characteristic polynomial (operator)
In Vol I you computed the characteristic polynomial of a matrix as det(xI - A). The danger is that A lives in coordinates — so does this polynomial somehow depend on the basis you happened to pick? Reassuringly, no. Similar matrices have the same characteristic polynomial, so the formula attaches to the operator T itself, not to any particular matrix for it.
Define it basis-free as chi_T(x) = det(xI - T), where det is the coordinate-free determinant of an operator. It is a monic polynomial of degree exactly n = dim V. Its roots, counted with multiplicity, are the eigenvalues of T. Two pieces fall out of its coefficients for free: the coefficient of x^(n-1) is minus the trace, and the constant term is (-1)^n det(T).
It matters because it manufactures eigenvalues. Setting chi_T(x) = 0 is the characteristic equation, and solving it is how you find the spectrum. Over an algebraically closed field like C it always factors completely into linear pieces (x - lambda_1)...(x - lambda_n), so the operator always has n eigenvalues counted with multiplicity — the multiplicity of a root here is the algebraic multiplicity.
One honest limitation: the characteristic polynomial knows the eigenvalues and their algebraic multiplicities, but it can be FOOLED about the operator's finer structure. The matrices [2, 1; 0, 2] and 2I share chi(x) = (x-2)^2 yet behave completely differently. To see past that you need the minimal polynomial, which is why these two travel together.
The trace and determinant read straight off the characteristic polynomial's coefficients, and its roots 1 and 2 are the eigenvalues.
Convention warning: some texts define the characteristic polynomial as det(A - xI), which differs from det(xI - A) by a factor of (-1)^n. The det(xI - A) convention keeps it monic, which is why most modern operator-theory texts prefer it.