Download MATH 490 Section 1.1 1. Let c be a number and assume c0 = 0

MATH 490
S PRING 2012
H OMEWORK #1 S ELECTED S OLUTIONS
Section 1.1
1. Let c be a number and assume c0 = 0. Then:
c0 = c(0 + 0) by VS2
= c0 + c0 by VS5.
By VS3, there exists −c0 which we now add to both sides to get: c0 − c0 = c0 + c0 − c0. This simplifies
to: 0 = c0.
8c. Consider the set of all pairs (x, y) such that x + 4y = 0. Thus, x = −4y. To show this set is a
subspace we must show it is closed under addition, closed under scalar multiplication, and contains 0.
(a) Let (x, y) and (x0 , y 0 ) be pairs in the set. Then, by definition, x = −4y and x0 = −4y 0 . Thus,
(x, y) + (x0 , y 0 ) = (x + x0 , y + y 0 ) = (−4y − 4y 0 , y + y 0 ) = (−4(y + y 0 ), y + y 0 ) and consequently the sum
is also in the set.
(b) Let c be a number and (x, y) be an element in the set. Then c(x, y) = (cx, cy) = (−4cy, y) which is
an element in the set. Thus the set is closed under scalar multiplication.
(c) Finally (0, 0) = (−4 · 0, 0) and thus is in the set.
So by checking the three criteria, we know the set is a subspace.
10. I will only prove the part of the problem that some had trouble with.
(closed under addition.) Let x, y ∈ U +W. Then, by the definition of a sum of vector spaces, x = u+w
and y = u0 + w0 where u, u0 ∈ U and w, w0 ∈ W. So, x + y = u + w + u0 + w0 = (u + u0 ) + (w + w0 ).
Since U is a subspace, then u + u0 ∈ U . Since W is a subspace, w + w0 ∈ W. So x + y ∈ U + W. So
U + W is closed under addition.
Section 2
1h. To show the vectors are linearly independent we need to the only solution to the equation below is
a = b = c = 0.
Using the given equations we get: a(0, 1, 1) + b(0, 2, 1) + c(1, 5, 3) = (0, 0, 0). The first coordinate forces
c = 0. Using this fact and the last two coordinates we get the system of equations: (1) a + 2b = 0 and
(2) a + b = 0. Subtracting (2) from (1) we get: b = 0. This forces a = 0. Thus, the vectors are linearly
independent.
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3c We need to find the constants a, b, c such that:
a(1, 1, 1) + b(−1, 1, 0) + c(1, 0, −1) = (0, 0, 1). This immediately translates to a system of equations:
a − b + c = 0, a + b = 0 and a − c = 1.
The solution to this system is: (a, b, c) = (1/3, −1/3, −2/3).
4. Note that if a = b = c = d = 0, then certainly both ad − bc = 0 and (a, b) and (c, d) are linearly
dependent. Thus we can assume without loss of generality that a 6= 0. Thus, given the equation:
x(a, b) + y(c, d) = (0, 0)
we get the following system:
xa + yc = 0 and xb + yd = 0.
Since a 6= 0, we can solve for x and plug in to get: x = −a−1 yc and −a−1 ycb + yd = 0. Thus,
y(ad − bc) = 0.
Now, if ad − bc = 0, then there exist multiple solutions for y, so the vectors are linearly dependent.
On the other hand, if ad − bc 6= 0, then the only solution for y is 0. Thus, x = −a−1 c · 0 = 0. Since
only the trivial solution is possible, in this case, the vectors are linearly independent.
10. Assume v 6= 0. If w = 0, then let a = 0 and certainly av = w. So assume w 6= 0. Since v and w are
linearly dependent, we know there exist nontrivial solutions to av + bw = 0. Since neither v nor w are the
zero vector, we know neither a nor b can be zero. So, we can solve for w to get: w = −b−1 av.
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