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Chapter Four Exercises
September 30, 2008
1. Use the vector space axioms to show (x1 + x2 ) + x3 = (x1 + x3 ) + x2 : Identify the
axiom used at each step. Answer: (x1 + x2 ) + x3 = (assoc:)x1 + (x2 + x3 ) =
(comm:)x1 + (x3 + x2 ) = (assoc:)(x1 + x3 + x2
2. Let V = R4 . Evaluate the following:
a)
b)
c)
d)
e)
(2; 1; 3; 1) + (3; 1; 1; 1). Answer: (5; 2; 4; 0)
(2; 1; 5; 1) (3; 1; 2; 2).Answer: ( 1; 0; 3; 1)
10 (2; 0; 1; 1). Answer: (20; 10; 10)
(1; 2; 3; 1) + 10 (1; 1; 0; 1) 3 (0; 2; 1; 2). Answer: (11; 18; 0; 17)
x1 (1; 0; 0; 0)+x2 (0; 1; 0; 0)+x3 (0; 0; 1; 0)+x4 (0; 0; 0; 1). Answer: (x1 ; x2 ; x3 ; x4 )
3. Determine whether the given subset of R is a subspace or not (Explain):
n
a) W = f(x; y ) 2 R2 j xy = 0g. Answer: Not closed under sum (e.g. (0; 1) +
(1; 0) = (1; 1) is not a solution of xy = 0
b) W = f(x; y; z ) 2 R3 j 3x + 2y 2 + z = 0g. Answer: Not closed under sum (e.g.
(0; 1; 2) + (0; 1; 2) is not a solution.
c) W = f(x; y; z ) 2 R3 j 2x + 3y z = 0g. Answer: Yes, given two solutions
(x1 ; y1 ; z1 ) and (x2 ; y2 ; z2 ) then expand 2(x1 + x2 ) + 3(y1 + y2 ) (z1 + z2 ) and
2(r x1 ) + 3(r y1 ) (r z1 ) and to show zero (& hence solutions). You will need
the fact the given vectors are solutions.
d) The set of all vectors (x1 ; x2 ; x3 ) satisfying
2x3 = x1
Answer:
10x2 :
Yes, same reasoning as in previous item.
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e) The set of all vectors in R4 satisfying the system of linear equations
2x1 + 3x2 + 5x4 = 0
x1 + x2 3x3 = 0
Yes, same reasoning as in previous item.
f) The set of all points (x1 ; x2 ; x3 ; x4 ) 2 R4 satisfying
Answer:
x1 + 2x2 + 3x3 + x4 = 1:
Answer:
No, the zero vector is not in the set.
4. Write the product
(2[0 3]
;
3[3 6] )([0 2] + 5[2 4]
;
;
7[4 6] )
;
;
as a linear combination of the indicator functions of intervals. Answer: Use distributivity and the fact that = to show the expression equals:
A
B
A\B
= 2[0 3] [0 2] + 10[0 3] [2 4] 14[0 3] [4 6] 3[3 6] [0 2] 15[3 6] [2 4] + 21[3 6] [4 6] )
;
;
;
;
;
= 2[0 2] + 10[2 3]
;
;
;
14(0)
;
3(0)
;
;
;
;
;
15[3 4] + 21[4 6]
;
;
5. Determine whether the given subset of the vector space of polynomials with real
coecients,P , is a subspace or not (Explain):
n
a) All polynomials of the form p(t) = at2 for a in R. Answer: Yes, the sum and
scalar products are in the set.
b) All polynomials of the form p(t) = a + t2 for a in R. Answer: No, t2 + 1 + t2 =
1 + 2t2 which is not in the set.
c) All polynomials of degree less than three with integer coecients. Answer:
Not closed under scalar multiplication by real numbers, e.g. 12 t
6. Let A be the set of all functions f (x) on [0; 1] which are dierentiable and satisfy
f (x) + 2f (x) = 0. Is this a subspace of the space of all functions? Answer: Yes, use
the facts that the derivative of a sum is the sum of the derivatives and the derivative
of a scalar multiple of a function is the scalar multiple of the derivative to show that
the sum and scalar multiple of solutions is also a solution.
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