Math 20F Homework Assignments, Winter 2009 

HW9, due March 10:

5.2:#3,9,16
5.3:#4,17,19 and 

Problem A:  Let A denote the 2 by 2 matrix
.9   .2
.1   .8
and let  x be the column vector (.7, .3).
Find the limit of A^k x  as k  tends to infinity.

ANSWER: Col vector (2/3, 1/3).

6.2: #10,12,14,16

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HW8, due March 3:

5.1: #24-29, #31-32

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HW7, Due Feb 24

Problem A:
Let v1, v2  be a basis for a two-dimensional vector space V.
Suppose that T is a linear transformation from V to V defined by
T(v1) = a v1 + b v2,  T(v2) = c v1 + d v2.
(i)  Write down the 2 by 2 matrix that represents T wrt the basis v1, v2.
(ii) Write down the four entries in the 2 by 2 matrix that represents T wrt the basis w1, w2,
where w1 = v1 + v2, w2 = v1 - v2.  

ANSWER to Problem A:
(i)  
a  c
b  d
(ii)
B^-1 A B  where B is the matrix taking old to new, i.e.,  B = 
1   1
1  -1
Multiplying this out, we get  B^-1 A B = 
(a+b+c+d)/2   (a-c+b-d)/2
(a+c-b-d)/2   (a-c-b+d)/2


3.1:  #7, 17
3.2:  #22, 24, 31, 32, 33, 34, 35

Answer to #22:  Invertibility is equivalent to nonzero determinant, so this matrix
is NOT invertible. 
Answer to #24: The matrix with these columns has nonzero determinant (namely, 11), so it is invertible.
Thus the column vectors are linearly independent.
3.3:  #1, 19, 30


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HW6, Due Feb 17

Problem A:  Let A be an  m by n matrix with n > m.
Explain why the set of column vectors of A cannot be independent.

Problem B:  Let A be an m by n matrix with m > n.
Is it possible for the vector space CS(A) to have dimension m?

Problem C:  Let A be the 4 by 3 matrix
1  1  0
1  2  1
1  3  0
1  4  2
(i) Give an LU decomposition for A.
(ii)  Show the set of three columns vectors of A is independent.
(iii) Find a 4 by 1 vector which is not in CS(A).
(iv) Answer (i) and (ii) and (iii) for the 4 by 3 matrix PA in place of A,
where P is the 4 by 4 matrix 
0  0  1  0
0  0  0  1
1  0  0  0 
0  1  0  0 


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HW5, Due Feb 10

2.5: #10, 19
Hint: During row reduction to obtain LU, do not change pivots to 1,
keep them as they are--this will give U in echelon form
that isn't reduced, but that gives the simplest LU factorization.
For example, if the first row of A is 2 * * * * and the second row
is 3 * * * *,  you could begin by subtracting 3/2 times row 1 from row 2.

4.1: #2, 5, 8, 13, 16, 18
4.2: Problem A:
Let A be the 3 by 3 matrix
 2  1   3
-4  3  -1
-6  4  -2.
(i)Show that the col vector b = (b1, b2, b3)
is in CS(A) if and only if b3 = (7/5)b2 - (1/5)b1. 
(ii)  True or False: CS(A) is spanned by the two col vectors
(1,2,-1/5) and (0,1,7/5).
(iii) Use elementary row matrices to find the LU decomposition of A.
(iv) For any fixed vector b in CS(A) (see (i)), solve Ax = b for the vector
x=(x1,x2,x3), using x3 as a free variable.
Hint: The first entry in the answer x  will be x1 = (3/10)b1 - (1/10)b2.
The third entry in x will of course be the free variable x3.
(v)  True or False: NS(A) can be viewed as a straight line through the origin.

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HW4, Due Feb 3

2.3: #14,20,28

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HW3, Due Jan 27

1.7: #1, 5, 21, 31, 33, 35, 37, 39
1.8: #7, 19, 21, 22, 27, 31
1.9: #23, 25, 27, 31, 32, 35

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HW2, Due Jan 20

1.4: #13, 14, 15, 21, 22
1.5: #6, 8, 15, 26, 29, 30, 31, 32
1.6: #6 (balance the equation)

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HW1, Due Jan 13

1.1: #9, 12, 14, 17, 18
1.2: #9,11,14,23,25,29,31
1.3: #26, 29

(No Matlab hw is due on Jan 13.)
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