= Solution
{c}
Choose an <ordered basis> $\mathcal B$ of $V$ and let $A=[T]_{\mathcal B}$. Define the <characteristic polynomial> by
$$
\boxed{\chi_T(x)=\det(xI-A).}
$$
If a different <basis> gives $B=P^{-1}AP$ for an <invertible matrix> $P$, multiplicativity of the <determinant> gives
$$
\det(xI-B)=\det\bigl(P^{-1}(xI-A)P\bigr)
=\det(P^{-1})\det(xI-A)\det P
=\det(xI-A).
$$
Thus the definition is independent of the chosen <basis>. The alternate convention $\det(A-xI)$ differs by the constant factor $(-1)^{\dim V}$ and gives the same roots.
Back to article page
