# Functions in Hom(V,F)

## 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
Staff Emeritus
Gold Member
(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.

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 $$f(v)=\alpha\neq 0,g(v)=\beta\neq 0,\;\text{then}\; g=\lambda f \text{ where }\lambda=\frac {\beta}{\alpha}$$.

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

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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 $$/lambda=0$$.

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