# Having trouble understanding dual basis?

In my linear algebra textbook it says that a dual basis for $V^{*}$ is {${f_{1}, ..., f_{n}}$} where $f_{i}(x)$ is the function that takes a vector in your vector space and writes it as a linear combination of your basis vectors and its value is equal to the coefficient in front of the ith basis vector. My book gave a proof about why this is the case. But I am having trouble conceptually understanding why this is the case. I was wondering whether someone could intuitively explain why this forms a basis for the dual space. Also I'm having trouble understanding the elements in the dual basis. How do you evaluate $f_{i}$ without being given a vector.

• Check something called the Kronecker delta function. Apr 17, 2016 at 20:53
• I know what that is it just says if i isn't equal to j then the value is 0, but if i =j then the value is 1. Apr 17, 2016 at 20:54
• "the function that takes a vector in your vector space and writes it as a linear combination of your basis vectors and its value is equal to the coefficient in front of the ith basis vector. " -- I think if you carefully write down in symbols the definition in your book, this will become more clear. Apr 17, 2016 at 20:55
• I was wondering how would you know what $f_{i}$ is without being given a vector from the vector space? Apr 17, 2016 at 20:58
• Maybe an explicit example would help make it clear what the basis is supposed to look like. Apr 17, 2016 at 20:59

Say we have a finite dimensional vector space, $V$. (In the infinite dimensional case everything I say below is still true, but other things get much, much weirder.) We usually picture elements of $V$ as being fairly concrete - e.g. an arrow pointing in a certain direction.

By contrast, elements of the dual space are much more abstract objects: the dual space consists of all linear maps from $V$ to $\mathbb{R}$ (or whatever field $V$ is a vector space over). On the face of it, this is a totally different kind of object. An element of the dual space is a function - so, e.g., we might think of the dual space of $\mathbb{R}^2$ (as a $\mathbb{R}$-vector space) as the set of all linear functions in two variables with no constant term, for example:

• $x+y$,

• $x-2y$,

• $\pi x$ (that is, $\pi x+0y$),

• etc.

(Of course, this is shorthand: I should really be writing e.g. $\langle x, y\rangle\mapsto x+y$, etc.) The idea here is that the two variables in the expressions above correspond to the two coordinates of the input vector.

On the face of it, an element of the dual space is a totally abstract thing! And so on the face of it, we shouldn't expect any connection between the dual space and the original space.

However, if you think about it for a minute you'll notice that in the example above, all I really need are two linear maps:

• $x+0y$, and

• $0x + y$.

That is, the maps $\langle x, y\rangle\mapsto x$ and $\langle x, y\rangle\mapsto y$. With these maps in hand, I can build more complicated maps, like $\langle x, y\rangle\mapsto \pi x-17ey$.

These are exactly the "basis elements" you're talking about. I think if you think through carefully this example of the dual space of $\mathbb{R}^2$, things will make a lot more sense.

As a short and sweet answer, $V^*$ is a set of linear maps from $V \rightarrow \Bbb{F}$, where $\Bbb{F}$ is the field the vector space is over. This is also sometimes denoted $\operatorname{Hom}(V,\Bbb{F})$.

Given a basis $\{e_i\}$ of $V$, the dual basis is the unique basis $\{f_i\}$ of $V^*$ such that $$f_i(e_j)=\delta_{ij},$$ which of course makes sense because a field has distinct elements $1,0$ always.

I actually think this is explained quite well in Halmos' Finite Dimensional Vector Spaces, but as a linear algebra book it can be tougher to read than some.