In Section 6.1 we discussed how to decide whether a given number is an eigenvalue of a matrix, and if so, how to find all of the associated eigenvectors. In this section, we will give a method for computing all of the eigenvalues of a matrix. This does not reduce to solving a system of linear equations: indeed, it requires solving a nonlinear equation in one variable, namely, finding the roots of the characteristic polynomial.
Let be an matrix. The characteristic polynomial of is the function
We will see below that the characteristic polynomial is in fact a polynomial. Finding the characterestic polynomial means computing the determinant of the matrix whose entries contain the unknown
If is an matrix, then the characteristic polynomial has degree by the above theorem. When one can use the quadratic formula to find the roots of There exist algebraic formulas for the roots of cubic and quartic polynomials, but these are generally too cumbersome to apply by hand. Even worse, it is known that there is no algebraic formula for the roots of a general polynomial of degree at least
In practice, the roots of the characteristic polynomial are found numerically by computer. That said, there do exist methods for finding roots by hand. For instance, we have the following consequence of the rational root theorem (which we also call the rational root theorem):
Rational Root Theorem
Suppose that is an matrix whose characteristic polynomial has integer (whole-number) entries. Then all rational roots of its characteristic polynomial are integer divisors of
For example, if has integer entries, then its characteristic polynomial has integer coefficients. This gives us one way to find a root by hand, if has an eigenvalue that is a rational number. Once we have found one root, then we can reduce the degree by polynomial long division.
In the above example, we could have expanded cofactors along the second column to obtain
Since was the only nonzero entry in its column, this expression already has the term factored out: the rational root theorem was not needed. The determinant in the above expression is the characteristic polynomial of the matrix so we can compute it using the trace and determinant: