This last equation presents a challenge. Anything multiplied by the zero
vector yields the zero vector, but clearly that is a trivial solution
for that we aren't interested in. Thus, our assumption that
exists must be wrong. The eigenvalue
equation
therefore has non-trivial solutions
only when
does not exist. In
our previous discussion of determinants, we noted that a matrix
does not have an inverse if its determinant is zero. Thus, we
can satisfy the eigenvalue equation for those special values of
such that
. This is called the
*secular determinant*, and expanding the determinant gives
an -th degree polynomial in called the *secular equation*
or the *characteristic equation*. Once the roots of this equation are
determined to give eigenvalues , these eigenvalues may be
inserted into the eigenvalue equation, one at a time, to yield
eigenvectors.

As an example, let us find the eigenvalues and eigenvectors for the matrix

(25) |

We begin with the secular determinant , which in this case becomes

(26) |

Expanding out this determinant using the rules given above for the determinants of matrices, we obtain the following characteristic equation:

(27) | |||

which has solutions . These are the three eigenvalues of . What are the corresponsing eigenvectors? We substitute each of these eigenvalues, one at a time, into the eigenvalue equation, and solve for the system of equations that result.

Let us begin with the eigenvalue . Substituting this into
, we obtain

(28) |

This is three equations in three unknowns, which we may rewrite as

(29) | |||

(30) | |||

(31) |

The middle equation is, of course, not particularly useful. However, we know that the components and of the eigenvector corresponding to are both zero, and there is no equation governing the choice of . We are therefore free to chose

(32) |

(33) |

that is, multiplication of times the eigenvector yields the eigenvector again times a constant (the eigenvalue, ).

By a similar procedure, one can obtain the other two eigenvectors, which,
when normalized, are

(34) |

Although an matrix has eigenvalues, they are
not necessarily distinct. That is, one or more of the roots of
the characteristic equation may be identical. In this case, we say that
those eigenvalues are *degenerate*. Determination of eigenvectors is
somewhat more complicated in such a case, because there will be additional
flexibility in selecting them. Consider the matrix

(35) |

which has the characteristic equation

(36) |

(37) |

or

(38) | |||

In this case, the first and third equations are equivalent (one equation is just the minus of the other), and so they are not independent. Additionally, the second equation gives no information. We therefore have only one equation to determine the three coefficients of the eigenvector. Recall that only two valid equations resulted above for our previous non-degenerate case, because any multiple of an eigenvector still yields a valid eigenvector. Here, a double degeneracy has lost us one of our equations. Hence, we only know that and is arbitrary. Any eigenvector satisfying these rules will be satisfactory, and clearly there are an infinite number of ways to chose them, even if we require them to be normalized. Let us, somewhat arbitrarily, pick the normalized vector

(39) |

(40) |