# Question involving annihilators and solution spaces

#### britatuni

##### New member
I was hoping to get help on a question that has been bugging me, I goes like this:

V is a vector space with a dual space V* and U is a subspace of V and W a subspace of V*

The question ask to show that:
'the solution space of W intersected with U' is a subspace of 'the solution space of (W + the annihilator of U)'.

Now, looking at the left hand side I see that an element, 'x', within U must be satisfy f(x)=0 for all functions, 'f', within W.
I realise that the above is barely a start on the question at all. But after looking at eh definitions I just don't see where I am expected to go next.

#### Fantini

MHB Math Helper
Hello britatuni! I'm not familiar with your definitions of solution space and annihilator. Can you write them for me? English is not my native language and I think I know what is being asked, but I cannot be sure until the terms are defined.

Cheers!

#### britatuni

##### New member
Hi, sorry for the late response since I was on holiday without internet.

I have the solution space of W defined as the set of elements (v) in V that satisfy the condition that f(v) = 0 for every function f in the dual space of W.
And the annihilator is defined as the set of functions (g) in V*such that g(u) = 0 for every element u in the subspace U.

#### Opalg

##### MHB Oldtimer
Staff member
I was hoping to get help on a question that has been bugging me, I goes like this:

V is a vector space with a dual space V* and U is a subspace of V and W a subspace of V*

The question ask to show that:
'the solution space of W intersected with U' is a subspace of 'the solution space of (W + the annihilator of U)'.

Now, looking at the left hand side I see that an element, 'x', within U must be satisfy f(x)=0 for all functions, 'f', within W.
I realise that the above is barely a start on the question at all. But after looking at eh definitions I just don't see where I am expected to go next.
So it appears that "the solution space of $W\,$" is what I would call the pre-annihilator of $W$, namely the space $W_{\perp} \stackrel {\text{def}}{=} \{x\in X : f(x) = 0\ \forall f\in W\}$. You are asked to show that $W_{\perp} \cap U \subseteq \bigl(W + U^{\perp}\bigr)_{\perp}$ (where $U^{\perp}\stackrel {\text{def}}{=} \{f\in W : f(u)=0\ \forall u\in U\}$ is the annihilator of $U$).

Let $x\in W_{\perp} \cap U$. You correctly say that this implies $f(x) = 0$ for all $f$ in $W$. You are asked to show that $f(x) = 0$ for all $f$ in $W + U^{\perp}$. An element of $W + U^{\perp}$ is by definition the sum of an element in $W$ and an element in $U^{\perp}$. Show that both those elements of $V^*$ must vanish at $x$, and you have completed the proof.

[I think that this problem belongs to linear algebra rather than analysis, so I have transferred it to the Linear and Abstract Algebra section.]

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