Page 236 - 35Linear Algebra

236 Eigenvalues and Eigenvectors

λ = 1: Again we set up an augmented matrix and find the solution set:

   
1 1 −1 0 1 1 −1 0
1 1 −1 0 ∼ 0 0 0 0 .
   
−1 −1 1 0 0 0 0 0
Then the solution set has two free parameters, s and t, such that z = z =: t,
y = y =: s, and x = −s + t. Thus L leaves invariant the set:

  
  
−1 1
 

s  1  + t   s, t ∈ R .
0

0 1
 
This set is a plane through the origin. So the multiplicity two eigenvalue has
   
−1 1
0
two independent eigenvectors,  1  and   that determine an invariant
0 1
plane.
Example 129 Let V be the vector space of smooth (i.e. infinitely differentiable)
functions f : R → R. Then the derivative is a linear operator d : V → V . What are
dx
the eigenvectors of the derivative? In this case, we don’t have a matrix to work with,
so we have to make do.
A function f is an eigenvector of d if there exists some number λ such that
dx
d
f = λf .
dx
λx
λx
An obvious candidate is the exponential function, e ; indeed, d e λx = λe . The
dx
operator d has an eigenvector e λx for every λ ∈ R.
dx
12.3 Eigenspaces

In the previous example, we found two eigenvectors
   
−1 1
1 and 0
   
0 1

for L, both with eigenvalue 1. Notice that
     
−1 1 0
1 + 0 = 1
     
0 1 1

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