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Characteristic Polynomial of a Matrix
Let $A$ be a $n\times n$ matrix over some field $k$. The characteristic polynomial $p_{A}(x)$ of $A$ in an indeterminate $x$ is defined by the determinant:
$p_{A}(x):=\det(AxI)=\left\begin{matrix}a_{{11}}x&a_{{12}}&\cdots&a_{{1n}}\\ a_{{21}}&a_{{22}}x&\cdots&a_{{2n}}\\ \vdots&\vdots&\ddots&\vdots\\ a_{{n1}}&a_{{n2}}&\cdots&a_{{nn}}x\end{matrix}\right$ 
Remarks

The polynomial $p_{A}(x)$ is an $n$thdegree polynomial over $k$.

If $A$ and $B$ are similar matrices, then $p_{A}(x)=p_{B}(x)$, because
$\displaystyle p_{A}(x)$ $\displaystyle=$ $\displaystyle\det(AxI)=\det(P^{{1}}BPxI)$ $\displaystyle=$ $\displaystyle\det(P^{{1}}BPP^{{1}}xIP)=\det(P^{{1}})\det(BxI)\det(P)$ $\displaystyle=$ $\displaystyle\det(P)^{{1}}\det(BxI)\det(P)=\det(BxI)=p_{B}(x)$ for some invertible matrix $P$.

The characteristic equation of $A$ is the equation $p_{A}(x)=0$, and the solutions to which are the eigenvalues of $A$.
Characteristic Polynomial of a Linear Operator
Now, let $T$ be a linear operator on a vector space $V$ of dimension $n<\infty$. Let $\alpha$ and $\beta$ be any two ordered bases for $V$. Then we may form the matrices $[T]_{{\alpha}}$ and $[T]_{{\beta}}$. The two matrix representations of $T$ are similar matrices, related by a change of bases matrix. Therefore, by the second remark above, we define the characteristic polynomial of $T$, denoted by $p_{T}(x)$, in the indeterminate $x$, by
$p_{T}(x):=p_{{[T]_{{\alpha}}}}(x).$ 
The characteristic equation of $T$ is defined accordingly.
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