practice midterm solutions
1.
V
is not a vector space because it violates VS8, distributivity of multiplication over scalar addi
tion. We just need one counterexample. Choose
α
= 1
, β
= 1
, v
= (0
,
1). Then
(
α
+
β
)
·
v
= (1 + 1)
·
(0
,
1)
= 2
·
(0
,
1)
=
parenleftbigg
2
·
0
,
1
2
parenrightbigg
=
parenleftbigg
0
,
1
2
parenrightbigg
,
whereas
α
·
v
+
β
·
v
= 1
·
(0
,
1) + 1
·
(0
,
1)
=
parenleftbigg
1
·
0
,
1
1
parenrightbigg
+
parenleftbigg
1
·
0
,
1
1
parenrightbigg
= (0
,
1) + (0
,
1)
= (0
,
2)
.
In fact, you could have chosen any
α, β
such that
α
negationslash
= 0
, β
negationslash
= 0
, α
+
β
negationslash
= 0, and the second
component of
v
negationslash
= 0, and you would have demonstrated the same contradiction.
2. Because we know the set
W
is contained in the vector space
R
3
, then we do not have to test
VS18; instead we only have to test closure under addition and scalar multiplication, in which
case
W
will be a
subspace
of
R
3
. If we did not know
W
lived in some larger vector space, we
would have to verify VS18.
Right away we can tell this will be a subspace because we are given a linear relation between the
components, and this relation has no constant terms (i.e. it is set equal to zero, not a nonzero
constant).
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 Summer '09
 Sprouse
 Linear Algebra, Addition, Multiplication, Scalar, Vector Space, linearly independent set

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