7—Operators and Matrices 190Write the problem in terms of components, and of course you aren’t yet in the basis where the matrix is
diagonal. If you were, you’re already done. The defining equation isf(~v) =λ~v, and in components this reads
∑
fkivi=λvk, or
f 11 f 12 f 13
f 21 f 22 f 23
f 31 f 32 f 33
v 1
v 2
v 3
=λ
v 1
v 2
v 3
Here I arbitrarily wrote the equation for three dimensions. That will change with the problem. Put everything on
the left side and insert the components of the identity, the unit matrix.
f 11 f 12 f 13
f 21 f 22 f 23
f 31 f 32 f 33
−λ
1 0 0
0 1 0
0 0 1
v 1
v 2
v 3
=
0
0
0
The only way that this has a non-zero solution for the vector~vis for the determinant of the left-hand side to
be zero. In the case as written, that’s a cubic equation inλ. This equation is called thecharacteristic equation
of the matrix. If it has all distinct roots, no double roots, then you’re guaranteed that this procedure will work.
If this equation has a multiple root then there is no guarantee. It may work, but is may not; you have to look
closer. See section7.11. Also, if the operator has certain symmetry properties then it’s guaranteed to work.
Example of Eigenvectors
To keep the algebra to a minimum, I’ll work in two dimensions and will specify an arbitrary but simple example:
f(~e 1 ) = 2~e 1 +~e 2 , f(~e 2 ) = 2~e 2 +~e 1 with components M=(
2 1
1 2
)
(29)
The eigenvalue equation is, in component form
(
2 1
1 2
)(
v 1
v 2)
=λ(
v 1
v 2)
or[(
2 1
1 2
)
−λ(
1 0
0 1
)](
v 1
v 2)
= 0
The condition that there be a non-zero solution to this is
det[(
2 1
1 2
)
−λ(
1 0
0 1
)]
= 0 = (2−λ)^2 − 1