Page 278 - A Course in Linear Algebra with Applications

262 Chapter Eight: Eigenvectors and Eigenvalues

which, by 3.1.5, equals (an — x)(a,22 — x) ... (a nn — x). The
eigenvalues of the matrix are therefore just the diagonal entries
^11) <^22) • • • i^nn-

Example 8.1.3
Consider the 3 x 3 matrix

The characteristic polynomial of this matrix is

2-x - 1 - 1
- 1 2-x - 1 = -x 6 + 4x 2 - x - 6.
- 1 - 1 -x

Fortunately one can guess a root of this cubic polynomial,
namely x = — 1. Dividing the polynomial by x + 1 using long
2
division, we obtain the quotient — x + 5x — 6 = — (x — 2)(x — 3).
Hence the characteristic polynomial can be factorized com-
pletely as — (x + l)(x — 2)(x — 3), and the eigenvalues of A are
— 1, 2 and 3.
To find the corresponding eigenvectors, we have to solve
the three linear systems (A + I 3)X ~ 0, (A - 2I 3)X = 0 and
(A — 3/s)X = 0. On solving these, we find that the respective
eigenvectors are the non-zero scalar multiples of the vectors

The eigenspaces are generated by these three vectors and so
each has dimension 1.