# Linear Algebra - Dual Spaces

1. Feb 15, 2008

### smithg86

[SOLVED] Linear Algebra - Dual Spaces

1. The problem statement, all variables and given/known data

(V and W are vector spaces. F is a field)
"The space L(V,W) of linear maps from V to W is always a vector space. Take W = F. We then get the space V* := L(V,F) of F-linear maps V --> F. This is called the dual space of V."

1. Let V = F$$^{2}$$ with basis e$$_{1}$$, e$$_{2}$$. Define elements e*$$_{1}$$, e*$$_{2}$$ $$\in$$ V* by:

e*$$_{1}$$ (e$$_{1}$$) = 1
e*$$_{1}$$ (e$$_{2}$$) = 0

and

e*$$_{2}$$ (e$$_{1}$$) = 0
e*$$_{2}$$ (e$$_{2}$$) = 1

Show that e*$$_{1}$$, e*$$_{2}$$ form a basis for V*. Deduce that dim (F$$^{2}$$)* = 2.

2. Relevant equations

1. In F$$^{2}$$, e$$_{1}$$ = (1,0) and e$$_{2}$$ = (0,1).

2. if vectors v$$_{1}$$, ...v$$_{n}$$ $$\in$$ V and the list (v$$_{1}$$, ...v$$_{n}$$) is linearly independent, then:

a$$_{1}$$v$$_{1}$$ + ... + a$$_{n}$$v$$_{n}$$ =0 $$\Rightarrow$$ all a's = 0, such that a $$\in$$ F

3. The definition of a linear map.

3. The attempt at a solution

If e$$_{1}$$, e$$_{2}$$ form a basis for V, then the list (e$$_{1}$$, e$$_{2}$$) is linearly independent and spans V. If I am to show that e*$$_{1}$$, e*$$_{2}$$ form a basis for V*, then the list (e*$$_{1}$$, e*$$_{2}$$) must be linearly independent and span V*. (?)

Here is where I get confused. From what I understand, e*$$_{1}$$ 'keeps' the first component of the vector and discards all others; e*$$_{2}$$ 'keeps' the second component of the vector and discards all others. Since the vector space V*has elements that are linear maps, e*$$_{1}$$ and e*$$_{2}$$ must be linear maps. I'm not sure how to 'work with' linear maps as vectors.

Since e$$_{1}$$, e$$_{2}$$ span V, any v$$\in$$V can be written as a linear combination of them:

v = a$$_{1}$$ e$$_{1}$$ + a$$_{2}$$ e$$_{2}$$

To show that e*$$_{1}$$,e*$$_{2}$$ span, I have to show that any T$$\in$$V* can be written as:

T = b$$_{1}$$ e*$$_{1}$$ + b$$_{2}$$ e*$$_{2}$$

But what are the a's? Are they in F (i.e., real numbers) or are they in V (are they vectors)? (I'm not sure how to proceed.)

To show that the list is linearly independent, I need to show:

c$$_{1}$$ e*$$_{1}$$ + c$$_{2}$$ e*$$_{2}$$ = 0 $$\Rightarrow$$ all c's = 0 (again, what are the c's?)

Can anyone help me? I know the second part of the problem is easy: from the definition of dimension, if V* has a 2-vector-long basis, then dim V* = 2.

Last edited: Feb 15, 2008
2. Feb 16, 2008

### CompuChip

Basically, it's just plugging in definitions in a convenient way.
You have already said that if you have a vector v in V, then you can write it as $\vec v = v_1 e_1 + v_2 e_2$, where the vi are numbers in F (that is, the field over which V is a vector space, for example real numbers).
Now take a linear map in the dual space and suppose it is zero. That is, consider
$$T \in V^*, T : V \to F, v \mapsto T(v)$$
such that T(v) = 0 for all v in V. You know that T is a linear map, so if you decompose v you can write
$$T(v) = T(a_1 e_1 + a_2 e_2) = a_1 T(e_1) + a_2 T(e_2)$$
(you can split T over the additions and take the numbers ai out).

Since V* is a vector space as well, you can again decompose T into basis vectors,
$$T = b_1 e^*_1 + b_2 e^*_2$$.

Now I'll let you try to write out what T(v) is by plugging in the formulas. Then if you use that on the other hand it should be equal to zero, you will get an equation in terms of the $a_i$ and $b_i$. Then you should try to conclude from that that all $b_i = 0$.

3. Feb 16, 2008

### HallsofIvy

Staff Emeritus
The trouble is, Smithg86, you don't seem to know the definitions! Also, why are you assuming that V itself is 2 dimensional? You say,
"Since e, e span V, any v$\in$V can be written as a linear combination of them:

v = a1 e1 + a2 e2

But what are the a's? Are they in F (i.e., real numbers) or are they in V (are they vectors)? (I'm not sure how to proceed.)"

You know, I hope that the basis vectors, e1, etc. are vectors, they are in V. And that multiplication of vectors is not defined- the only multiplication is "scalar" multiplication, multiplication of real numbers and vectors. The "a"s must be in F for "$a^1e^1+ a^2e^2+ \cdot\cdot\cdot+ a^ne^n[itex]" to make sense! Suppose f is in the dual space. Any vector in V can be written [itex]v= a^1e^1+ a^2e^2+ \cdot\cdot\cdot+ a^nv^n$. Since f is linear, $f(v)= a^1 f(e^1)+ a^2f(e^2)+ \cdot\cdot\cdot+ a^nf(v^n)$. That is, what f "does" to any vector v depends only on what it does to the basis vectors.

But if we write $f= b^1e*^1+ b^2e*^2+ /cdot/cdot/cdot+ b^ne*^n$,then $f(e^1)= b^1 e*^1(e^1)+ b^3e*^2(e^1)+ \cdot\cdot\cdot+ b^ne*^n(e^1)= b^1$. Put that into the general formula and show that there always exist bi that will work.

Last edited: Feb 16, 2008
4. Feb 16, 2008

### smithg86

HallsofIvy,

Thanks for clarifying what the a's were.

CompuChip,

I tried doing what you suggested:

For some v in V, $$T(v) = T(a_{1} e_{1} + a_{2} e_{2})$$, since $$e_{1}, e_{2}$$ span V. But don't I have to show that V* is spanned by $$e*_{1}, e*_{2}$$ before I decompose T into a linear combination of them? (I'll continue, assuming I've proved that before I did this part.) Then, $$T = b_1 e^*_1 + b_2 e^*_2$$ implies:

$$T(v) = T(a_{1} e_{1} + a_{2} e_{2}) = b_1 e^*_1 (a_{1} e_{1} +a_{2} e_{2}) + b_2 e^*_2 (a_{1} e_{1} +a_{2} e_{2}) = b_{1} e*_{1} (a_{1} e_{1}) + 0 + 0 + b_{2} e*_{2} (a_{2} e_{2}) = b_{1} a_{1} + b_{2} a_{2} = 0$$.

I want to show that all b's are zero is the only possibility. But what if $$v = a_{1} e_{1} + a_{2} e_{2} = 0$$? That means that $$a_{1} = a_{2} = 0$$ Then the b's can be anything. Is this a counter-example? Assuming $$v \neq 0$$, then: $$a_{1} b_{1} = -a_{2} b_{2}$$. But a and b are in F: they are just scalars. Scalars can be multiples of other scalars, can't they? I don't see why this relationship leads me to all b's=0.

Also, I'm not sure how to show that V* is spanned by $$e*_{1}, e*_{2}$$.

5. Feb 16, 2008

### CompuChip

Indeed, you get $a_1 b_1 + a_2 b_2 = 0$.
The trick is that this must hold for this particular b1 and b2, for any a1 or a2. So no matter what numbers a1, a2 you plug in (that is: on which v you apply T), it must be zero. This can only happen if both b1 and b2 are zero (for example, take v = e1 and v = e2).

To see why V* is 2-dimensional in the first place, you need to think about the linear maps a bit. You know you have two linear maps (which are vectors of the dual space) $e^*_1, e^*_2 \in V^*$. They are linearly independent (why) so V* is at least 2-dimensional. Now look at what you need to define a linear map $V -> F$. You know that it is enough to define it on the basis elements of V*. So the question is: if you define it on $e^*_1, e^*_2$, do you get a unique linear map?

By the way, HallsOfIvy,
He said: take V = F^2, and I guess $F = \mathbb{R}$

I actually have to go now, so I'll let you figure this out for a bit. I'm not sure it is entirely clear what I wrote though, so I'll read it back later and clarify a bit if I think it's necessary. In the mean time, if I confused you, please shout!

6. Feb 17, 2008

### smithg86

Thanks for your help so far. I came up with something to show that $$span(e^{*}_{1}, e^{*}_{2}) = \textbf{V}^*$$:

$$\forall T \in \textbf{V}^*, T:\textbf{F}^2 \rightarrow \textbf{F}$$, define $$T(e_{1}) = a_{1}, T(e_{2}) = a_{2}$$, for some $$a_{1},a_{2} \in \textbf{F}$$. Since $$\forall v \in \textbf{F}^2, v = b_{1} e_{1} + b_{2} e_{2}$$, and because $$T$$ is a linear map, it follows: $$T(v) = T(b_{1} e_{1} + b_{2} e_{2}) = a_{1} b_{1} + a_{2} b_{2} = a_{1} b_{1} (1) + a_{2} b_{2} (1) = a_{1} b_{1} e^*_{1}(e_{1}) + a_{2} b_{2} e^*_{2}(e_{2}) = a_{1} e^*_{1}(b_{1} e_{1}) + a_{2} e^*_{2}(b_{2} e_{2})$$.
So, $$\forall T(v) \in \textbf{V}^*, T(v) = a_{1} e^*_{1}(b_{1} e_{1}) + a_{2} e^*_{2}(b_{2} e_{2}),$$ therefore $$span(e^*_{1}, e^*_{2}) = \textbf{V}^*$$.

Since $$span(e^*_{1},e^*_{2}) = \textbf{V}^*$$ and the list $$(e^*_{1},e^*_{2})$$ is linearly independent, $$(e^*_{1},e^*_{2})$$ is a basis for $$\textbf{V}^*$$, as required. It follows that $$dim (\textbf{V}^*) = 2$$.

How does it look?

7. Feb 17, 2008

### CompuChip

Overall, it looks good. There is just one gap that is maybe a little too large. So I took advantage of the opportunity to give some (overly) critic remarks, just to get the rigorous proof right. It is almost complete though, and you definitely understand it now (as far as I can tell, of course ). So don't feel bad about my somewhat pedantic remarks, they're meant for the best.

Let me reformulate that:
$$\forall T \in \textbf{V}^* \ \text{ (i.e. } T:\textbf{F}^2 \rightarrow \textbf{F}), \exist a_1, a_2 \in \textbf{F} \text{ such that } T(e_{1}) = a_{1}, T(e_{2}) = a_{2}$$.
Note how I used the "for all ... exists ...", making more clear that the a's depend on which map you take (for different maps, you have different a's).
Also, you might want to assert something about the a's uniquely determining T here.

Again, I'd insert something like: "for all v, there exist b1, b2 such that ...".

You also used a property of $e_i^*$...

Technically, you are still looking at some (randomly chosen, but specific) map, so you don't need the for all again.
I am still missing a step here. I assume you are trying to show that you can write T as a linear combination of the $e_i^*$, so you want something like
$$T(v) = (a_1 e_1^* + a_2 e_2^*)(v)$$
right?

Actually you haven't shown that they are linearly independent, but it's completely obvious of course

Last edited: Feb 17, 2008
8. Feb 17, 2008

### smithg86

$$T(v) = a_{1} e^*_{1}(b_{1} e_{1}) + a_{2} e^*_{2}(b_{2} e_{2})$$
$$= a_{1} e^*_{1}(b_{1} e_{1}) + 0 + a_{2} e^*_{2}(b_{2} e_{2}) + 0$$
$$= a_{1} e^*_{1}(b_{1} e_{1}) + a_{1} e^{*}_{1}(b_{2} e_{2}) + a_{2} e^*_{2}(b_{2} e_{2}) + a_{2} e^{*}_{2}(b_{2} e_{2})$$
$$= a_{1} e^{*}_{1}(b_{1} e_{1} + b{2} e_{2}) + a_{2} e^{*}_{2}(b_{1} e_{1} + b{2} e_{2})$$
$$= (a_{1} e^{*}_{1} + a_{2} e^{*}_{2})(b_{1} e_{1} + b_{2} e_{2})$$
$$= (a_{1} e^{*}_{1} + a_{2} e^{*}_{2})(v)$$

This was only supposed to be a proof of $$span(e^*_{1},e^*_{2}) = \textbf{V}^*$$. I would still have to include the proof of linear independence, which you helped me with before.

Thanks again for your help CompuChip and HallsofIvy. I'm closing this thread now.