3

My lecturer gave us this result today in class, but he didn't give a proof, he said we can prove it ourselves, only I'm really struggling to see how to do it.

Let $E$ be a normed space with dual $E'$. Then $E$ is finite dimensional if and only if $E'$ is finite dimensional, and in fact $\dim{E} =\dim{E'}$.

Would appreciate any input!

Victoria
  • 307
  • Do you know that if $E$ is finite-dimensional then $E'$ is finite-dimensional with $\dim E' = \dim E$? – Daniel Fischer Oct 29 '14 at 16:55
  • @DanielFischer, no, why is that true? – Victoria Oct 29 '14 at 17:00
  • 1
    Because on a finite-dimensional normed space, every linear functional is continuous. For the algebraic dual [disregarding continuity, since that is generally not even defined in that setting], that is proved in linear algebra. Since the space of continuous linear functionals is a subspace of the space of all linear functionals, you know $\dim E' \leqslant \dim E^\ast = \dim E$ (where $E^\ast$ denotes the algebraic dual), and $\dim E' = \dim E^\ast$ is equivalent to the continuity of all linear functionals. – Daniel Fischer Oct 29 '14 at 17:07
  • @DanielFischer OK, but I still wouldn't know how to show that $\dim{E^*} = \dim{E}$? Am I missing something obvious? – Victoria Oct 29 '14 at 17:15
  • Think back to your linear algebra. And go over your notes again, without a solid grounding in linear algebra you're screwed in functional analysis. What was the relation between bases of a vector space and linear mappings from there to other vector spaces? – Daniel Fischer Oct 29 '14 at 17:18
  • @DanielFischer I can't think what relation you're referring to, and can't find anything in my notes – Victoria Oct 29 '14 at 19:08
  • If you haven't been taught that every linear map is determined by its values on a basis, and that every map from a basis of $E$ into a vector space $V$ can be (uniquely) extended to a linear map $E\to V$ in linear algebra, change the university. A direct consequence of that is that the algebraic dual of a finite-dimensional vector space $E$ (over any field) is finite-dimensional and has the same dimension as $E$. – Daniel Fischer Oct 29 '14 at 19:21

2 Answers2

6

Suppose that $E$ is finite dimensional. Let $x_1, \ldots, x_n$ be a basis for $E$. Consider the coordinate functionals $f_1, \ldots, f_n$ given by

$$\langle f_i, \sum_{k=1}^n \lambda_k x_k \rangle = \lambda_i\quad (i\leqslant n)$$

Prove that $f_1, \ldots, f_n$ are linearly independent. Having this done, note that they span $E^\prime$. Indeed, if $f$ is any functional on $E$, then

$$\langle f, \sum_{k=1}^n \lambda_k x_k\rangle = \sum_{i=1}^n \lambda_k\langle f,x_k\rangle $$

so $f = \sum_{i=1}^n \langle f, x_k \rangle f_i$ because

$$\langle \sum_{i=1}^n \langle f, x_k \rangle f_i, \sum_{k=1}^n \lambda_k x_k\rangle = \sum_{k=1}^n\sum_{i=1}^n \lambda_k \langle f_i, x_k\rangle \langle f, x_k \rangle $$

Note that $\langle f_i, x_k\rangle = 0$ unless $i=k$ in which case this is equal to 1. Consequently, the right hand side is equal to $\sum_{i=1}^n \lambda_k\langle f,x_k\rangle$.

We thus proved that $f_1, \ldots, f_n$ form a basis for $E^\prime$. So $E$ and $E^\prime$ have the same dimension.

Tomasz Kania
  • 16,361
3

See if you can figure it out from the following hint:

If $E$ is finite dim, by defn. it has a finite basis. Say $x_1, x_2, \dots x_n$ is a basis of $E$.

For any $x \in E$, x can be written uniquely as $x = a_1x_1 + a_2x_2 + \dots a_n x_n$. Try to prove the following:

1) For $j \in \{1, \dots n \}$, the maps $x_j^*: X \to \mathbb{R}$ defined by $$ x_j^*(a_1x_1 + a_2x_2 + \dots a_n x_n) = a_j $$ are in the dual space of $E$. 2) Can you show the $x_j^*$ form a basis of the $E'$?

Matthew25
  • 326
  • I'm not sure how to do this? for the linear independence, I have to show that $\lambda_1 x_1^(x) + \dots + \lambda_n x_n^(x) = 0$ means that the $\lambda_i$'s are $0$? but I've never shown linear independence of anything other than really easy vectors before, so I'm getting a little muddled with what's actually going on. – Victoria Oct 29 '14 at 18:45
  • What you've said is almost correct - to see the vectors $x_i^$ in $E'$ are linearly independent, we need to show that if $\lambda_1 x_1^ + \dots \lambda_n x_n^* = 0$ then all $\lambda_i$ are zero. What happens if you apply the $x_i$ to the previous equation? – Matthew25 Oct 29 '14 at 20:35