Characteristic Polynomial of a Matrix
Definition
For a square matrix $A \in M_{n\times n}(F)$ over a field $F$, the following polynomial in the variable $\lambda$ is called the characteristic polynomial of $A$.
$$ p(\lambda) := \det(A - \lambda I) $$
Explanation
Since the roots of $p(\lambda) = 0$ are precisely the eigenvalues of $A$, finding the eigenvalues amounts to finding the roots of the characteristic polynomial. $p$ is a polynomial of degree $n$ in $\lambda$, and its leading term is $(-1)^{n}\lambda^{n}$.
For $n \ge 3$, the coefficients are not determined solely by the trace and the determinant as in (a) below.
Properties
(a) The characteristic polynomial of a $2 \times 2$ matrix $A$ is as follows.
$$ \det(A - \lambda I) = \lambda^{2} - (\operatorname{tr} A)\lambda + \det A $$
Here $\det$ is the determinant and $\operatorname{tr}$ is the trace.
(b) The characteristic polynomial of an $n \times n$ matrix $A$ with leading coefficient $1$ is as follows.
$$ (-1)^{n}\det(A - \lambda I) = \det(\lambda I - A) = \lambda^{n} - (\tr A) \lambda^{n-1} + \cdots + (-1)^{n}\det A $$
Proof
(a)
Let $A$ be as follows.
$$ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} $$
Then a direct computation gives the following.
$$ \det(A - \lambda I) = \begin{vmatrix} a - \lambda & b \\ c & d - \lambda \end{vmatrix} = (a - \lambda)(d - \lambda) - bc = \lambda^{2} - (a + d)\lambda + (ad - bc) $$
Since $a + d = \operatorname{tr} A$ and $ad - bc = \det A$, the conclusion follows.
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(b)
Let the eigenvalues of the $n \times n$ matrix $A$, counted with algebraic multiplicity, be $\lambda_{1}, \dots, \lambda_{n}$. Since the eigenvalues are the roots of the characteristic equation $\det(\lambda I - A) = 0$, the characteristic polynomial factors as follows.
$$ \det(\lambda I - A) = (\lambda - \lambda_{1})(\lambda - \lambda_{2}) \cdots (\lambda - \lambda_{n}) $$
When expanding the right-hand side, the $\lambda^{n-1}$ term is the sum of the products obtained by choosing $-\lambda_{i}$ from one of the $n$ factors and $\lambda$ from the rest, and the constant term is obtained by choosing $-\lambda_{i}$ from every factor. Therefore we have the following.
$$ \det(\lambda I - A) = \lambda^{n} - \left( \sum_{i=1}^{n} \lambda_{i} \right) \lambda^{n-1} + \cdots + (-1)^{n} \prod_{i=1}^{n} \lambda_{i} $$
Meanwhile, since $\tr(A) = \sum_{i=1}^{n}\lambda_{i}$, $\det(A) = \prod_{i=1}^{n}\lambda_{i}$ hold, we obtain the following.
$$ (-1)^{n}\det(A - \lambda I) = \det(\lambda I - A) = \lambda^{n} - (\tr A) \lambda^{n-1} + \cdots + (-1)^{n}\det A $$
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