# Dual vector spaces and linear maps

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1. Dec 4, 2014

### "Don't panic!"

Hi all.

I was hoping I could clarify my understanding on some basic notions of dual spaces.
Suppose I have a vector space $V$ along with a basis $\lbrace\mathbf{e}_{i}\rbrace$, then there is a unique linear map $\tilde{e}^{i}: V\rightarrow \mathbb{F}$ defined by $$\tilde{e}^{i}(\mathbf{v})=v^{i} \Rightarrow \tilde{e}^{i}(\mathbf{e}_{j})= \delta^{i}_{j}$$
which maps each vector $\mathbf{v}\in V$ to its $i^{th}$ component $v^{i}\in \mathbb{F}$ with respect to the basis vector $\mathbf{e}_{i}$. The set of linear maps $\lbrace\tilde{e}^{i}\rbrace$ form a basis for the dual space, $V^{\ast}$, of $V$.

To prove that this map is unique suppose that we have some other linear map $\tilde{f}^{i}$ which also satisfies $\tilde{f}^{i}(\mathbf{e}_{j})= \delta^{i}_{j}$, then as $\lbrace\mathbf{e}_{i}\rbrace$ is a basis for $V$ we can express a vector $\mathbf{v}\in V$ as a unique linear combination $\sum_{j}v^{j}\mathbf{e}_{j}$, and so

$\tilde{e}^{i}\left(\mathbf{v}\right)= \tilde{e}^{i}(\sum_{j}v^{j}\mathbf{e}_{j})= \sum_{j}v^{j} \tilde{e}^{i}(\mathbf{e}_{j})= \sum_{j}v^{j}\delta^{i}_{j} = \sum_{j}v^{j}\tilde{f}^{i}(\mathbf{e}_{j}) = \tilde{f}^{i}\left(\sum_{j}v^{j}\mathbf{e}_{j}\right) = \tilde{f}^{i}(\mathbf{v})$

Hence, as $\mathbf{v}$ was chosen arbitrarily, we conclude that $\tilde{e}^{i}=\tilde{f}^{i}$ and as such the mapping $\tilde{e}^{i}: V\rightarrow V^{\ast}$ defined by $$\tilde{e}^{i}(\mathbf{v})=v^{i}$$ is unique.

To prove that such a mapping exists, let $\mathbf{v}$, and as above, express it in terms of the basis $\lbrace\mathbf{e}_{i}\rbrace$, $\sum_{j}v^{j}\mathbf{e}_{j}$. Now, as the scalars $v^{i}$ are uniquely determined by $\mathbf{v}$ and therefore, $v^{i}$ is a uniquely determined element of $\mathbb{F}$. This gives us a well defined rule for obtaining an element of $\mathbb{F}$ from $V$, i.e. a function from $V$ to $\mathbb{F}$. Thus, there is a function $\tilde{e}^{i}:V \rightarrow \mathbb{F}$ satisfying $$\tilde{e}^{i}\left(\mathbf{v}\right)= v^{i}.$$

Now we show that this is a linear map. Let $\mathbf{u}, \mathbf{v}\in V$ and $\alpha, \beta \in\mathbb{F}$. Let $\mathbf{u}=\sum_{i}a^{i}\mathbf{e}_{i}$ and $\mathbf{v}=\sum_{i}b^{i}\mathbf{e}_{i}$, with respect to the basis $\lbrace\mathbf{e}_{i}\rbrace$. Then

$$\alpha\mathbf{u} + \beta\mathbf{v} = \sum_{i}\left(\alpha a^{i} + \beta b^{i}\right)\mathbf{e}_{i},$$
and the definition of $\tilde{e}^{i}$ gives
$$\tilde{e}^{i}\left(\alpha\mathbf{u} + \beta\mathbf{v}\right) = \left(\alpha a^{i} + \beta b^{i}\right) = \alpha a^{i} + \beta b^{i}= \alpha\tilde{e}^{i}\left(\mathbf{u}\right) + \beta\tilde{e}^{i}\left(\mathbf{v}\right)$$
and so $\tilde{e}^{i}$ is linear.

Sorry for the long-windedness of this post, but I just wanted to check whether my understanding is correct?

2. Dec 6, 2014

### Stephen Tashi

It's extremely hard for a person to check mathematics that he already believes! These results assume a finite dimensional vector space, correct? Which result establishes that the ${\tilde{e}^{i} }$ are a basis for $V^*$? (i.e. given an arbitrary linear functional $\tilde{v}$, where is it shown that $\tilde{v}$ can be expessed as a linear combination of the $\tilde{e}^{i}$ ?

3. Dec 7, 2014

### Fredrik

Staff Emeritus
It is.

Not sure if you're asking for yourself or if you intended it to be an exercise for the OP. Either way, here's a hint:
Let $\mathbf v\in V$ be arbitrary. What is $\tilde e^i(\mathbf v)$?

4. Dec 8, 2014

### "Don't panic!"

Ok cool, thanks for taking a look over it. Apologies for the lack of rigour in parts (particularly forgetting to mention that it is a finite dimensional vector space), I was just trying to sort it out in my head really.

Also, why is the dual space to a particular basis for a given vector space $V$ introduced? One thought I had is that it allows one to uniquely determine the components of each of the vectors $v\in V$ with respect to the chosen basis $\lbrace\mathbf{e}_{i} \rbrace$, when there is no notion of an inner product (and as such orthogonality of vectors), etc?!

5. Dec 8, 2014

### Fredrik

Staff Emeritus
You don't need the dual space for that. The uniqueness of the components, i.e. the numbers $x_i$ in the expansion $x=\sum_{i=1}^n x_i e_i$, follows from the linear independence of the set $\{e_1,\dots,e_n\}$. (Suppose that we also have $x=\sum_{i=1}^n y_i e_i$. Then we have $0=\sum_i (x_i-y_i)e_i$, and since $\{e_1,\dots,e_n\}$ is linearly independent, that implies that $x_i-y_i=0$ for all i).

Edit: I see now that you weren't talking about their uniqueness in general, but rather about having a simple formula for them.

Dual spaces are used mainly to define tensors in a way that's much more elegant and much less confusing than the old-fashioned "transforms as" definition. The concept also has its uses in functional analysis, in particular the mathematics of quantum mechanics. For example, it shows up in the rigorous definition of the adjoint operation ($A\mapsto A^*$ or $A\mapsto A^\dagger$).

6. Dec 8, 2014

### "Don't panic!"

Yes, sorry I was referring to the particular form of the components rather than their uniqueness in general.

Is that through using differential forms and relating them to covariant vector field components (in this case one-forms, $dx^{\mu}$), as using tangent vectors as a basis (e.g. $\frac{\partial}{\partial x^{i}}$) only enables one to describe contravariant components?

7. Dec 8, 2014

### Fredrik

Staff Emeritus
Yes (if I understand the question correctly). It's sometimes convenient to associate something like an n-tuple of numbers with each coordinate system, in a way that makes the n-tuple transform under a change of coordinates, in the "opposite" way of how tangent vector component n-tuples transform. The dual space gives us an easy way to do this.

8. Dec 8, 2014

### Stephen Tashi

Properly speaking, the dual space $V^*$ of a vector space $V$ is "the" dual space of $V$, it isn't merely a dual space to a particular basis for $V$.

The $\tilde{e}^i$ are one particular basis for $V^*$ and it's fair to say they are defined relative to a particular basis for $V$.

9. Dec 9, 2014

### "Don't panic!"

Ok, thanks for your help Fredrik!

Yes, sorry that's what I meant really, just didn't think about the wording when I posted my comment.