Mathematical Tools for Physics

7—Operators and Matrices 191

The solutions to this quadratic areλ= 1, 3. For these values then, the apparently two equation for the two
unknownsv 1 andv 2 are really one equation. The other is not independent. Solve this single equation in each
case. Take the first of the two linear equations forv 1 andv 2 as defined by Eq. ( 29 ).

2 v 1 +v 2 =λv 1

λ= 1implies v 2 =−v 1 , λ= 3implies v 2 =v 1

The two new basis vectors are then

~e′ 1 = (~e 1 −~e 2 ) and ~e′ 2 = (~e 1 +~e 2 ) (30)

and in this basis the matrix of components is the diagonal matrix of eigenvalues.
(
1 0
0 3

)

If you like to keep your basis vectors normalized, you may prefer to say that the new basis is(~e 1 −~e 2 )/

√

2 and
(~e 1 +~e 2 )/

√

2. The eigenvalues are the same, so the new matrix is the same.

Example: Coupled Oscillators
Another example drawn from physics: Two masses are connected to a set of springs and fastened between two
rigid walls. This is a problem that appeared in chapter 4, Eq. (4.27).

m 1 d^2 x 1 /dt^2 =−k 1 x 1 −k 3 (x 1 −x 2 ), and m 2 d^2 x 2 /dt^2 =−k 2 x 2 −k 3 (x 2 −x 1 )

The exponential form of the solution that I used was

x 1 (t) =Aeiωt, x 2 (t) =Beiωt

The algebraic equations that you get by substituting these into the differential equations are a pair of linear
equations forAandB, Eq. (4.28). In matrix form these equations are, after rearranging some minus signs,

(

k 1 +k 3 −k 3

−k 3 k 2 +k 3

)(

A

B

)

=ω^2

(

m 1 0

0 m 2

)(

A

B

)