Mathematical Tools for Physics

7—Operators and Matrices 192

You can make it look more like the previous example with some further arrangement

[(

k 1 +k 3 −k 3

−k 3 k 2 +k 3

)

−ω^2

(

m 1 0

0 m 2

)](

A

B

)

=

(

0

0

)

The matrix on the left side maps the column matrix to zero. That can happen only if the matrix has zero
determinant (or the column matrix is zero). If your write out the determinant of this 2 × 2 matrix you have a
quadratic equation inω^2. It’s simple but messy, so rather than looking first at the general case, look at a special
case with more symmetry. Takem 1 =m 2 =mandk 1 =k 2.

det

[(

k 1 +k 3 −k 3

−k 3 k 1 +k 3

)

−ω^2 m

(

1 0

0 1

)]

= 0 =

(

k 1 +k 3 −mω^2

) 2

−k 32

This is now so simple that you don’t even need the quadratic formula; it factors directly.

(

k 1 +k 3 −mω^2 −k 3

)(

k 1 +k 3 −mω^2 +k 3

)

= 0

The only way that the product of two numbers is zero is if one of the numbers is zero, so either

k 1 −mω^2 = 0 or k 1 + 2k 3 −mω^2 = 0

This determines two possible frequencies of oscillation.

ω 1 =

√

k 1

m

and ω 2 =

√

k 1 + 2k 3

m

You’re not done yet; these are only the eigenvalues. You still have to find the eigenvectors andthengo back to
apply them to the original problem. This isF~=m~aafter all. Look back to section4.8for the development of
the solutions.

7.9 Change of Basis
In many problems in physics and mathematics, the correct choice of basis can enormously simplify a problem.
Sometimes the obvious choice of a basis turns out in the end not to be the best choice, and you then face the