# Linear functionals: Φ(u)=0 implies Φ(v)=0, then u=kv.

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1. Apr 12, 2017

1. The problem statement, all variables and given/known data
Suppose u,v ∈ V and that Φ(u)=0 implies Φ(v)=0 for all Φ ∈ V* (the duel space). Show that v=ku for some scalar k.

2. Relevant equations
N/A

3. The attempt at a solution
I've managed to solve the problem when V is of finite dimension by assuming u,v are linearly independent, expanding them to a basis of V and defining a function such that Φ(v)=0, Φ(u)=1 and Φ(wi)=0 where wi are all the other basis vectors. This contradicted the original statement in the problem (since Φ(v)=0 but Φ(u)=1) which meant u,v must be linearly dependent, which implies u is a scalar multiple of v.

However I don't know how to prove it in the general case where V could also be of infinite dimension, the intuitive solution I had was an argument of linear dependence since we have a scalar multiple but I cannot seem to use it in the general case, so this is where I got stuck. Any help would be greatly appreciated.

2. Apr 12, 2017

### Staff: Mentor

What did you need a basis for?

3. Apr 12, 2017

I assume you mean in my proof for a vector space of finite dimension. I needed to expand u,v to a basis to show that Φ is indeed an element of the duel space by defining its action on a basis.

4. Apr 12, 2017

### Staff: Mentor

But you didn't use finiteness, nor did you use the entire basis. All you used, is a complementary space of $\mathbb{F}\cdot u$ on which you defined a functional (that isn't identically zero) to vanish.

5. Apr 12, 2017

I don't recall defining any complimentary space where Φ vanishes. If V is of dimension n, I merely expanded {v,u} to a basis S={v,u,w1,w2,...,wn-2} and defined a function Φ from V to its field K defined by Φ(v)=0, Φ(u)=1, Φ(w1)=0, Φ(w2)=0,..., Φ(wn-2)=0.
This function is obviously a linear functional since it's a part of the basis of V* duel to the basis S, and this conflicted with the original statement.

All I did here was show that if u,v are linearly independent, there exists a function which contradicts the conditions of the question, and so u,v must be linearly dependent.

6. Apr 14, 2017

### zwierz

There is very important and simple theorem. Let $X,Y,Z$ be vector spaces say over field $\mathbb{R}$ or $\mathbb{C}$. Perhaps some of these spaces are infinite dimensional. And let $A:X\to Y,\quad B:X\to Z$ be linear operators such that $A(X)=Y$ and $\mathrm{ker}\,A\subset\mathrm{ker}\,B$.

Theorem 1. There is a linear operator $\Lambda:Y\to Z$ such that $B=\Lambda A$.
This theorem particularly implies thee followin proposition.

Theorem 2. If $f,f_1,\ldots, f_n\in X^*$ and $\bigcap_{i=1}^n\mathrm{ker}\, f_i\subseteq \mathrm{ker}\,f$ then there exist numbers $\lambda_1,\ldots,\lambda_n$ such that $f=\lambda_1f_1+\ldots+\lambda_nf_n$.

By the way this idea is a source of all Lagrange multipliers theorems.

Now turn to the problem of OP. Define linear functionals $\tilde u,\tilde v:V^*\to\mathbb{R}$ as follows $\tilde u(\Phi)=\Phi(u)$ and the same for $\tilde v$. From conditions of the OP's problem it follows that $\mathrm{ker}\,\tilde u\subseteq\mathrm{ker}\,\tilde v$. By Theorem 2 it follows that $\tilde v=k\tilde u$. That is $\Phi(v)=k\Phi(u)$ for all $\Phi\in V^*$. Thus $v=ku$.

This prompts that there is no need to employ the whole space $V^*$ it is sufficient to use only its total subspace

Last edited: Apr 14, 2017