Page 409 - 35Linear Algebra

G.11 Eigenvalues and Eigenvectors 409

   
1 0 1 0 0 0 1 0 1 0 0 0
0 1 1 0 3 0 0 1 1 0 3 0
   
∼   ∼  
 0 0 0 1 1 0   0 0 0 1 1 0 
0 0 0 0 2 1 0 0 0 0 1 3
 
1 0 1 0 0 0
 0 1 1 0 0 1 
∼  
 0 0 0 1 0 2 
0 0 0 0 1 3
The pivots are underlined. The columns corresponding to non-pivot variables
are the ones that can be eliminated--their coefficients (the α’s) will be
arbitrary, so set them all to zero save for the one next to the vector you are
solving for which can be taken to be unity. Thus that vector can certainly be
expressed in terms of previous ones. Hence, altogether, our basis is
       
 1 0 0 0 
 
3
2
0
1
 
       
  ,   ,   ,   .
 3   2   0   1  

 
4 1 0 0
 
Finally, as a check, note that e 1 = v 1 + v 2 which explains why we had to throw
it away.
Hint for Review Problem 2
n
n
Since there are two possible values for each entry, we have |B | = 2 . We note
1
n
that dim B = n as well. Explicitly we have B = {(0), (1)} so there is only 1
1
basis for B . Similarly we have

0 1 0 1
2
B = , , ,
0 0 1 1
and so choosing any two non-zero vectors will form a basis. Now in general we
note that we can build up a basis {e i } by arbitrarily (independently) choosing
the first i−1 entries, then setting the i-th entry to 1 and all higher entries
to 0.
G.11 Eigenvalues and Eigenvectors

2 × 2 Example

Here is an example of how to find the eigenvalues and eigenvectors of a 2 × 2
matrix.

4 2
M = .
1 3
409