Basis of a Dual Space

**Showcase_22** · November 21st, 2009, 02:55 AM

If I have a vector space $U$ and a dual space $U^*$ , with $e_1,...,e_n$ a basis of $U$ and $e^1,....,e^n$ a basis of $U^*$ , does the basis of $U^*$ always have to have this form:

$e^i(e_j)=\delta_{ij}=\begin{cases} 1 \ \ \ \ if \ i=j\\ 0 \ \ \ \ if \ i \neq j \end{cases}$

?

It's easy to prove that this is a basis, but i'm wondering if the basis can only be like this.

If it is, how can I prove it?

**tonio** · November 21st, 2009, 03:09 AM

Originally Posted by Showcase_22

If I have a vector space $U$ and a dual space $U^*$ , with $e_1,...,e_n$ a basis of $U$ and $e^1,....,e^n$ a basis of $U^*$ , does the basis of $U^*$ always have to have this form:

$e^i(e_j)=\delta_{ij}=\begin{cases} 1 \ \ \ \ if \ i=j\\ 0 \ \ \ \ if \ i \neq j \end{cases}$

?

It's easy to prove that this is a basis, but i'm wondering if the basis can only be like this.

If it is, how can I prove it?

You can't because it isn't true: simple, for a given basis in V just choose a basis in V* which is dual to ANOTHER basis in V!

Tonio

**aman_cc** · November 21st, 2009, 03:09 AM

Originally Posted by Showcase_22

If I have a vector space $U$ and a dual space $U^*$ , with $e_1,...,e_n$ a basis of $U$ and $e^1,....,e^n$ a basis of $U^*$ , does the basis of $U^*$ always have to have this form:

$e^i(e_j)=\delta_{ij}=\begin{cases} 1 \ \ \ \ if \ i=j\\ 0 \ \ \ \ if \ i \neq j \end{cases}$

?

It's easy to prove that this is a basis, but i'm wondering if the basis can only be like this.

If it is, how can I prove it?

No, I don' think so.
Proof: We have this theorem that any set of independednt vectors can be expanded to find the basis of a finite-dim vector space. So start with a linear operator which is not in the above format. This can then be expanded to include other linear operators to form the basis of a dual space. And this bases will not be the form you suggested (as the staring operator in not in that form)

[Edit] After seeing Tonio's post. I would rather issue a warning for my post. It might not be correct. But I think as long as $e_1,...,e_n$ a 'fixed' basis of $U$ (for the context of this question) I should still be right.

**Showcase_22** · November 21st, 2009, 04:15 AM

Originally Posted by tonio

You can't because it isn't true: simple, for a given basis in V just choose a basis in V* which is dual to ANOTHER basis in V!

Tonio

umm, what do you mean by "dual to another basis in V"?

Also aman_cc, does a linear operator exist which is not in that form?

$T_U$ sends a vector to a scalar. I'll choose the scalar product and see if that gives me a basis.

ie. $e^i (e_j)=<e_j,e_j>=||e_j||^2$

This is linearly independent (since it's the only vector we have in our prospective basis) so we can extend it:

Basis: $\{ e^i, f^1,.....,f^{n-1} \}$ (and it doesn't matter what $f^p$ is $p=1,...,n-1$ since all we want is a basis where at least one basis vector is not in the form outlined in my previous post).

We have a basis, but we know that $e^i$ is not in the form in my previous post. Hence we have found a basis for $V*$ that is not in the form in my previous post.

Is this how the argument goes (but better worded of course!)?

**aman_cc** · November 21st, 2009, 06:41 AM

Originally Posted by Showcase_22

umm, what do you mean by "dual to another basis in V"?

Also aman_cc, does a linear operator exist which is not in that form?

$T_U$ sends a vector to a scalar. I'll choose the scalar product and see if that gives me a basis.

ie. $e^i (e_j)=<e_j,e_j>=||e_j||^2$

This is linearly independent (since it's the only vector we have in our prospective basis) so we can extend it:

Basis: $\{ e^i, f^1,.....,f^{n-1} \}$ (and it doesn't matter what $f^p$ is $p=1,...,n-1$ since all we want is a basis where at least one basis vector is not in the form outlined in my previous post).

We have a basis, but we know that $e^i$ is not in the form in my previous post. Hence we have found a basis for $V*$ that is not in the form in my previous post.

Is this how the argument goes (but better worded of course!)?

Yes - That's exactly the argument I was trying to give. But as Tonio said, given $\{ e^i, f^1,.....,f^{n-1} \}$ , we can always find a basis of U where these linear operators operate on that basis of U as per your original form.
@Tonio-It would be great if you can plz validate our understanding here.

**tonio** · November 21st, 2009, 07:50 AM

Originally Posted by aman_cc

Yes - That's exactly the argument I was trying to give. But as Tonio said, given $\{ e^i, f^1,.....,f^{n-1} \}$ , we can always find a basis of U where these linear operators operate on that basis of U as per your original form.
@Tonio-It would be great if you can plz validate our understanding here.

Well, think of it as follows: let $\{v_1,...,v_n\}$ be any basis of $V$ , and let $\{f_1,...,f_n\}$ be the dual basis in $V^{*}$ of the basis $\{v_2,v_1,v_3,...,v_n\}$ , thus:

$f_1v_i=\left\{\begin{array}{cc}1&\,\mbox{ if }\,i=2\\0&\,\mbox{ if }\,i=1,3,...,n\end{array}\right.$ , $f_2v_i=\left\{\begin{array}{cc}1&\,\mbox{ if }\,i=1\\0&\,\mbox{ if }\,i=2,3,...,n\end{array}\right.$ , ..., $f_jv_i=\delta_{ij}\,,\,1\leq i\leq n\,,\,3\leq j\leq n$

Well, now forget the basis $\{v_2,v_1,v_3,...,v_n\}$ and concentrate only on $\{v_1,...,v_n\}$ : then, the basis $\{f_1,...,f_n\}\,\mbox{ in }\,V^{*}$ is NOT of the proposed form.

Tonio

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