Proving Vector Space Axioms for f(x) = ax+b, a,b Real Numbers

[THarper]

Question: Show that the set of all functions of the form f(x) = ax+b, with a and b real numbers forms a vector space, but that the same set of functions with a > 2 does not.

Equations: the axioms for vector spaces

Attempt:
I think that the axiom about the zero vector is the one I need to use, but I can't figure out how to show this, or how it only affects a>2 functions. Think I may not be visualising the vector space correctly.

 

[Mark44]

Question: Show that the set of all functions of the form f(x) = ax+b, with a and b real numbers forms a vector space, but that the same set of functions with a > 2 does not.

Equations: the axioms for vector spaces

Attempt:
I think that the axiom about the zero vector is the one I need to use, but I can't figure out how to show this, or how it only affects a>2 functions. Think I may not be visualising the vector space correctly.

There are several vector space axioms that aren't satisfied if a > 2, not just the one about the zero vector.

 

[vela]

Question: Show that the set of all functions of the form f(x) = ax+b, with a and b real numbers forms a vector space, but that the same set of functions with a > 2 does not.

Equations: the axioms for vector spaces

Attempt:
I think that the axiom about the zero vector is the one I need to use, but I can't figure out how to show this, or how it only affects a>2 functions. Think I may not be visualising the vector space correctly.

And you need to use all the axioms for this problem because you're first asked to verify that the original set is a vector space.

 

[RUber]

Take the advice above. Work through the axioms one at a time, and you will see where it breaks. As far as visualizing the space, the function space of linear functions ax+b is the set of lines with slope a and y-intercept b. The zero function in this space is the horizontal line at zero. That is, f(x) = 0x+0. So I think your intuition is correct that you can show that the zero is not in the space. That is not the same as saying there is no function in the space that can touch the origin, but instead that there is no one function in the space which is identically zero for all x .

 

[THarper]

I've worked through the axioms, and have found that all of them hold apart from two in the case of a<2.

These two are:
There exists a zero vector such that f(x) + 0(x) = f(x) which can't work for a>2, since no zero function can exists as f(x) = (+ve integer)x + b

And that for any f(x), there exists a function -f(x), such that f(x) + (-f(x)) = 0(x).
This can't be true for a>2 since the x-coefficient of (-f(x)) couldn't possibly cancel the positive coefficient of f(x), since both can only be positive, so the zero function (which has an x coefficient of 0) could never be achieved.

Is this correct? All of the other rules seem to hold, with or without the constraint on a.

 

[Mark44]

I've worked through the axioms, and have found that all of them hold apart from two in the case of a<2.

Did you mean a > 2?

These two are:
There exists a zero vector such that f(x) + 0(x) = f(x) which can't work for a>2, since no zero function can exists as f(x) = (+ve integer)x + b

I get your reasoning, but this isn't the cleanest way to say it. You need to show what the "zero" function has to look like, and show why f(x) + 0(x) can't be equal to f(x).

And that for any f(x), there exists a function -f(x), such that f(x) + (-f(x)) = 0(x).
This can't be true for a>2 since the x-coefficient of (-f(x)) couldn't possibly cancel the positive coefficient of f(x), since both can only be positive, so the zero function (which has an x coefficient of 0) could never be achieved.

Is this correct? All of the other rules seem to hold, with or without the constraint on a.

Let g(x) = 3x + b, which is clearly a member of the set. For any choice of a constant k, is kg also a member of the set?

 

[vela]

IThere exists a zero vector such that f(x) + 0(x) = f(x) which can't work for a>2, since no zero function can exists as f(x) = (+ve integer)x + b

I don't know if this was just a typo, but the coefficient of x doesn't have to be an integer.

 

[Fredrik]

Since that second set is a subset of the first, you only need to check the three usual things to determine if it's a subspace.

But you still need to know how to check if the vector axioms are satisfied, so please continue along that path. It's only a little harder.