Confused about the concepts of dual spaces, dual bases, reflexivity and annihilators

[Fredrik]

Injectivity:

The definition of f implies that for every x in V, there is an f(x) in V'' (I think this is all that is required)

This is what's required to show that f is a function. To show that it's injective, you need to show this:

for arbitrary x₁,x₂ in V, if f(x₁)=f(x₂), then x₁=x₂:

If y is an arbitrary vector in V' and z₁ and z₂ are vectors in V'' such that z₁(y)=y(x₁) and z₂(y)=y(x₂), then if z₁=z₂, then y(x₁)=y(x₂), which implies that x₁=x₂.
I think this makes sense, but I don't know if I'm justified in the last part where I say that y(x₁)=y(x₂) implies x₁=x₂.

You have the right idea, but you're expressing it in a way that makes it hard to see that. This is how I would do it.

Suppose that f(x_1)=f(x_2). Then for all y\in V', we have f(x_1)(y)=f(x_2)(y). (The reason is this: f(x_1) and f(x_2) are members of V'', so they are functions from V' into ℝ. Two functions are equal if and only if they have the same domain and the same value at each point in the domain. These two functions are equal and have the domain V', so they must have the same value at each point in V'). This means that for all y\in V', we have y(x_1)=y(x_2). This implies that x_1=x_2.

The last step is not obvious. It's not true that y(x_1)=y(x_2) implies that x_1=x_2. That would mean that y is injective, and it's certainly not true that each y in V' is injective. For example, if V=\mathbb R^2 and y is defined so that for all x, y(x) is the projection of x onto the 1 axis, then y is linear but not injective. However, the statement "for all y in V', y(x_1)=y(x_2)" does imply that x_1=x_2. I don't see an easy way to show it right now. I can see a hard way, because by coincidence I read about how to do this in the context of infinite dimensional normed spaces today. But I feel like there must be an easier way. We can think about that tomorrow. I don't have time to answer everything today anyway.

Also, doesn't showing this prove bijectivity, not just injectivity? We know that V and V'' have the same dimension...

It doesn't. It's true that you can show that dim V=dim V'=dim V'', and (assuming that V is finite-dimensional) this means that they're all isomorphic. But this doesn't mean that the specific function f that we defined is an isomorphism, or even surjective.

I'll answer the rest tomorrow.

 

[micromass]

It doesn't. It's true that you can show that dim V=dim V'=dim V'', and (assuming that V is finite-dimensional) this means that they're all isomorphic. But this doesn't mean that the specific function f that we defined is an isomorphism, or even surjective.

I don't want to confuse anybody, but I just wanted to mention that an important theorem (= alternative theorem) states that any injective linear function between finite-dimensional spaces is in fact an isomorphism. So our f is indeed an isomorphism! But it only follows from the alternative theorem, not from anything else.

 

[Fredrik]

I think this will be easier if we first work through a few easy facts about basis vectors. Let \{e_i\} be a basis for V. Let \{e^i\} be the dual basis of \{e_i\}. Let \{\bar e_i\} be the dual basis of \{e^i\}. These are bases of V, V' and V'' respectively. (Suppose that we have already proved that, and that we have also already proved that dim V=dim V'=dim V'').
\begin{align}
x &=x^ie_i\\
y &=y_ie^i\\
z &=z^i\bar e_i
\end{align}
The first thing you need to know is what you get when you have one of these things act on a basis vector y(e_i)=y_j e^j(e_i)=y_j\delta^j_i=y_i and when a member of a dual basis acts on an arbitrary vector. <br /> \begin{align}e^i(x) &amp;=e^i(x^je_j)=x^je^i(e_j)=x^j\delta^i_j=x^i\\<br /> \bar e_i(y) &amp;=\bar e_i(y_j e^j)=y_j\bar e_i(e^j) =y_j\delta^j_i=y_i.<br /> \end{align} You also need to know about the relationship between \{\bar e_i\} and \{e_i\}. For all y in V, we have f(e_i)(y)=y(e_i)=y_i=\bar e_i(y), so \bar e_i=f(e_i) for all i.

Now let's return to the thing I didn't prove last night. We found that if f(x)=f(x&#039;), then for all y in V', we have y(x)=y(x&#039;). We need to show that this implies that x=x&#039;. The easiest way to do that is to note that since the equality y(x)=y(x') holds for all y in V', we have e^i(x)=e^i(x&#039;) for all i. This means that x^i=x&#039;^i for all i, and that means that x=x'.

Surjectivity:

Let z be an arbitrary member of V''
Since z=f(x), by the definition of f,

Here you're making a statement that involves x, but you haven't said what x is. Yes, I understand that your notation is to use x for members of V, y for members of V', and z for members of V'', but even if you include that information in the proof, you still have to explain if you mean that the equality z=f(x) holds for all x in V, or that there exists an x in V such that z=f(z). I'm guessing that you're trying to say that "the definition of f implies that there exists an x in V such that z=f(x)". The thing is, this is the statement we're trying to prove! So you haven't actually proved anything here.

Let z be an arbitrary member of V''. We need to show that there exists an x in V such that f(x)=z. By definition of f, such a z would satisfy z(y)=y(x) for all y. This means that it would satisfy z(e^i)=e^i(x) for all i. This equality is equivalent to z^i=x^i. So if there's an x with the desired property, it has the same components in the basis \{e_i\} as z has in the basis \{\bar e_i\}. This doesn't prove that f is surjective, but it tells us what we should try to do. We should try to prove that f(z^ie_i)=z. This is the same thing as proving that for all y in V', we have f(z^ie_i)(y)=z(y), and this is very easy if we use the results above.

Linearity:

Let x₁, x₂ be arbitrary members of V
Let y be an arbitrary member of V'
If f is a linear functional on V, then it must be shown that f(ax₁+bx₂)=af(x₁)+bf(x₂)
f(ax₁+bx₂)(y)=y(ax₁+bx₂)
and since y is linear...
y(ax₁+bx₂)=ay(x₁)+by(x₂)=af(x₁)(y)+bf(x₂)(y)
Combining this result with the first line...
f(ax₁+bx₂)(y)=af(x₁)(y)+bf(x₂)(y)
f(ax₁+bx₂)=af(x₁)+bf(x₂)

I don't see a "thumbs up" smiley, so I'll just use the "approve" smiley.

Step 1. Let z in M⁰⁰ be arbitrary. Show that z is in f(M), i.e. that there's an x in M such that f(x)=z.

By the definition of M⁰⁰ we know that M⁰⁰\subseteqM''. Since z is therefore a member of M'', there must be an x in M such that f(x) is a member of M⁰⁰.

The surjectivity of f implies that there must be an x in the domain of f such that f(x)=z, but the domain is V, not M. So how do you know that this x is in M?

Step 2. Let z in f(M) be arbitrary. Show that z is in M⁰⁰.

If x is an arbitrary member of M and y is an arbitrary member of M⁰ (and M') then, by reflexivity:
z(y)=f(x)(y)=y(x)=0
Since z(y)=0, it follows that z is in M⁰⁰

Not bad, but x is not arbitrary in this calculation. It's specifically the x in M such that f(x)=z. (Apart from that, the proof is fine).

Note that what you've proved here is that f(M)\subset M^{00}. The other part of the problem was to show that M^{00}\subset f(M). Together, these two "inequalities" imply that f(M)=M^{00}.