2
$\begingroup$

The idea is to prove that this is a vector space based upon the following axioms:

  1. $\mathbf{u}+\mathbf{v}$ is in $V$. Closure under addition.
  2. $\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}$. Commutative property.
  3. $\mathbf{u}+(\mathbf{v}+\mathbf{w}) = (\mathbf{u}+\mathbf{v})+\mathbf{w}$. Associative property.
  4. $V$ has a zero vector $\mathbf{0}$ such that for every $\mathbf{u}\in V$, $\mathbf{u}+\mathbf{0}=\mathbf{u}$. Additive identity.
  5. For every $\mathbf{u}\in V$, there is a vector in $V$ denoted by $-\mathbf{u}$ such that $\mathbf{u}+(-\mathbf{u}) = \mathbf{0}$. Additive inverse.
  6. $c\mathbf{u}$ is in $V$. Closure under scalar multiplication.
  7. $c(\mathbf{u}+\mathbf{v}) = c\mathbf{u}+c\mathbf{v}$. Distributive property.
  8. $(c+d)\mathbf{u}=c\mathbf{u}+d\mathbf{u}$. Distributive property.
  9. $c(d\mathbf{u})= (cd)\mathbf{u}$. Associative property.
  10. $1(\mathbf{u}) =\mathbf{u}$. Scalar identity.

The exact problem: prove that the set of all quadratic functions whose graphs pass through the origin with standard operations is a vector space. I specifically have two questions in addition to proving the vector space. One, how does passing through the origin affect the supposed vector space - my assumption is that it affects axiom one? And two, how would the vectors be described?

Any help is appreciated.

$\endgroup$
9
  • 4
    $\begingroup$ Not quite true, unless you think of $3x$ as quadratic function. This is because for example $x^2+x$ is a quadratic that passes through the origin, as is $-x^2+2x$. Their sum is $3x$. So we will have to allow all things of shape $ax^2+bx$, where $a$ and/or $b$ may be $0$. $\endgroup$
    – André Nicolas
    Oct 2, 2012 at 2:21
  • $\begingroup$ That would be that this is not a vector space, as $3x$ is not quadratic. Am I correct in saying that? $\endgroup$
    – Rick
    Oct 2, 2012 at 2:24
  • 2
    $\begingroup$ The set of all polynomials of degree $\le 2$, under usual addition, is also a vector space over the reals. Just a different vector space than yours. Added Yes, if you keep $3x$ out, then the closure under addition condition is violated. If you keep the $0$ polynomial out, you lose closure under multiplication by scalars. $\endgroup$
    – André Nicolas
    Oct 2, 2012 at 2:24
  • 5
    $\begingroup$ The problem may have been carelessly posed. Your answer could be: If quadratic functions are really intended, meaning coefficient of $x^2$ cannot be $0$, then we do not have a vector space because $\dots$. If we allow all $ax^2+bx$, then we do get a vector space because $\dots$. Then you need to verify all the axioms. All the verifications are easy. $\endgroup$
    – André Nicolas
    Oct 2, 2012 at 2:29
  • 2
    $\begingroup$ OK, but it may be a trick question, so you should point out that if we don't allow $a=0$, or even $a=b=0$, we don't have vector space. $\endgroup$
    – André Nicolas
    Oct 2, 2012 at 2:35

1 Answer 1

Reset to default
1
$\begingroup$

Whether or not we have a vector space depends on how you interpret "quadratic function whose graph passes through the origin."

In order for us to have a vector space, we will need, for example, to think of $3x$ as a quadratic function. For certainly $x^2+x$ is a quadratic function that passes through the origin, as is $-x^2+2x$. So if we are to have closure under addition, the function $(x^2+x)+(-x^2+2x)$, that is, $3x$, will have to be in the collection.

Indeed the identically $0$ function has to be in our collection, for two reasons. If we do not allow it, then closure under addition can fail, since $(x^2-x)+(-x^2+x)=0$. Closure under multiplication by the constant $0$ also fails.

It is not difficult, however, to show that the set of functions of the form $ax^2+bx$, where $a$ and $b$ range over all the reals, is a vector space over the reals. Quite a number of axioms need to be verified, but each verification is easy.

As to your question about the "passing through $0$" part, one can also show that the set of all polynomial functions of degree $\le 2$ also is a vector space over the reals. It just is a different vector space than the one under consideration.

Remark: $1.$ In answering the homework question, it may be useful to be cautious. I would suggest doing the following: (i) Observe that if we interpret "quadratic" as meaning that the coefficient of $x^2$ is non-zero, then we do not have a vector space. (Of course one should explain why.) (ii) Show in detail that the collection of all functions of the shape $ax^2+bx$ is a vector space. Many of the steps can be possibly omitted. The key properties that have to be verified are closure under addition and under multiplication by scalars.

$2.$ You asked how to describe the vectors. Might as well use the standard polynomial notation, as in the answer above. The "pass through the origin" part just means the constant term is $0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .