Download Math 110 Homework 1 Solutions

Math 110 Homework 1 Solutions
January 30, 2013
3.) Prove that −(−v) = v for every v ∈ V .
Solution 1 : Given w ∈ V , the definition of −w (according to page 11 in the text) is
the unique vector (by proposition 1.3) −w such that w + (−w) = 0. Therefore by
definition we have −v + −(−v) = 0. Adding v to both sides of the equation gives
v + (−v) + −(−v) = 0 + v = v. We know that v + (−v) = 0 by definition of −v. So
the left hand side of the equation becomes
v + (−v) + −(−v) = (v + (−v)) + −(−v) = 0 + −(−v) = −(−v)
The right hand side of the equation was v so we conclude that v = −(−v) as desired.
Solution 2 : By proposition 1.6, (−1)w = −w for all w ∈ V . Therefore,
−(−v) = (−1) · (−v) = −1 · (−1 · v) = (−1 · −1) · v = 1 · v = v,
giving the desired equality. While this is a perfectly acceptable solution, it does not
readily use the definition of −w.
4.) Prove that if a ∈ F, v ∈ V , and av = 0 then either v = 0 or a = 0.
Solution: Clearly, a can either be 0 or not so all we need to do is assume that a 6= 0
and prove that v = 0. Since a 6= 0 it has a multiplicative inverse, a1 . Therefore,
multiplying both sides of the equation av = 0 by a1 gives:
1
1
· (a · v) = · 0
a
a
The right hand side of the above equation is 0 by Proposition 1.5. and the left hand
side of the equation is
1
1
· (a · v) = ( · a) · v = 1 · v = v
a
a
so we conclude v = 0.
1
(Remark : This approach does not work if we assume v 6= 0 and try to prove that
a = 0. Can you see why?)
5.) For each of the following subsets of F3 , determine whether it is a subspace of F3 :
(a) {(x1 , x2 , x3 ) ∈ F3 : x1 + 2x2 + 3x3 = 0}
(b) {(x1 , x2 , x3 ) ∈ F3 : x1 + 2x2 + 3x3 = 4}
(c) {(x1 , x2 , x3 ) ∈ F3 : x1 x2 x3 = 0}
(d) {(x1 , x2 , x3 ) ∈ F3 : x1 = 5x3 }
Solution: (a) : This is a subspace of F3 . To see this, we must check that this set
contains 0, and is closed under addition and scalar multiplication.
To see that the set contains 0, we note that 0 ∈ F3 is the tuple (0, 0, 0). We see that
if (x1 , x2 , x3 ) = (0, 0, 0) then x1 + 2x2 + 3x3 = 0 + 2(0) + 3(0) = 0 which means 0 is
in the set.
To see closure under addition, this means that we need to show x + y is in the
set whenever x and y both are. By definition, we can write x = (x1 , x2 , x3 ) and
y = (y1 , y2 , y3 ) with
x1 + 2x2 + 3x3 = 0 and y1 + 2y2 + 3y3 = 0
The element x + y is the vector (x1 + y1 , x2 + y2 , x3 + y3 ) and we see that
(x1 + y1 ) + 2(x2 + y2 ) + 3(x3 + y3 ) = x1 + 2x2 + 3x3 + y1 + 2y2 + 3y3 = 0 + 0 = 0
which proves closure under addition.
To see closure under scalar multiplication, Let λ ∈ F and x = (x1 , x2 , x3 ) a vector in
the given set. Then λ · x = (λx1 , λx2 , λx3 ) and we have:
λx1 + 2λx2 + 3λx3 = λ(x1 + 2x2 + 3x3 ) = λ · 0 = 0
so we have closure under scalar multiplication and therefore this set is a subspace of
F3 .
(b) : This is not a subspace of F3 . The easiest way to see this is that it does not
contain 0 = (0, 0, 0). Indeed, the coordinates (x1 , x2 , x3 ) of the zero vector satisfy
x1 + 2x2 + 3x3 = 0 6= 4 as seen in part (a).
(c) : This is not a subspace of F3 . We note that x = (1, 0, 0) and y = (0, 1, 1) are
both in the set (as each has at least one coordinate and the requirement for being in
this set is that the product of the coordinates is zero). The vector x + y is (1, 1, 1).
The product of the three coordinates is 1 so it is not in the set. Therefore the set is
not closed under addition so it is not a subspace of F3 .
(d) : This is a subspace of F3 . First, it contains 0 = (0, 0, 0) since the first coordinate
is clearly five times the third.
Second, let x and y be in the set, meaning that if we write x = (x1 , x2 , x3 ) and
y = (y1 , y2 , y3 ) then x1 = 5x3 and y1 = 5y3 . The vector x+y is (x1 +y1 , x2 +y2 , x3 +y3 )
and we have x1 + y1 = 5x3 + 5y3 = 5(x3 + y3 ), meaning that x + y is in the set.
Finally, let x be in the set and λ ∈ F. Then if we write x = (x1 , x2 , x3 ), we have
x1 = 5x3 . The vector λ · x is (λ · x1 , λ · x2 , λ · x3 ) and we have λx1 = λ(5x3 ) = 5(λx3 )
so λx is in the set, proving closure under scalar multiplication. Therefore, the set is
a subspace of F3 .
6.) Give an example of a nonempty set U of R2 such that U is closed under addition
and under additive inverses but U is not a subspace of R2 .
Solution: Consider the set U = {(n, 0) : n ∈ Z} (Z denotes the set of integers). Let
v and w ∈ A. Then there are integers n and m such that v = (n, 0) and w = (m, 0).
We have v + w = (n + m, 0) and n + m is an integer if n and m both are so the set
is closed under addition so U is closed under addition. Also, −v = (−m, 0) and −m
is an integer if m is, so U is closed under additive inverses. However, U is not closed
under scalar multiplication. Indeed, 21 · (1, 0) = ( 21 , 0) 6∈ U even though (1, 0) ∈ U .
Therefore, U has the desired properties.
7.) Give an example of a nonempty subset U of R2 such that U is closed under scalar
multiplication but not addition.
Solution: Let U consist of both coordinate axes, i.e.
U = {(x, 0) : x ∈ R} ∪ {(y, 0) : y ∈ R} =
{z ∈ R2 : z = (x, 0) with x ∈ R or z = (0, y) with y ∈ R}
To see that U is closed under multiplication, let z ∈ U and λ ∈ R. Then either
z = (x, 0) for some real x or z = (0, y) for some real y. In the first case, λ · z =
(λx, λ · 0) = (λx, 0) which is in U . In the second case, λ · z = (λ · 0, λy) = (0, λy)
which is in U . U is not closed under addition since the vectors (1, 0) and (0, 1) are in
U but their sum, (1, 1) is not.