Solution
= Solution
Choose a <basis> and let $A$ represent $T$. Define the <characteristic polynomial> by
$$
\boxed{\chi_T(t)=\det(tI-A).}
$$
Under a change of basis with invertible change matrix $S$, the new matrix is $S^{-1}AS$. Hence
$$
\det(tI-S^{-1}AS)=\det(S^{-1}(tI-A)S)
=\det(S)^{-1}\det(tI-A)\det(S)=\det(tI-A).
$$
Thus the polynomial is independent of the chosen coordinates. It is monic of degree $\dim V$.