Mathematical Tools for Physics

7—Operators and Matrices 189

This is called aneigenvalueequation. It says that for any one of these special vectors, the operatorfon it returns
a scalar multiple of that same vector. These multiples are called the eigenvalues, and the corresponding vectors
are called the eigenvectors. The eigenvalues are then the diagonal elements of the matrix in this basis.
The inertia tensor relates the angular momentum of a rigid body to its angular velocity, and if the angular
momentum isn’t in the same direction as the angular velocity, the angular momentum vector will be spinning
about the~ωaxis. There will then be a torque necessary to keep it going,~τ =dL/dt~ , and becauseL~is rotating
about~ω, this will cause a vibration of the axis at this rotation frequency. If on the other hand the angular
momentum is parallel to the angular velocity, the angular momentum will not be changing,dL/dt~ = 0, and the

torque~τ=d~L/dtwill be zero, meaning the vibrations will be absent. Have you ever taken your car in and asked
the mechanic to align the angular momentum and the angular velocity vectors of the tires? I’ve done it a number
of times; it’s called wheel-balancing.
How do you compute these eigenvectors? Just move everything to the left side of the equation.

f(~ei)−fii~ei= 0, or (f−fiiI)~ei= 0

Iis the identity operator, output equals input. This notation is cumbersome. I’ll change it.

(f−λI)~v= 0 (28)

λis the eigenvalue and~vis the eigenvector. This operator(f−λI)takes some non-zero vector into the zero
vector. In two dimensions then it will squeeze an area down to a line or a point. In three dimensions it will
squeeze a volume down to an area (or a line or a point). In any case its determinant is zero, and that’s the key
to computing the eigenvectors. Figure out whichλ’s will make this determinant vanish.
Look back at section4.7and you’ll see that the analysis there closely parallels what I’m doing here. In
that case I didn’t use the language of matrices or operators, but was asking about the possible solutions of two
simultaneous linear equations.

ax+by= 0 and cx+dy= 0, or

(

a b

c d

)(

x

y

)

=

(

0

0

)

The explicit algebra there led to the conclusion that there can be a non-zero solution to the two equations only
if the determinant of the coefficients vanishes,ad−bc= 0, and that’s the same thing that I’m looking for here:
a non-zero vector solution to Eq. ( 28 ).