Page 411 - 35Linear Algebra

G.11 Eigenvalues and Eigenvectors 411

and we note that we can just read off the eigenvector e 1 with eigenvalue λ.
2
(µ) = (µ − λ) so the only
However the characteristic polynomial of J 2 is P J 2
possible eigenvalue is λ, but we claim it does not have a second eigenvector
v. To see this, we require that
2
1
λv + v = λv 1
2
λv = λv 2
2
which clearly implies that v = 0. This is known as a Jordan 2-cell, and in
general, a Jordan n-cell with eigenvalue λ is (similar to) the n × n matrix
 
λ 1 0 · · · 0
.
 . 
0 λ 1 . 0
 
 
J n = . . . . . . . . . . .
.
.
.
.
 
 0 · · · 0 λ 1 
0 · · · 0 0 λ
which has a single eigenvector e 1 .
Now consider the following matrix
 
3 1 0
M =  0 3 1 
0 0 2
2
and we see that P M (λ) = (λ − 3) (λ − 2). Therefore for λ = 3 we need to find the
solutions to (M − 3I 3 )v = 0 or in equation form:
2
v = 0
3
v = 0
3
−v = 0,
and we immediately see that we must have V = e 1 . Next for λ = 2, we need to
solve (M − 2I 3 )v = 0 or
1
2
v + v = 0
3
2
v + v = 0
0 = 0,
1 2 3
and thus we choose v = 1, which implies v = −1 and v = 1. Hence this is the
only other eigenvector for M.
This is a specific case of Problem 13.7.

Eigenvalues

Eigenvalues and eigenvectors are extremely important. In this video we review
the theory of eigenvalues. Consider a linear transformation
L : V −→ V

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