Dual Bases

Jen · #1 $Old$ November 3rd, 2009, 06:02 PM

I am having a really hard time understanding what a dual basis is.

I understand that $V^*=\mathcal{L}(V, F)$ is the vector space of linear transformations from a vector space V onto its field of scalars F, also known as linear functionals.

I get this. What I am struggling on is how to find the dual basis and why it is defined the way it is.

We call the ordered basis $\beta ^*= \{ f_1, f_2, ..., f_n \}$ of $V^*$ that satisfies $f_i (x_j)=\delta_{ij}$ , $1 \le i,j \le n$ , the dual basis of $V^*$

The book only gave on example using $\mathbb{R}^2$ . I followed what they were doing but it doesn't help me understand the dual basis any better.

Based on the way we set up bases for any other vector space, I quess I figured that the dual basis would somehow be a set of linear functionals such that every linear functional could be written as a linear combination of these linear functionals.

I don't know, I am lost. Can someone explain this to me better or possibly refer me to a good website to read in more depth about dual spaces and dual bases?

Jen · #2 $Old$ November 3rd, 2009, 08:40 PM

Perhaps I should be more specific in my confusion.

In a problem that I am working on I am asked a question involving the vector space $V=P( \mathbb{R} )$ , the vector space of polynomials with coefficients from $\mathbb{R}$ .

I know that $\beta = \{ 1, x, x^2, ... \}$ is a basis for $V$ .

If I blindly apply the definition of a Dual Basis (which I mentioned previously) then I know,

$f_1(1)=1$
$f_1(x)=0$
...

and $f_2(1)=0$
$f_2(x)=1$
$f_2 \left( x^2 \right) = 0$

and so on...

What does this mean?

I don't even know if knowing what $V^*$ is will help me with this problem but I want to know.

Does this mean that $f_1$ is some linear functional that takes the first vector in my basis and maps it to 1 and every other vector in the basis to zero?

What does $V^*$ look like??

Please help.

Thank you in advance too!

tonio · #3 $Old$ November 3rd, 2009, 08:57 PM

Quote:

Originally Posted by Jen $View Post$

I am having a really hard time understanding what a dual basis is.

I understand that $V^*=\mathcal{L}(V, F)$ is the vector space of linear transformations from a vector space V onto its field of scalars F, also known as linear functionals.

I get this. What I am struggling on is how to find the dual basis and why it is defined the way it is.

We call the ordered basis $\beta ^*= \{ f_1, f_2, ..., f_n \}$ of $V^*$ that satisfies $f_i (x_j)=\delta_{ij}$ , $1 \le i,j \le n$ , the dual basis of $V^*$

No: if $X:=\{x_i\}_{i=1}^n$ is a basis of V, then $\{f_i\}_{i=1}^n$ as defined above is THE dual basis of X in $V^*$ , or wrt the basis X.
Its importance resides,among other possible things, on the fact that it is one simple way to show that when we're dealing with finite dimensional vector spaces, then $V\cong V^*$

The book only gave on example using $\mathbb{R}^2$ . I followed what they were doing but it doesn't help me understand the dual basis any better.

Based on the way we set up bases for any other vector space, I quess I figured that the dual basis would somehow be a set of linear functionals such that every linear functional could be written as a linear combination of these linear functionals.

Exactly, but NOT only: every element in V* indeed can be expressed as a lin. comb. of a dual basis, WHERE the elements of this basis get very simple and explicit values on some given basis of V.

I don't know, I am lost. Can someone explain this to me better or possibly refer me to a good website to read in more depth about dual spaces and dual bases?

Any decent linear algebra book covers all this subject, with dual and double dual spaces (V** is canonically isomorphic to V, for example), etc.

Tonio

tonio · #4 $Old$ November 3rd, 2009, 09:00 PM

Quote:

Originally Posted by Jen $View Post$

Perhaps I should be more specific in my confusion.

In a problem that I am working on I am asked a question involving the vector space $V=P( \mathbb{R} )$ , the vector space of polynomials with coefficients from $\mathbb{R}$ .

I know that $\beta = \{ 1, x, x^2, ... \}$ is a basis for $V$ .

If I blindly apply the definition of a Dual Basis (which I mentioned previously) then I know,

$f_1(1)=1$
$f_1(x)=0$
...

and $f_2(1)=0$
$f_2(x)=1$
$f_2 \left( x^2 \right) = 0$

and so on...

What does this mean?

I don't even know if knowing what $V^*$ is will help me with this problem but I want to know.

Does this mean that $f_1$ is some linear functional that takes the first vector in my basis and maps it to 1 and every other vector in the basis to zero?

What does $V^*$ look like??

Please help.

Thank you in advance too!

It's hard to know if dual basis will help you if you are not explicit about the problem you're dealing with.
V* is just a set of functions with a binary operation on it and a multiplication with scalars from a field which makes it into a linear space.

Tonio

Jen · #5 $Old$ November 3rd, 2009, 09:46 PM

Yeah, I know that you have no way of knowing if understanding the dual basis will help in a specific problem if I don't offer it up, but I don't want to post the full questions because somtimes people just answer the whole thing rather than giving hints and that doesn't help me, so I just post questions about the question.

Sorry, I asked previously what $V^*$ looked like. I meant to ask what the basis for $V^*$ looked like.

Is there a way to come up with an explicit basis for $\left( P( \mathbb{R} )\right)^*$ ?

I mean I know that it is an infinite dimensional basis so obviously we couldn't list every element...