Spectral Theorem

Spectral Theorem

Theorem 1. Let $T \in \mathcal{L}(V)$ where $V$ is a finite-dimensional inner product space over $F$ with the distinct eigenvalues $\lambda_1, \cdots, \lambda_k$. Assume that $T$ is normal if $F = \mathbb{C}$ and that $T$ is hermitian if $F = \mathbb{R}$. For each $i (1 \leq i \leq k)$, let $W_i$ be the eigenspace of $T$ corresponding to the eigenvalue $\lambda_i$, and that $T_i$ be the orthogonal projection of $V$ on $W_i$. Then the following statements are true.
(a) $V = \bigoplus_{i=1}^k W_i$.
(b) If we denote $W'_i = \bigoplus_{j \neq i} W'_j$, then $W^{\perp}_i = W'_i$.
(c) $T_iT_j = \delta_{ij}T_i$ for $1 \leq i, j \leq k$.
(d) $I = T_1 + \cdots + T_k$.
(e) $T = \lambda_1 T_1 + \cdots + \lambda_k T_k$.

Proof. Let $x$ denote $x = x_1 + \cdots + x_k$ for some $x_i \in W_i (1 \leq i \leq k)$.
(a) By Theorem 2 and Theorem 1, $T$ is diagonalizable. Then by Theorem 3 and Theorem 1, $T = \bigoplus_{i=1}^k W_i$.
(b) Since each $T_i$ is the orthogonal projection of $V$, $W_i = R(T)^{\perp} = N(T) = \bigoplus_{j \neq i} W_i$ by Theorem 1. 
(c) Note that $T_i T_j(x) = T_i (x_j) = \delta_{ij} x_i = \delta_{ij}T_i(x)$. Thus $T_iT_j = \delta_{ij}T_i$.
(d) Note that $\forall x \in V$, $(T_1 + \cdots + T_k)(x) = T_1(x) + \cdots + T_k(x) = x_1 + \cdots + x_k = x = I(x)$. Thus $I = T_1 + \cdots + T_k$.
(e) Note that $T(x)$ $= T(x_1) + \cdots + T(x_k)$ $= \lambda_1 x_1 + \cdots$ $+ \lambda_k$ $x_k = \lambda_1 T_1(x)$$+ \cdots + \lambda_k T_k(x) =$ $(\lambda_1 T_1 + \cdots + \lambda_k T_k)(x)$. Thus $T = \lambda_1 T_1 + \cdots + \lambda_k T_k$. $\blacksquare$

Definition 1

Definition 1. Using the notation of above theorem,
(a) the set $\{\lambda_1, ..., \lambda_k \}$ of eigenvalues of $T$ is called the spectrum of $T$.
(b) the sum $I = T_1 + \cdots + T_k$ is called the resolution of the identity operator induced by $T$.
(c) the sum $T = \lambda_1 T_1 + \cdots + \lambda_k T_k$ is called the spectral decomposition of $T$.

Remark

Remark. If $T = \lambda_1 T_1 + \cdots + \lambda_k T_k$ is the spectral decomposition of $T$, then $g(T) = g(\lambda_1)T_1 + \cdots + g(\lambda_k)T_k, \forall g \in P(F)$. 
$((\because) \,\, g(T)(x) = a_nT^n(x) + \cdots + a_1T(x) + a_0 I(x) \\ = a_n(T^n(x_1) + \cdots + T^n(x_k)) + \cdots + a_1(T(x_1) + \cdots + T(x_k)) + a_0(x_1 + \cdots + x_k) \\ = (a_n \lambda^n_1 x_1 + \cdots + a_1 \lambda_1 x_1 + a_0 x_1) + \cdots + (a_n \lambda^n_k x_k + \cdots + a_1 \lambda_1 x_1 + a_0 x_k) \\ = g(\lambda_1)x_1 + \cdots + g(\lambda_k)x_k = g(\lambda_1)T_1(x) + \cdots + g(\lambda_k)T_k(x).)$

Corollary 1

Corollary 1. If $F = \mathbb{C}$, then $T$ is normal $\iff$ $T^* = g(T)$ for some $g \in P(F)$.

Proof. $(\Longrightarrow)$
Let $T = \lambda_1 T_1 + \cdots + \lambda_k T_k$ be the spectral decomposition of $T$. Then $T^* = \overline{\lambda_1} T_1 + \cdots + \overline{\lambda_k} T_k$ by Theorem 1. By the Lagrange Interpolation formula, we can choose a polynomial $g$ such that $g(\lambda_i) = \overline{\lambda_i}$ for $1 \leq i \leq k$. Then $T^* = g(\lambda_1)T_1 + \cdots + g(\lambda_k) T_k = g(T)$ by remark.
($\Longleftarrow$)
By Theorem 1 (c), $TT^* = g(T)^*g(T) = |g(\lambda_1)|^2 T_1 + \cdots + |g(\lambda_k)|^2 T_k = g(T)g(T)^* = T^*T$. $\blacksquare$

Corollary 2

Corollary 2. If $F = \mathbb{C}$, then $T$ is unitary $\iff$ T is normal and $|\lambda| = 1$ for every eigenvalue $\lambda$ of $T$.

Proof. ($\Longrightarrow$)
It is immediate from Corollary 2.
($\Longleftarrow$)
Let $T = \lambda_1 T_1 + \cdots +\lambda_k T_k$ be the spectral decomposition of $T$. Then $TT^* = T^*T = |\lambda_1|^2 T_1 + \cdots + |\lambda_k|^2 T_k = T_1 + \cdots + T_k = I$ by Theorem 1. $\blacksquare$

Corollary 3

Corollary 3. If $F = \mathbb{C}$ and $T$ is normal, then $T$ is hermitian $\iff$ every eigenvalue of $T$ is real.

Proof. ($\Longrightarrow$)
It is immediate from Lemma.
($\Longleftarrow$)
Let $T = \lambda_1 T_1 + \cdots + \lambda_k T_k$ be the spectral decomposition of $T$. Then $T^* = \overline{\lambda_1} T_1 + \cdots + \overline{\lambda_k} T_k = \lambda_1 T_1 + \cdots + \lambda_k T_k = T$. $\blacksquare$

Corollary 4

Corollary 4. Let $T$ as in the spectral theorem with spectral decomposition $T = \lambda_1 T_1 + \cdots + \lambda_k T_k$. Then each $T_j$ is a polynomial in $T$. 

Proof.