangle = \langle A^*y, x \rangle + \langle A^*x, y \rangle$ $\langle Ax, y \rangle - \langle A^*x, y \rangle = \langle A^*y, x \rangle - \langle Ay, x \rangle$ $\langle (A - A^*)x, y \rangle = \langle A^*y, x \rangle - \langle Ay, x \rangle = \langle x, Ay \rangle - \langle x, A^*y \rangle = \langle x, (A-A^*)y \rangle = - \langle (A - A^*)x, y \rangle$ Thus $\langle (A - A^*)x, y \rangle = - \langle (A - A^*)x, y \rangle$ $2 \langle (A - A^*)x, y \rangle = 0$ $\langle (A - A^*)x, y \rangle = 0$ Therefore $A - A^* = 0$ and $A = A^*$. ## Quick Note If $A$ is a self-adjoint operator, then all of its eigenvalues are real.

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