Chapter 7. The Eigenvalue Problem

7. The Eigenvalue Problem, December 17, 2009 20
where α 1 , α 2 , α 3 are suitably chosen numbers.
The answer will be yes if the three vectors are linearly independent, which
means that there are no numbers β 1 , β 2 , β 3 such that
β 1 v(1) + β 2 v(2) + β 3 v(3) = 0 (97)
(the selection β 1 = β 2 = β 3 = 0 does not count). Another way to put this is
that none of v(1), v(2), v(3) can be written as a linear combination of the
others. How do we establish that this is true Calculating the left-hand side
of Eq. 97 for all possible numbers would take too long. Fortunately, there
is a short-cut we can use, based on the theory of determinants. Form the
matrix
⎛
⎞
B =
⎜
⎝
v 1 (1) v 2 (1) v 3 (1)
v 1 (2) v 2 (2) v 3 (2)
v 1 (3) v 2 (3) v 3 (1)
⎟
⎠ (98)
where v i (α) isthei-thcomponentofthevectorv(α).
If a row can be written as a linear combination of the other rows, then
the determinant of B is zero (and vice versa). Therefore
det B = 0 (99)
ensures that the three vectors are linearly independent. Straightforward calculation
(or Mathematica) tells me that in this case
det B = −3 (100)
We conclude that the three eigenvectors are linearly independent.
Exercise 7 Suppose three 3-dimensional vectors lie in the same plane. Are
they linearly independent What about three vectors for which two are in
the same plane but not parallel and the third is perpendicular to that plane
Because these three vectors are linearly independent, any three-dimensional
vector x can be written as
x = μ 1 v(1) + μ 2 v(2) + μ 3 v(3) (101)
where μ 1 , μ 2 ,andμ 3 are numbers. So v(1), v(2), v(3) do form a complete
basis set. But the three vectors making up the basis are not orthonormal.