Download 1 Why is a parabola not a vector space

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Why is a parabola not a vector space
x
x
is
a
real
number
inside R2 . This is the
x2
parabola y = x2 in R2 . We shall prove that this is NOT a vector space. A
vector space has to satisfy three properties Consider the set A =
1. The zero vector is in the vector space.
2. Closed under addition
3. Closed under scalar multiplication
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Notice that the zero vector
is in the set A.
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We shall show that A is NOT closed under addition or scalar multiplication.
Our strategy wil be producing a specific counterexample.
Not closed under addition
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2
Let us look at the two vectors ~v1 =
and ~v2 =
. These vectors are
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obtained by plugging in x = 1 andx = 2 in the parabola equation y = x2 .
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Consider the vector ~v1 + ~v2 =
. Is this vector in A ? Let us suppose
5
for the sake of contradiction
that
~v1 +~v2 was in A. If the vector ~v1 + ~v2 was in
3
x
A, then we have that
=
, for some real number x. Solving for x,
5
x2
the first row gives us that x = 3 and the second row gives us that x2 = 5. Both
of these cannot be simultaneously true. Hence, the only possible conclusion is
that ~v1 + ~v2 is NOT in A.
We have displayed two vectors ~v1 and ~v2 such that their sum is NOT in A.
Hence, we have shown that A is NOT closed under addition.
Not closed under scalar multiplication
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Let us look at the vector ~v1 =
. Consider the scalar k = 2. Look at
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2
k · ~v1 =
. Is this vector in A ? Again, for the sake of contradiction, let
2
2
x
us suppose that it was in A. Then,
=
, for some real number x.
2
x2
Solving, the first
√ row gives us that x = 2 and solving the second number gives
us that x = ± 2. Both of these cannot be true simultaneously. Hence, the only
posssible conclusion is that the vector 2 · ~v1 is NOT in A. Hence the set A is not
closed under scalar multiplication.
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Remark : Negation of statements (optional)
“Every student in MATH 308L is a girl”. What is the negation of this statement? Is it the statement - “Every student in MATH 308L is a boy ”? No! If you
think about it, if you want to negate the statement “Every student in MATH
308L is a girl”, it is enough for one student in the class NOT to be a girl. Hence,
the correct negation would be “Atleast one student in MATH 308L is not a girl”.
A similar reasoning applies in the case when we are trying to prove that the
set A is NOT a vector space. To show that the property ‘closed under addition’
does not hold, it is enough to provide one pair of vectors that is NOT closed
under addition. Similarly, it is enough to provide one counterexample to show
that the property ‘closed under scalar multiplication’ does not hold.
If the MATH 308L class had 47 girls and 1 boy, the statement “Every student
in MATH 308L is a girl” would still be FALSE. It could happen that some of
the students in the MATH 308L class are girls, but since there is one boy, the
statement would be false.
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Indeed, in our set A, if you look at the following vectors in A - ~v0 =
0
1
1
and ~v1 =
, then ~v0 + ~v1 =
is also in A. So, there are some pairs
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1
of vectors whose sum is also in A.
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