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Solutions to Tutorial 3
Simon Rose
March 21, 2013
Recall that a subset W ⊂ V is a subspace if
1. The zero vector ~0 ∈ W .
2. For every w, v ∈ W we have that w + v ∈ W .
3. For every w ∈ W and for every α ∈ R, we have that α · w ∈ W
Remember also that the zero vector of R2 is ~0 = (0, 0).
Problem 1 (a): S = {(x, y) ∈ R2 | 6x + 8y = 0}.
Let us check the three conditions:
(a) Let us check that (x, y) = (0, 0) satisfies the given condition.
6x + 8y = 6 · 0 + 8 · 0 = 0 + 0 = 0
and so ~0 ∈ S.
(b) Suppose that w, v ∈ S. Then
w = (x1 , y1 )
such that 6x1 + 8y1 = 0
v = (x2 , y2 )
such that 6x2 + 8y2 = 0
Then w + v = (x1 + x2 , y1 + y2 ) (by definition of addition). Let us
consider the defining condition for S: we want that 6(x1 + x2 ) +
8(y1 + y2 ) = 0. Let us compute this:
6(x1 + x2 ) + 8(y1 + y2 ) = 6x1 + 6x2 + 8y1 + 8y2
= 6x1 + 8y1 + 6x2 + 8y2
| {z } | {z }
=0
=0
=0+0=0
and so w + v ∈ S, meaning that this also satisfies condition 2.
(c) Lastly, let us consider multiplication. Again, consider some w =
(x1 , y1 ) ∈ S. Because w, ∈ S, we have that 6x1 + 8y1 = 0. So choose
α ∈ R, and consider the product
α · w = α · (x1 , y1 ) = (αx1 , αy1 )
1
Then we have that
6(αx1 ) + 8(αy1 ) = α(6x1 ) + α(8y1 )
= α(6x1 + 8y1 )
| {z }
=0
=α·0=0
and so α · w satisfies the defining condition of the subset S.
It follows that S satisfies all of the conditions to be a subspace of V .
Problem 1 (g): S = {(x, y) ∈ R2 | x2 + y 2 = 1}
This is not a subspace, since the very first condition fails: 02 + 02 6= 1,
and so ~0 ∈
/ S.
Problem 1 (h): S = {(x, y) ∈ R2 | xy = 0}
The first and the third conditions are satisfied (This implies that this set
is what mathematicians call a cone), which you can and should verify for
practice.
The second condition, however, fails. To show this, we want to choose
w, v ∈ S such that w + v ∈
/ S. So let us consider w = (1, 0) and v = (0, 2).
Note that both of these satisfy the condition that xy = 0. Then
w + v = (1, 0) + (0, 2) = (1, 2)
Now, the defining condition for S is that the product of the two coordinates
is 0. However, 1 · 2 = 1 6= 0, and so this does not satisfy that condition.
Hence it is not in S as claimed, and so S is not a subspace of R2 .
Problem 4: Show that if W is a subspace of (V, +, ·), then (W, +, ·) is also a vector
space.
Proof We first note that since addition and multiplication are simply
those from V , they automatically satisfy the following properties (why
should this be so? Think about this and convince yourself!):
(a) Associativity of +
(b) Commutativity of +
(c) Associativity of ·
(d) Distributivity of +, · (both versions)
(e) 1 · x = x
Thus it only remains to verify the following two properties.
(a) There exists a zero element in W
(b) For all v ∈ W , there is an element (−v) ∈ W such that v + (−v) = ~0
2
The first property is true by assumption that W is a subspace; in particular, it included the assumption that ~0 ∈ W .
For the last, choose some v ∈ W . Since W is a subspace, and since
(−1) ∈ R, it follows that (−1) · v ∈ W . But
v + (−1) · v = 1 + (−1) · v = 0 · v = 0
as we have seen earlier, and so this last condition is satisfied. Thus (W, +, ·)
is a vector space as claimed.
Q.E.D.
Remark Alternatively, we could have used the earlier Theorem from
class which stated that
(−1) · v = (−v)
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