Linear Algebra MA237
Fall 2009 Dr. Byrne
Homework for Section 5.1
Due 11/16
1. Determine whether the given vectors v1, v2, v3, v4 are eigenvectors of the given
matrix. If they are, state the corresponding eigenvalue.
1 2
A
;
 2 4
1 
 2
v1 =   ; v2 =   ; v3 =
1 
2
1
0 
1 ; v4 = 0 

 
1
v1: A  v1  
2
1
v2: A  v2  
2
2  1   5 
Eigenvector: =5

4 2 10
2  2 0
Eigenvector: =0

4  1  0
1
v3: A  v3  
2
1
v4: A  v4  
2
2 1 3

4 1 6
2  0  0 

4 0 0
Not an eigenvector.
Not an eigenvector. (!)
For the matrices below find (a) the characteristic polynomial, (b) the real eigenvalues,
(c) the eigenvector and (d) eigenspace corresponding to each eigenvalue. (e) Describe
the eigenspace geometrically (i.e., as two lines or something else).
5 8
2. A  

4 1
(a) characteristic polynomial:
8 
5 8
1 0 5  
A  I  
 

0


1   
4 1
0 1  4
5   1     32  0
2  6  27  0
(b) real eigenvalues: 2  6  27    3  9  0 so =-3,9
(c)eigenvectors:
1
2
-3-eigenvector is   , 9-eigenvector is  
1
1 
(d) eigenspaces:
  1 
s    and
 1 
 2 
s   
 1 
(e) eigenspaces geometrically are 2 lines
0
1
4

3. A   1  6  2 (Hint: 5 is an eigenvalue.)


 5
0
0 
(a) characteristic polynomial: 3  22  29  30  0
(b) real eigenvalues:  = 5, -6, -1
 1 
(c) eigenvectors:
5-eigenvector =  3 
 11
 1 
0 
-6-eigenvector = 1 
 
0 
 1 
5

-1-eigenvector =  9 
 25
 1 
(Hint: use row reduction to solve the linear systems, the eigenvectors are
the spanning vectors of the solution, the next problem is shown in more
detail.)
(d) eigenspaces:
  1 
  3  
s  11  and
  1 

 
 0  
  
s 1  and
 0  
  
(e) eigenspaces geometrically are 3 lines.
   1 
5 
 


9
s  25 
  1 
 
 
0 1 1 
4. A  1 0 1


1 1 0
(a) characteristic polynomial:
 
det(A-I)=det(  1

 1
1

1
1 
1  )=0
  
(-)*(2-1) - 1*(--1)+1*(1+) = -3+++1+1+=-3+3+2=3-3-2
(b) eigenvalues:
To factor polynomials of degree 3, first check possible rational roots: 1, 2
try +1: (-1) 3-3-2 … doesn’t work
try -1: (+1) 3-3-2 … works … yields 2--2 = (-2)*(+1)
roots are -1,+2 and -1
(c) eigenvectors:
=-1: Solve (A-I)x=0
1 1 1
(A-I)x = 1 1 1 x=0 


1 1 1
1 1 1
0 0 0 x+y+z=0  x=-y-z but y and z are free


0 0 0
 x   t  s   1  1
 y   t   t  1   s 0 
Two eigenvectors!
  
    
 z   s   0   1 
=2: Solve (A-I)x=0
1
 2 1

(A-I)x =  1  2 1  x=0 
 1
1  2
1 0  1
0 1  1  x-z=0, y-z=0  only z is free


0 0 0 
 x  s 
1
 y    s    s 1
   

 z   s 
1
 1 


(d) two distinct eigenspaces:  s 1  and
 1 
  
  1  1 
    
t  1   s  0  
  0   1 
    
(e) the geometry of the eigenspaces is the union of a line and a plane