Linear algebra - Practice problems for final

 
3 0 0
1. Diagonalize the matrix −3 4 9.
0 0 3

Solution.
To find the eigenvalues, compute
 
3−λ 0 0
det  −3 4 − λ 9  = (3 − λ)(4 − λ)(3 − λ).
0 0 3−λ

So the eigenvalues are λ = 3 and λ = 4.    

3 1
We can find two linearly independent eigenvectors 0 , 3 corresponding to the eigenvalue 3, and one
1 0
 
0
eigenvector 1 with eigenvalue 4. The diagonalized form of the matrix is
0
     
3 0 0 3 1 0 3 0 0 0 0 1
−3 4 9 = 0 3 1 0 3 0  1 0 −3 .
0 0 3 1 0 0 0 0 4 −3 1 9

Note that if you chose different eigenvectors, your matrices will be different. The middle matrix should have
entries 3, 3, 4 in some order, and you should multiply out the product to make sure you have the right answer.
 k
1 −6
2. Find a formula for by diagonalizing the matrix.
2 −6

Solution. The eigenvalues are −3, −2, and the diagonalized form of the matrix is
     
1 −6 3 2 −3 0 −1 2
= .
2 −6 2 1 0 −2 2 −3

It follows that k 
3 2 (−3)k
   
1 −6 0 −1 2
= .
2 −6 2 1 0 (−2)k 2 −3

3. Let B = {b1 , b2 } and C = {c1 , c2 } be two bases for R2 such that b1 = 6c1 − 2c2 and b2 = 9c1 − 4c2 .

(a) Find the change of coordinates matrix from B to C.

 
−3
(b) If the vector v has coordinate vector vB = , find vC .
2

Solution.

1
  
6 9
(a) The change of coordinate matrix from B to C is . Note that for instance the fact that
−2 −4
    
6 9 1 6
=
−2 −4 0 −2
 
1
corresponds to the statement that the coordinate vector of b1 relative to the basis B is , whereas
0
 
6
relative to the basis C it is .
−2
    
6 9 −3 0
(b) vC = = .
−2 −4 2 −2
     
1 0 0
0 1 0 4
4. Find the projection of b = 
−1 onto the subspace W = sp 1 , 1 of R .
    

1 0 1
 
0 1 1 0
Solution. First find the orthogonal complement of W . This is the nullspace of the matrix . A
0 0 1 1
basis for this nullspace (found by row reduction) is
   

 1 0 
   
0
 , 1  .

 0 −1
 
0 1
 

Now we need to find the coordinate vector of b relative to the basis

       

 0 0 1 0 
       
1 , 0 , 0 ,  1 

 1 1 0 −1
 
0 1 0 1
 
 
−2/3
 1/3 
Again, by row reduction, we find the coordinate vector   1 . To find the projection of b onto W we only


2/3
take the part of the coordinate vector that corresponds to basis elements in W . We get
     
0 0 0
−2 
1 + 1 0 = −2/3 .
    
bW =
3 1  3 1
  −1/3
0 1 1/3

   
1 −1
2  1  4
1 ,  0  of R . Use this to compute the
5. Find the projection matrix onto the subspace W = sp     

1 1
 
1
1
projection of the vector 
2 onto W .


2
  
1 −1
2 1  T −1 T
1 0 . The projection matrix is A(A A) A . Computing this we get
Solution. Let A =  

1 1
 
14 1 5 −4
1 
 1 11 4 7  .
A(AT A)−1 AT =

17 5
 4 3 1
−4 7 1 6
 
1
1
The projection of 
2 onto W is then


1
    
14 1 5 −4 1 21
 1 11 4 7  1
   = 1 27
 .
5 4 3 1  2 17 16
−4 7 1 6 1 11

6. Find the best fit line y = mx + c through the data points (0, 0), (1, 1), (2, 3).

Solution. Finding a line y = mx + c through those points amounts to solving the system
   
0 1   0
1 1 m = 1 .
c
2 1 3

This system is inconsistent (since

 there is no line through those
points),
  and
 to find the best fit line, we need
0 0 1
to project the vector b = 1 onto the column space W = sp 1 , 1 of the matrix. The projection is
3 2 1
     
0 1 −1/6
3   −1   
bW = 1 + 1 = 8/6  .
2 6
2 1 17/6

The best fit line is given by the solution of

   
0 1   −1/6
1 1 m =  8/6  ,
c
2 1 17/6

which is m = 3/2, c = −1/6.