Mathematical Tools for Physics

7—Operators and Matrices 197

Take the matrix (

1 2

0 1

)

You can’t diagonalize this. If you try the standard procedure, here is what happens:
(
1 2
0 1

)(

v 1

v 2

)

=λ

(

v 1

v 2

)

then det

(

1 −λ 2

0 1−λ

)

= 0 = (1−λ)^2

The resulting equations you get forλ= 1are

0 v 1 + 2v 2 = 0 and 0 = 0

This provides only one eigenvector, a multiple of

(

1

0

)

. You need two for a basis.
Change this matrix in any convenient way to make the two roots of the characteristic equation different
from each other. For example,

M=

(

1 + 2

0 1

)

The eigenvalue equation is now
(1 +−λ)(1−λ) = 0

and the resulting equations for the eigenvectors are

λ= 1 : v 1 + 2v 2 = 0, 0 = 0 λ= 1 +: 0v 1 + 2v 2 = 0, v 2 = 0

Now you have two distinct eigenvectors,

λ= 1 :

(

1

−/ 2

)

, and λ= 1 +:

(

1

0

)

Differential Equations at Critical
The problem4.11was to solve the damped harmonic oscillator for the critical case thatb^2 − 4 km= 0.

m

d^2 x

dt^2

=−kx−b

dx

dt

(34)