어떤 선형 연산자 $T$가 주어졌을 때 대각화가능한지 결정하고, 가능하다면 대각화하도록 고유벡터들로 이루어진 기저 $\beta$를 찾는 것이 우리의 목표이다. $T$의 고유값은 특성 다항식 $f(t) = \det (T - tI)$를 풀어서 구할 수 있다. 만약 이를 통해 서로 다른 고유값 $\lambda_1, ..., \lambda_k$를 구했을 때, 이 고유값들에 대응되는 고유벡터들은 $v \in E_{\lambda}$을 이용해서 구할 수 있다. 이제 이 고유벡터들로 기저를 구성해야 하고, 그 방법을 아래의 정리들이 제시해준다.
Theorem 1
Theorem 1. Let $T \in \mathcal{L}(V)$, and let $\lambda_1, ..., \lambda_k$ be distinct eigenvalues of $T$. If $v_1, ..., v_k$ are eigenvectors of $T$ such that $\lambda_i$ corresponds to $v_i (1 \leq i \leq k)$, then $\{v_1, ..., v_k\}$ is linearly independent.
Proof. The proof is by mathematical induction on $k$. If $k = 1$, then clearly $\{v_1\}$ is linearly independent.
Suppose that the theorem is ture for $k-1$, where $k-1 \geq 1$. If $\sum_{i=1}^n a_iv_i = \mathbf{0}$, then $$(T - \lambda_k I_V)(\sum_{i=1}^n a_iv_i) = \sum_{i=1}^n a_i\lambda_iv_i - \lambda_k(\sum_{i=1}^n a_iv_i) \\ = (\lambda_1 - \lambda_k)a_1v_1 + \cdots + (\lambda_{k-1} - \lambda_k)a_{k-1}v_{k-1} = \mathbf{0}.$$ By the induction hypothesis, each $(\lambda_i - \lambda_k)a_i = 0$ for $1 \leq i \leq k-1$. Since each eigenvalues are distinct, $a_i = 0$ for $1 \leq i \leq k-1$. Then $a_kv_k = \mathbf{0} \Longrightarrow a_k = 0$. Thus $\{v_1, ..., v_k\}$ is linearly independent. $\blacksquare$
즉 특성다항식을 풀어서 고유벡터들을 얻었다면, 이들의 집합은 적어도 선형 독림임이 보장된다. 이는 고유값의 대수적 중복도는 기하적 중복도보다 작기 때문인데, 따라서 두 중복도의 값이 같아져서 각 고유값들에 대응하는 고유벡터들을 모두 모은 집합이 $T$를 대각화하는 기저이다. Theorem 1을 확장한 것이 Theorem 2이고, 상술한 내용을 바탕으로 $T$의 대각화가능 조건과 대각화하는 기저를 구하는 방법을 제시하는 것이 Theorem 3이다.
Corollary
Corollary. Let $T \in \mathcal{L}(V)$, and let $\dim(V) = n$. If $T$ has $n$ distinct eigenvalues, then $T$ is diagonalizable.
Proof. By Corollary 3 - 2 (b), it is clear. $\blacksquare$
Lemma
Lemma. Let $T \in \mathcal{L}(V)$, and let $\lambda_1, ..., \lambda_k$ be distinct eigenvalues of $T$. Let $v_i \in E_{\lambda_i}$ for each $i (1 \leq i \leq k)$. If $v_1 + \cdots + v_k = \mathbf{0}$, then each $v_i = \mathbf{0}$.
Proof. Suppose there is $m \in \mathbb{N}$ such that for $1 \leq i \leq m$, $v_i \neq \mathbf{0}$, and for $i > m$, $v_i = \mathbf{0}$. Then $v_1 + \cdots + v_k = v_1 + \cdots + v_m = \mathbf{0}$. But by Theorem 1, $\{v_1, ..., v_m\}$ is linearly independent. $\bigotimes$ Thus each $v_i = \mathbf{0}$. $\blacksquare$
Theorem 2
Theorem 2. Let $T \in \mathcal{L}(V)$, and let $\lambda_1, ..., \lambda_k$ be distinct eigenvalues of $T$. For each $i = 1, ..., k$, let $S_i$ be a finite linearly independent subset of $E_{\lambda_i}$. Then $$S = \bigcup_{i=1}^{k} S_i$$ is a linearly independent subset of $V$.
Proof. Suppose that $S_i = \{v_{i1}, ..., v_{in_i}\}$ for each $i$. If $$\sum_{i=1}^k \sum_{j = 1}^{n_i} a_{ij}v_{ij} = \mathbf{0}$$ for some $a_{ij} \in F$, then denote $$u_i = \sum_{j=1}^{n_i} a_{ij}v_{ij}.$$ Then $\sum_{i=1}^k u_i = \mathbf{0}$. By Lemma, each $u_i = \sum_{j=1}^{n_i} a_{ij}v_{ij} = \mathbf{0}$. Since each $S_i$ is linearly indepdent, each $a_{ij} = 0$. Thus $S$ is linearly indepdent. $\blacksquare$
Theorem 3
Theorem 3. Let $T \in \mathcal{L}(V)$ such that the characteristic polynomial of $T$ splits, and let $V$ be finite-dimensional. Let $\lambda_1, ..., \lambda_k$ be the distinct eigenvalues of $T$. Then
(a) $T$ is diagonalizable $\iff$ the algebric multiplicity of $\lambda_i$ is equal to the geometric multiplicity of $\lambda_i$, i.e., $\dim(E_{\lambda_i})$, for each $i (1 \leq i \leq k)$.
(b) If $T$ is diagonalizable and $\beta_i$ is an ordered basis for $E_{\lambda_i}$ for each $i (1 \leq i \leq k)$, then $$\beta = \bigcup_{i=1}^k \beta_i$$ is an ordered basis for $V$ consisting of eigenvectors of $T$.
Proof. For each $i (1 \leq i \leq k)$, let $m_i$ denote the algebric multiplicity of $\lambda_i$, $d_i = \dim(E_{\lambda_i})$, and $n = \dim(V)$. Suppose that $T$ is diagonalizable. Let $\beta$ be a basis for $V$ consisting of eigenvectors of $T$. For each $i$, let $\beta_i = \beta \cap E_{\lambda_i}$, and let $n_i = |\beta_i|$.
Note that for each $i$, $n_i \leq d_i$, because $\beta_i$ is a linearly independent subset of $E_{\lambda_i}$, and $d_i \leq m_i$ by Theorem 1.
Suppose that u and v are the eigenvectors of $T$ corresponding to $\lambda_i$ and $\lambda_j$ for some $i, j (1 \leq i, j \leq k)$, respectively. If $u = v$, then $T(u) = \lambda_i u = \lambda_j v = T(v)$. Thus $(\lambda_i - \lambda_j)u = \mathbf{0} \Longrightarrow \lambda_i = \lambda_j \bigotimes$. Thus $\beta_i (1 \leq i \leq k)$ is pairwise disjoint. Since $\beta$ consists of eigenvectors of $T$, $\sum_{i=1}^k n_i = n$.
Then we have $$n = \sum_{i=1}^k n_i \leq \sum_{i=1}^k d_i \leq \sum_{i=1}^k m_i = n.$$ Thus $\sum_{i=1}^k d_i = n$. This means that $d_i = m_i, \forall i$.
As we showed above, $n_i = |\beta_i| = d_i = \dim(E_{\lambda_i})$. Since $\beta_i$ is linearly independent subset of $E_{\lambda_i}$, $\beta_i$ is an ordered basis for $E_{\lambda_i}$. Let denote $$\beta = \bigcup_{i=1}^k \beta_i.$$ Then $|\beta| = \sum_{i=1}^k d_i = n$. Hence $\beta$ is an ordered basis for $V$ consisting of eigenvectors of $T$. $\blacksquare$
따라서 각 고유공간의 기저를 구해 합집합을 시켜줌으로서 주어진 선형 연산자를 대각화할 수 있다. 이러한 대각화 과정을 요약하면 다음과 같다.
Test for Diagonalization
Let $T \in \mathcal{L}(V)$, and let $\dim(V) = n$.
1. Choose a convenient basis $\alpha$ for $V$ (For example, the standard ordered basis for $V$ can be chosen.)
2. Determine whether the characteristic polynomial of $T$ splits: Solve the equation $\det([T]_{\alpha} - tI) = 0$.
3. If it does, then write the distinct eigenvalue $\lambda$ of $T$ and check the algebric multiplicity of each $\lambda$.
4. Find an ordered basis for each $E_{\lambda}$ of $[T]_{\alpha}$.
5. Check that the algebric multiplicity of $\lambda$ equal to the geometric multiplicity of $\lambda$ for each $\lambda$.
6. If it does, then the union of these bases is a basis $\gamma$ for $F^n$ consisting of eigenvectors of $[T]_{\alpha}$.
7. Since each vector in $\gamma$ is $\phi_{\alpha}(v)$, where $v$ is an eigenvector of $T$, the set consisting of these eigenvectors is the desired basis $\beta$.
