An eigenvalue is a scalar, λ, such that
Ax = λx
where A is a square matrix of order n, and x is a vector with n components. The vector x cannot be null; that is, a vector with all zero components.
Let I represent the identity matrix of order n. Thus, it has 1's on its principal diagonal and zeros everywhere else.
Then
Ax = λx = λIx
Ax - λIx = 0
where 0 is the null vector, as described above.
So
(A -λI)x = 0
For our problem, we need only subtract λ from each of the diagonal elements of our given matrix and solve for x. Most math calculators such as a TI83/84 or a Casio CG50 can easily solve a 3 by 3 linear system.
| -3 8 -2 | | x1 | | 0 |
| 6 -16 4 | | x2 | = | 0 |
| -12 -4 -26 | | x3 | | 0 |
NOTE that we have subtracted 18 from each of the diagonal elements:
15-18=-3
2-18=-16
-8-18=-26
I used a Casio CG50. The answer was -2x3, -0.5x3, x3. There are infinitely many vectors of this form that will solve this equation. If we pick the coefficients from the x3's, we get ( -2 -0.5 1). If we then use this as the x vector, it is easy to verify that Ax = λx, and this is the eigenvector we sought.
