Mentor

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

I only have time for a short answer right now.
 Quote by Philmac how can there only be one zi in V'' for each xi in V?
Suppose that there are two. To be more precise, suppose that for all y in V',
\begin{align}
z(y) &=y(x)\\
w(y) &=y(x).
\end{align} Then z=w. (They have the same domain and we have z(y)=w(y) for each y in the domain. If you think about what a function is, it should be obvious that this means that z=w).

Mentor
 Quote by Philmac V' is the set of functionals that take x as their argument and V'' is the set of functionals that take y as their argument. I find this notation to be slightly misleading (I'm probably just misunderstanding something) -- is there a V'''?
V' is the set of linear functions from V into ℝ. V'' is defined as (V')', so it's the set of linear functions from V' into ℝ. V''' is defined as (V'')' so it's the set of linear functions from V'' into ℝ. The sequence goes on forever. Each "V with lots of primes" is isomorphic to V if the number of primes is even, and isomorphic to V' if the number of primes is odd.

Actually, V is also isomorphic to V'. The difference between this isomorphism and the one between V and V'' is that to define an isomorphism between V and V', we need to use something like an inner product on V or a specific basis on V, but to define an isomorphism between V and V'', we just need to understand that members of V'' act on members of V' and that members of V' act on members of V.

 Quote by Philmac However, now that I think about it again, how can there only be one zi in V'' for each xi in V? Changing the value of yj will change the value of [xi,yj],
You should think about what you said here until you understand that it's like replying to the statement
For each integer n, define $f_n:\mathbb R\rightarrow\mathbb R$ by $f_n(x)=x^n$ for each $x\in\mathbb R$.
by saying
How can there be only one $f_n$ for each n? Changing the value of x will change the value of $f_n(x)$.

 Ohh, I get it now (I think)! The claim isn't that z0(y) is always the same value, the claim is simply that z0(y) will always be the same as y(x0). Which should be obvious just by looking at the equation provided, but, oh well. I think I'm ready for annihilators now.

Mentor
 Quote by Philmac I think I'm ready for annihilators now.
I don't understand your definitions. Can you post them as they appear in the book?

 Quote by Fredrik I don't understand your definitions. Can you post them as they appear in the book?
Here it is:
 The annihilator S0 of any subset S of a vector space V (S need not be a subspace) is the set of all vectors y in V' such that [x, y] is identically zero for all x in S. Thus 00 = V' and V0 = 0 ($\subset$V'). If V is finite-dimensional and S contains a non-zero vector, so that S ≠ 0, then [reference to another theorem] shows that S0 ≠ V'.

Mentor
 Quote by Philmac The problem I have here is "By definition, M00 is the set of all vectors x such that [x,y]=0 for all y in M0". Shouldn't it be the set of all vectors z (or whatever letter you like) in V'?
M00 isn't equal to M, it's isomorphic to it. What you're supposed to prove is that f(M)=M00, where $f:V\rightarrow V''$ is the isomorphism defined in a previous post. M00 is the vector space of all z in V'' such that z(y)=0 for all y in M0. Now use the definition of the isomorphism f to show that x is in M if and only if f(x) is in M00.

 Quote by Fredrik M00 isn't equal to M, it's isomorphic to it.
Are you saying that the textbook is wrong? The symbol in the book is =, not $\approx$ or $\cong$. (Although, equivalency implies isometry)

 Quote by Fredrik What you're supposed to prove is that f(M)=M00, where $f:V\rightarrow V''$ is the isomorphism defined in a previous post. M00 is the vector space of all z in V'' such that z(y)=0 for all y in M0. Now use the definition of the isomorphism f to show that x is in M if and only if f(x) is in M00.
Not quite sure I follow here. Does this have something to do with reflexivity? That is to say, for every z in M00, z(y)=0, so y(x)=0? Which means that M00 is isomorphic to M0?

Mentor
 Quote by Philmac Are you saying that the textbook is wrong? The symbol in the book is =, not $\approx$ or $\cong$. (Although, equivalency implies isometry)
I'm saying that the author is a bit careless with the notation/terminology. When he writes M00=M, he means M00=f(M). It's obvious from the definitions (assuming that I understood them) that M00 can't be equal to M. They are subspaces of two different vector spaces.

 Quote by Philmac Not quite sure I follow here.
I left out the details on purpose because I think it's time that you start doing some of these things on your own instead of having us do everything for you. I actually typed the proof that f(M)=M00 and then deleted it before I submitted the post.

 Quote by Philmac Does this have something to do with reflexivity? ... Which means that M00 is isomorphic to M0?
That's not what "reflexive" means.

 Quote by Philmac for every z in M00, z(y)=0
This statement isn't OK as part of a proof. What is y?

Please try to prove that M00=f(M). Recall that two sets are equal if and only if they have the same members. This means that you can break up the proof into these two steps:

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.

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

I also recommend that you do the exercise I suggested earlier: Prove that f is an isomorphism.

I'm exceptionally terrible at proving things, but I'll try.

First, proving that f is an isomorphism:
 Quote by Fredrik You're looking for an isomorphism from V into V''. This is very easy, because it turns out that the first function we can think of from V into V'' (except for constant functions of course) is an isomorphism. We want to define a function f:V→V'', so we must specify a member of V'' for each x. This member of V'' will of course be denoted by f(x). A member of V'' is defined by specifying what we get when it acts on an arbitrary member of V'. So we must specify f(x)(ω) for each ω in V'. f(x)(ω) is supposed to be a real number, and ω(x) is a real number. So... For each x in V, we define f(x) in V'' by f(x)(ω)=ω(x) for all ω in V'. This defines a function f:V→V''. Now you just need to verify that this function is linear and bijective onto V''.
V is an n-dimensional vector space
V' is the dual space of V and is therefore n-dimensional
V''(=(V')') is the dual space of V' and is therefore n-dimensional
Since dimV=dimV'', V is isomorphic to V''.

or

x is an element of V
y is an element of V'
z is an element of V''
Define the function f:V$\rightarrow$V''
This means that f(x) is an element of V'': z
Since f is a function acting on x, f is an element of V': y
So we have y(x0)=z0
I believe that this implies that f is bijective, but I'm not sure how to show that it's linear. Don't we know that it's linear by definition?

 Quote by Fredrik Please try to prove that M00=f(M). Recall that two sets are equal if and only if they have the same members. This means that you can break up the proof into these two steps: 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. Step 2. Let z in f(M) be arbitrary. Show that z is in M00.
I'm really not sure at all what to do here. Don't we know that there is such a correspondence due to reflexivity?

 Quote by Fredrik That's not what "reflexive" means.
Sorry, I meant M, not M0.

 Quote by Fredrik This statement isn't OK as part of a proof. What is y?
y is an element of V'

Mentor
 Quote by Philmac I'm exceptionally terrible at proving things, but I'll try.
That's OK. We all start out that way.

 Quote by Philmac V is an n-dimensional vector space V' is the dual space of V and is therefore n-dimensional V''(=(V')') is the dual space of V' and is therefore n-dimensional Since dimV=dimV'', V is isomorphic to V''.
This proves that V and V'' are isomorphic, but it says nothing about whether f is an isomorphism. I was hoping you'd use the definition of f to prove that it's linear and bijective.

 Quote by Philmac x is an element of V
I understand that you make statements like this to explain the notation before you start the proof, but it's not necessary, because before you refer to x in the actual proof, you're going to have to do one of the following:

1. Specify the value that the variable represents. For example: Let x be the vector (1,1,0) in $\mathbb R^3$.
2. Say that it represents an arbitrary member of some set. "For example: Let x be an arbitrary real number".
3. Precede the statement with "for all x". For example: For all x in $\mathbb R$, we have $x^2\geq 0$.
4. Precede the statement with "there exists an x in ... such that". For example: There exists an x in $\mathbb C$ such that $x^2$=-1.

So I suggest that you don't include statements like "x is an element of V" in any proofs.

 Quote by Philmac Define the function f:V$\rightarrow$V'' This means that f(x) is an element of V'': z
A better way of saying that: The definition of f implies that for all x in V, f(x) is in V''.

There's no need to mention z, unless you think you can simplify the notation in the rest of the proof by saying "define z=f(x)". This only makes sense if you have specified the value of x or used the phrase "Let x in V be arbitrary" earlier.

 Quote by Philmac Since f is a function acting on x
I don't like phrases like this, because they suggest that it matters what variable symbols we're using. Say something like, "since the domain of f is V..." instead.

 Quote by Philmac f is an element of V': y
It's actually not. If the codomain of f had been ℝ, it might have been. But even then, you would have had to prove that f is linear before you can draw that conclusion. And the codomain of our f is V'', not ℝ.

 Quote by Philmac So we have y(x0)=z0 I believe that this implies that f is bijective,
Hm, you're saying that y(x)=f(x)? That doesn't make sense. Anyway, in a proof you have to show how the things you know imply the result you want to prove.

 Quote by Philmac but I'm not sure how to show that it's linear.
Let a,b be arbitrary real numbers, and let x,y be arbitrary members of V. Use the definition of f to rewrite f(ax+by).

 Quote by Philmac Don't we know that it's linear by definition?
No. We defined $f:V\rightarrow V''$ by saying that for each x in V, f(x) is the member of V'' such that for all ω in V', f(x)(ω)=ω(x). Since the definition doesn't say that f is linear, we have to prove it. You also have to prove that f is bijective. I suggest that you do it in two steps:

Injectivity: Show that for all x,y in V, if f(x)=f(y), then x=y.

Surjectivity: Let z in V'' be arbitrary and show that there exists an x in V such that f(x)=z.

 Quote by Philmac I'm really not sure at all what to do here. Don't we know that there is such a correspondence due to reflexivity? Sorry, I meant M, not M0.
Reflexivity ensures that there's an isomorphism from M into M'', not that there's an isomorphism from M into M00.

Reflexivity also ensures that there's an isomorphism from V into V''. Let f be such an isomorphism. M is a subspace of V. That makes f(M) a subspace of V''. M00 is a subspace of V''. You don't know that f(M) and M00 is the same subspace of V'' until you have proved that they have the same members.

 Mentor Did you finish these proofs or did you give up? Just curious.

 Quote by Fredrik Did you finish these proofs or did you give up? Just curious.
Sorry, a bunch of other issues came up and I haven't had time to properly sit down and think about it yet; I probably should have said something before. I'll finally have some time tomorrow afternoon, so I'll post what I come up with then :)

 Quote by Fredrik You're looking for an isomorphism from V into V''. This is very easy, because it turns out that the first function we can think of from V into V'' (except for constant functions of course) is an isomorphism. We want to define a function f:V→V'', so we must specify a member of V'' for each x. This member of V'' will of course be denoted by f(x). A member of V'' is defined by specifying what we get when it acts on an arbitrary member of V'. So we must specify f(x)(ω) for each ω in V'. f(x)(ω) is supposed to be a real number, and ω(x) is a real number. So... For each x in V, we define f(x) in V'' by f(x)(ω)=ω(x) for all ω in V'. This defines a function f:V→V''. Now you just need to verify that this function is linear and bijective onto V''.
 Quote by Fredrik This proves that V and V'' are isomorphic, but it says nothing about whether f is an isomorphism. I was hoping you'd use the definition of f to prove that it's linear and bijective. ... Before you refer to x in the actual proof, you're going to have to do one of the following: 1. Specify the value that the variable represents. For example: Let x be the vector (1,1,0) in $\mathbb R^3$. 2. Say that it represents an arbitrary member of some set. "For example: Let x be an arbitrary real number". 3. Precede the statement with "for all x". For example: For all x in $\mathbb R$, we have $x^2\geq 0$. 4. Precede the statement with "there exists an x in ... such that". For example: There exists an x in $\mathbb C$ such that $x^2$=-1. ... A better way of saying that: The definition of f implies that for all x in V, f(x) is in V''. ... There's no need to mention z, unless you think you can simplify the notation in the rest of the proof by saying "define z=f(x)". This only makes sense if you have specified the value of x or used the phrase "Let x in V be arbitrary" earlier. ... I don't like phrases like this, because they suggest that it matters what variable symbols we're using. Say something like, "since the domain of f is V..." instead. ... It's actually not. If the codomain of f had been ℝ, it might have been. But even then, you would have had to prove that f is linear before you can draw that conclusion. And the codomain of our f is V'', not ℝ. ... Hm, you're saying that y(x)=f(x)? That doesn't make sense. Anyway, in a proof you have to show how the things you know imply the result you want to prove. ... Let a,b be arbitrary real numbers, and let x,y be arbitrary members of V. Use the definition of f to rewrite f(ax+by). ... No. We defined $f:V\rightarrow V''$ by saying that for each x in V, f(x) is the member of V'' such that for all ω in V', f(x)(ω)=ω(x). Since the definition doesn't say that f is linear, we have to prove it. You also have to prove that f is bijective. I suggest that you do it in two steps: Injectivity: Show that for all x,y in V, if f(x)=f(y), then x=y. Surjectivity: Let z in V'' be arbitrary and show that there exists an x in V such that f(x)=z.
Define $f:V\rightarrow V''$ such that for each x in V, f(x) is the member of V'' such that for all y in V', f(x)(y)=y(x)

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)

or

Show that if 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.

Also, doesn't showing this prove bijectivity, not just injectivity? We know that V and V'' have the same dimension and we know that that if x1=x2 then z1=z2 (i.e. you cannot have x1=x2 and z1≠z2 and vice versa). This injectivity goes both ways, doesn't it? Since V and V'' have the same dimension, doesn't this mean that V injective to V'' and V'' injective to V implies bijectivity?

Surjectivity:

Let z be an arbitrary member of V''
Since z=f(x), by the definition of f, we know that z(y)=f(x)(y)=y(x). Therefore, for every z in V'', there exists an x in V.

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)

 Quote by Fredrik Prove that M00=f(M). Recall that two sets are equal if and only if they have the same members. This means that you can break up the proof into these two steps: 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. Step 2. Let z in f(M) be arbitrary. Show that z is in M00.
 Quote by Fredrik Reflexivity ensures that there's an isomorphism from M into M'', not that there's an isomorphism from M into M00. Reflexivity also ensures that there's an isomorphism from V into V''. Let f be such an isomorphism. M is a subspace of V. That makes f(M) a subspace of V''. M00 is a subspace of V''. You don't know that f(M) and M00 is the same subspace of V'' until you have proved that they have the same members.
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$\subseteq$M''. Since z is therefore a member of M'', there must be an x in M such that f(x) is a member of M00.

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

That's what I managed to come up with. I hope I got it right this time.

Mentor
 Quote by Philmac 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:
 Quote by Philmac for arbitrary x1,x2 in V, if f(x1)=f(x2), then x1=x2:
 Quote by Philmac 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.

 Quote by Philmac 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.

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 Quote by Fredrik 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.

Mentor
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. \begin{align}e^i(x) &=e^i(x^je_j)=x^je^i(e_j)=x^j\delta^i_j=x^i\\ \bar e_i(y) &=\bar e_i(y_j e^j)=y_j\bar e_i(e^j) =y_j\delta^j_i=y_i. \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 $f(x)=f(x')$, then for all y in V', we have $y(x)=y(x')$. We need to show that this implies that $x=x'$. 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 $e^i(x)=e^i(x')$ for all i. This means that $x^i=x'^i$ for all i, and that means that x=x'.

 Quote by Philmac 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.

 Quote by Philmac 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.

 Quote by Philmac 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$\subseteq$M''. 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?

 Quote by Philmac 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 $f(M)=M^{00}$.