Is f a scalar multiple of g in Hom(V,F)?

[Robert1986]

Homework Statement

Let V be a vector space over the field F. Let Hom(V,F) be the set of all homomorphisms of V into F (this is a pretty standard definition, noting new here.) Now, let f and g be functions in Hom(V,F). If f(v)=0 forces g(v)=0 then g = \lamda f for some \lamda in F.

Homework Equations

No relevant equations

The Attempt at a Solution

So, since we are talking about mapping stuff to 0, it seems that we should look at the kernel of the transformations. We see that ker(f) is a subset (and hence sub-vector space) of ker(g). By homomorphism theorems, the quotient spaces formed by these kernels are both isomorphic to F, when F is being considered as a Vector Space. So V\ker(f) and V\ker(g) are both 1-D spaces and are isomorphic. So everything in each of the quotient spaces can be written as a multiple of a basis vector, v+ker(f) for V\ker(f) and u+ker(g) for v\ker(g) for some u,v in V. Ok, so since ker(f) in ker(g) it seems that everything in V\ker(f) can be written as a multiple of u+ker(g).

Other than that, I am kind of stuck on this. I need hints or ideas. I think perhaps I have built some sort of mental road block here. This problem is problem 4.4.11 from Topics In Algebra (Hernstein).

 

[Hurkyl]

(note that you've been assuming that both f and g are nonzero)

Two things you don't really seem to have thought about:

By homomorphism theorems, the quotient spaces formed by these kernels are both isomorphic to F

And how does that relate to f and g?

g = \lamda f

That's the same thing as finding a homomorphism F --> F that makes a commutative triangle.

 

[jshtok]

You have confused the order of inclusion in your solution. You have $$K(g)\subset K(f)$$, where K(g) is the kernel, or nullspace, of the linear functional g (that's how elements in Hom (V,F) are called in the daily life).
Now, notice that the statement is wrong if f is identically zero; so we must assume $$f\neq 0$$. In this case, both K(f),K(g) are of dimension dim(V)-1, therefore $$K(g)\subset K(f)$$ implies $$K(g)=K(f)$$. Therefore, both f and g are determined by their action on some common element $$v\in V$$; if $$f(v)=\alpha\neq 0,g(v)=\beta\neq 0,\;\text{then}\; g=\lambda f \text{ where }\lambda=\frac {\beta}{\alpha}$$.

 

[Robert1986]

Hmm. f(v) = 0 implies g(v)=0 , doesn't this mean that Ker(f) \subset Ker(g) since anything that f maps to 0 g also maps to 0? But then we don't know from the problem statement alone that g(v)=0 implies f(v) = 0. Second, I don't think the problem is incorrect inf f is identically 0. All the problem states is that f(v) = 0 implies g(v) = 0 so if f is identically 0, then so is g.

I'm confused about what the rest of what you say, though. I see that the kernels are equal, but I don't get how it follows that g is a multiple of f.

Thanks,
Robert

 

[jshtok]

You are right, I confused the order of inclusion and this also led me to think there is a problem with f=0. Anyway, here is a more rigor version of my proof:
if g is not identically zero, then both kernels are of the same dimension and therefore they are the same space. So, the kernel K=K(g)=K(f) is of dimension n-1. The functions f, g are determined by their values on the one-dimensional quotient space V/K, spanned by, say, v+K for some v in V. Now consider the values of f and g on this generator v ($$\alpha,\beta$$, like in my previous post) and deduce the linear dependence as I did earlier.

If g=0, then the statement holds for $$/lambda=0$$.