이 포스트에서 $V$는 $n$차원 $F$-벡터공간으로 취급한다.
Theorem 1
Theorem 1.
(a) Let $T \in \mathcal{L}(V)$. Then a scalar $\lambda$ is an eigenvector of $T$ $\Longleftrightarrow$ $\det(T - \lambda I_V) = 0$.
(b) Let $A \in M_{n \times n}(F)$. Then a scalar $\lambda$ is an eigenvector of $A$ $\Longleftrightarrow$ $\det(A - \lambda I_n) = 0$.
Proof. (a) Since $\lambda$ is an eigenvector of $T$, there is an nonzero vector $v \in V$ such that $T(v) = \lambda v$. Then $T(v) = \lambda v = \lambda I_V(v) \Longrightarrow (T - \lambda I_V)(v) = \mathbf{0}$. By Theorem 2, $T - \lambda I_V$ is not invertible $\Longleftrightarrow$ $\det(T - \lambda I_V) = 0$.
(b) It is similar to (a). $\blacksquare$
즉 방정식 $f(t) = \det(T - t I_V) = 0$을 만족하는 $t$는 $T$의 고유값이 되므로 다항식 $f(t)$가 0이 되는 값을 구하면 된다. 이러한 다항식 $f(t)$를 $T$의 characteristic polynomial, 특성 다항식이라고 부른다.
The Characteristic Polynomial
Definition 1. Let $T \in \mathcal{L}(V)$. The polynomial $f(t) = \det(T - tI_V)$ is called the characteristic polynomial of $T$.
Let $A \in M_{n \times n}(F)$. The polynomial $f(t) = \det(A - tI_n)$ is called the characteristic polynomial of $A$.
Theorem 2
Theorem 2. Let $T \in \mathcal{L}(V)$, and let $A \in M_{n \times n}(F)$
(a) The characteristic polynomial of $T$ and $A$ is a polynomial of degree $n$ with leading coefficient $(-1)^n$.
(b) $T$ and $A$ has at most $n$ distinct eigenvalues.
Proof. (a) The proof is by the induction on $n$. If $n = 1$, then $f(t) = \det(T - tI_V) = \det([T]_{\beta} - t) = [T]_{\beta} - t$.
Assume that the theorem is true for $n-1$ where $n > 1$. Then $$f(t) = \det(T - tI_V) = \det([T]_{\beta} - tI_n) \\ = \sum_{j=1}^n (-1)^{1+j}([T]_{\beta} - tI_n)_{1j} \det(\widetilde{[T]_{\beta} - tI_n})_{1j}.$$ We only need to know the leading coefficient of the polynomial. By the induction hypothesis, $\det(\widetilde{[T]_{\beta} - tI_n})_{1j} = (-1)^{n-1}(t^{n-1} + \cdots + t_0)$. Then $f(t)$ is a polynomial of degree $n$ and the leading coefficient is $(-1)^{1+1}(-1)(-1)^{n-1} = (-1)^{n}$ because $(I_n)_{1j} = 0$ for $j \neq 1$.
(b) Suppose that $T$ has $n+1$ distinct eigenvalues, says $\lambda_1, ..., \lambda_n, \lambda_{n+1}$. Then $f(\lambda_i) = \det(T - \lambda_i I_V) = 0$ for $i = 1, ..., n, n+1$. Thus we have $f(t) = (-1)^n(t - \lambda_1) \cdots (t - \lambda_n)(t - \lambda_{n+1})$ because each $\lambda_{i}$ is distinct $\bigotimes$. Hence $T$ has at most $n$ distinct eigenvalues.
The case of $A$ is can be proved similarily. $\blacksquare$
Theorem 3
Theorem 3. (a) Similar matrices have the same characteristic polynomial.
(b) For any square matrix $A$, $A$ and $A^t$ have the same characteristic polynomial.
Proof. (a) Since $A$ and $B$ are similar, $\exists$ invertible $Q \in M_{n \times n}(F)$ such that $B = Q^{-1}AQ$. Then $f_B(t) = \det(B - tI) = \det(Q^{-1}AQ - tI)$ = $\det(Q^{-1}(A - tI)Q)$ = $\det(Q^{-1})\det(A-tI)\det(Q)$ = $\det(A - tI) = f_A(t)$.
(b) $f_A(t) = \det(A - tI) = \det((A - tI)^t) = \det(A^t - tI) = f_{A^t}(t)$. $\blacksquare$
Splitting of a Polynomial
Definition 2. A polynomial $f(t) \in P(F)$ splits over $F$ if $\exists c, a_0, ..., a_n \in F$ such that $f(t) = c(t - a_1) \cdots (t - a_n)$.
위와 같이 다항식 $f(t)$가 표현되면 분해된다고 표현한다. 자명하게 대각화 가능한 선형 연산자 혹은 행렬의 특성 다항식은 분해됨을 알 수 있고, 스칼라 $a_0, ..., a_n$은 고유값이 됨을 알 수 있다.
Theorem 4
Theorem 4. The characteristic polynomial of any diagonalizable linear operator splits.
Proof. Let $T \in \mathcal{L}(V)$ be diagonalizable. Then there is an ordered basis $\beta$ such that $[T]_{\beta}$ is a diagonal matrix. Then the characteristic polynomial $f(t) = \det(T - t I_V) = \det([T]_{\beta} - tI) = (-1)^n(t - \lambda_1) \cdots (t - \lambda_n)$ where $\lambda_1, ..., \lambda_n$ are eigenvalues of $T$. $\blacksquare$
