Page 410 - 35Linear Algebra

410 Movie Scripts

Remember that an eigenvector v with eigenvalue λ for M will be a vector such
~
that Mv = λv i.e. M(v) − λI(v) = 0. When we are talking about a nonzero v
then this means that det(M − λI) = 0. We will start by finding the eigenvalues
that make this statement true. First we compute

4 2 λ 0 4 − λ 2
det(M − λI) = det − = det
1 3 0 λ 1 3 − λ
so det(M − λI) = (4 − λ)(3 − λ) − 2 · 1. We set this equal to zero to find values
of λ that make this true:
2
(4 − λ)(3 − λ) − 2 · 1 = 10 − 7λ + λ = (2 − λ)(5 − λ) = 0 .
This means that λ = 2 and λ = 5 are solutions. Now if we want to find the
eigenvectors that correspond to these values we look at vectors v such that

4 − λ 2 ~
v = 0 .
1 3 − λ
For λ = 5

4 − 5 2 x −1 2 x
~
= = 0 .
1 3 − 5 y 1 −2 y
This gives us the equalities −x+2y = 0 and x−2y = 0 which both give the line

2
1
y = x. Any point on this line, so for example , is an eigenvector with
2 1
eigenvalue λ = 5.
Now lets find the eigenvector for λ = 2

4 − 2 2 x 2 2 x
~
= = 0,
1 3 − 2 y 1 1 y
which gives the equalities 2x + 2y = 0 and x + y = 0. (Notice that these equa-
tions are not independent of one another, so our eigenvalue must be correct.)

x 1
This means any vector v = where y = −x , such as , or any scalar
y −1
multiple of this vector , i.e. any vector on the line y = −x is an eigenvector
with eigenvalue 2. This solution could be written neatly as

2 1
λ 1 = 5, v 1 = and λ 2 = 2, v 2 = .
1 −1

Jordan Block Example

Consider the matrix

λ 1
J 2 = ,
0 λ
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