> [!definition] > > Let $\ch$ be a [[Hilbert Space|Hilbert space]]. A [[Bounded Linear Operator|bounded linear operator]] $T \in L(\ch, \ch)$ is self-adjoint if $T = T^*$. > [!theorem] > > Let $\cm \subset \ch$ be a closed $T$-invariant subspace, then $\cm^\perp$ is also a closed $T$-invariant subspace. > > *Proof*. Let $x \in \cm$ and $y \in \cm^\perp$, then > $ > \angles{x, Ty} = \angles{Tx, y} = 0 > $ > [!theorem] > > Let $\lambda \in \complex$, then > $ > \ol{(T - \lambda I)(\ch)} = \ker{(T^* - \lambda I)^\perp} > $ > [!theorem] > > Let $\lambda = a + bi \in \complex$, then > $ > \norm{(A - \lambda I)x}^2 \ge b^2 \norm{x}^2 > $ > Thus if $b \ne 0$, then > 1. $(A - \lambda I)$ is injective. > 2. $(A - \lambda I)(\ch)$ is closed (operator is "open" and satisfies the [[Method of Successive Approximations|method of successive approximations]]). > 3. $\lambda$ is not in the continuous spectrum. > > *Proof*. > $ > \begin{align*} > \norm{(T - \lambda I)x}^2 &= \angles{(T - \lambda I)x, (T - \lambda I)x} \\ > &= \angles{(T - (a + bi) I)x, (T - (a + bi)I)x} \\ > &= \norm{(T - a I)x}^2 + b^2\norm{x}^2 + i\angles{(A - a)x, bx} - i\angles{bx, (A - a)x} \\ > &= \norm{(T - aI)x}^2 + b^2\norm{x}^2 > \end{align*} > $ > If $b \ne 0$, then $\norm{(A - \lambda I)x}^2 \ge b^2\norm{x}^2$. > [!theorem] > > $ > \sigma(T) \subset [-\norm{T}, \norm{T}] > $ > > *Proof*. Let $\lambda \in \sigma(T)$, then either $\lambda$ is in the [[Resolvent|resolvent]] or it is in the residual spectrum. Suppose that $(T - \lambda I)(\ch) \subsetneq \ch$, then $\ol \lambda$ is in the pure point spectrum because $T$ is self-adjoint. This implies that $\lambda \in \real$ and $\lambda \in \sigma_p(T) \subset [-\norm{T}, \norm{T}]$.