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# Lecture 24: The Spectral Theorem Let $V$ be a vector space over $\mathbb{F}$, where $\mathbb{F} = \mathbb{R}$ or $\mathbb{C}$. ## Definition Let $T \in \mathcal{L}(V)$. We say that $T$ is **self-adjoint** if $T = T^*$. **Note:** If $\mathbb{F} = \mathbb{R}$, then $T$ is called **symmetric**....

# Lecture 24: The Spectral Theorem Let $V$ be a vector space over $\mathbb{F}$, where $\mathbb{F} = \mathbb{R}$ or $\mathbb{C}$. ## Definition Let $T \in \mathcal{L}(V)$. We say that $T$ is **self-adjoint** if $T = T^*$. **Note:** If $\mathbb{F} = \mathbb{R}$, then $T$ is called **symmetric**. **Example:** Let $A = \begin{bmatrix} 2 & i \\ -i & 5 \end{bmatrix}$. Then $A^* = \begin{bmatrix} 2 & i \\ -i & 5 \end{bmatrix}$, so $A = A^*$. **Example:** Let $A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$. Then $A^* = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$, so $A = A^*$. **Example:** Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$. Then $A^* = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$, so $A \neq A^*$. ## Theorem If $T$ is self-adjoint, then all eigenvalues of $T$ are real. **Proof:** Let $\lambda$ be an eigenvalue of $T$, and let $v$ be an eigenvector corresponding to $\lambda$. Then: $\qquad T(v) = \lambda v$, where $v \neq 0$. $\qquad \langle T(v), v \rangle = \langle \lambda v, v \rangle = \lambda \langle v, v \rangle$ Since $T = T^*$, we have $\qquad \langle T(v), v \rangle = \langle v, T^*(v) \rangle = \langle v, T(v) \rangle = \langle v, \lambda v \rangle = \overline{\lambda} \langle v, v \rangle$. So $\lambda \langle v, v \rangle = \overline{\lambda} \langle v, v \rangle$, which implies $(\lambda - \overline{\lambda}) \langle v, v \rangle = 0$. Since $v \neq 0$, $\langle v, v \rangle \neq 0$. Thus, $\lambda - \overline{\lambda} = 0$, so $\lambda = \overline{\lambda}$, meaning $\lambda \in \mathbb{R}$. ## Theorem Let $T \in \mathcal{L}(V)$ be self-adjoint. Then $N(T) = R(T)^{\perp}$. **Proof:** We want to show that $w \in N(T)$ if and only if $w \in R(T)^{\perp}$. ($\subseteq$) Let $w \in N(T)$. Then $T(w) = 0$. We want to show that $w \in R(T)^{\perp}$. That is, $\langle w, v \rangle = 0$ for all $v \in R(T)$. Let $v \in R(T)$. Then $v = T(u)$ for some $u \in V$. We have $\qquad \langle w, v \rangle = \langle w, T(u) \rangle = \langle T^*(w), u \rangle = \langle T(w), u \rangle = \langle 0, u \rangle = 0$. So $w \in R(T)^{\perp}$. ($\supseteq$) Let $w \in R(T)^{\perp}$. Then $\langle w, v \rangle = 0$ for all $v \in R(T)$. Since $\langle w, v \rangle = 0$ for all $v \in R(T)$, it follows that $\langle w, T(u) \rangle = 0$ for all $u \in V$. Since $T = T^*$, we have $\langle w, T(u) \rangle = \langle T^*(w), u \rangle = \langle T(w), u \rangle = 0$ for all $u \in V$. So $T(w) = 0$, because the only vector orthogonal to everything is 0. Thus $w \in N(T)$. ## Corollary If $T \in \mathcal{L}(V)$ is self-adjoint, then $V = N(T) \oplus R(T)$. **Proof:** Since $N(T) = R(T)^{\perp}$, we have $N(T) \oplus R(T) = R(T)^{\perp} \oplus R(T) = V$. ## Theorem If $T \in \mathcal{L}(V)$ is self-adjoint, then there exists an orthonormal basis for $V$ consisting of eigenvectors of $T$. **Note:** This is known as the **Spectral Theorem**.