# characteristic polynomial

## 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:

 $p_{A}(x):=\det(A-xI)=\left|\begin{matrix}a_{11}-x&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}-x&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn}-x\end{matrix}\right|$

Remarks

• The polynomial $p_{A}(x)$ is an $n$th-degree polynomial over $k$.

• If $A$ and $B$ are similar matrices, then $p_{A}(x)=p_{B}(x)$, because

 $\displaystyle p_{A}(x)$ $\displaystyle=$ $\displaystyle\det(A-xI)=\det(P^{-1}BP-xI)$ $\displaystyle=$ $\displaystyle\det(P^{-1}BP-P^{-1}xIP)=\det(P^{-1})\det(B-xI)\det(P)$ $\displaystyle=$ $\displaystyle\det(P)^{-1}\det(B-xI)\det(P)=\det(B-xI)=p_{B}(x)$

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.

Title characteristic polynomial CharacteristicPolynomial 2013-03-22 12:17:47 2013-03-22 12:17:47 CWoo (3771) CWoo (3771) 10 CWoo (3771) Definition msc 15A18 Equation characteristic equation