# If $E$ is finite-dimensional, then $\dim E =\dim E^\star$

I'm trying to prove this well-known result. Could you have a check on my attempt?

Let $$(E, | \cdot|)$$ be a normed linear space and $$E^\star$$ its continuous dual space. If $$E$$ is finite-dimensional, then $$\dim E =\dim E^\star$$.

My proof: Assume $$\{e_1, \ldots, e_n\}$$ is a basis of $$E$$. For $$k \in \{1, \ldots, n\}$$, we define $$f_k : E \to \mathbb R$$ that sends $$x$$ to its $$k$$-th coordinate, i.e., $$f_k \left ( \sum_{m=1}^n \lambda_m e_m \right ) := \lambda_k.$$ It's clear that $$f_1 \ldots, f_n \in E^\star$$. Take $$\alpha_1, \ldots, \alpha_n \in \mathbb R$$ such that $$\sum_{k=1}^n \alpha_k f_k =0_{E^\star}$$. We pick a particular $$\overline x := \sum_{m=1}^n \alpha_m e_m$$. Then $$\sum_{k=1}^n \alpha_k f_k(\overline x) =0$$ implies $$\sum_{k=1}^n \alpha^2_k = 0$$ and thus $$\alpha_1 = \cdots = \alpha_n=0$$. It follows that $$\{f_1 \ldots, f_n\}$$ is linearly independent. Notice that $$\bigcap_{k=1}^n \ker f_k = \{0_E\} \subseteq \ker f, \quad \forall f\in E^\star.$$

Lemma: Let $$f_1, \ldots,f_n,f$$ be linear functionals such that $$\bigcap_{k=1}^n \ker f_k \subseteq \ker f$$. Then $$f$$ is linearly dependent from $$f_1,\ldots,f_n$$.

By our lemma, each $$f\in E^\star$$ is linearly dependent of $$f_1, \ldots, f_n$$. Hence $$\{f_1 \ldots, f_n\}$$ is indeed a basis of $$E^\star$$. This completes the proof.

• Do you know the more general result that $\dim L(F,E) = \dim F \cdot \dim E$, where $L(E,F)$ is the space of linear maps from $E$ to $F$? If so, this yields an easier proof: $\dim E^* = \dim L(E, \mathbb{F}) = \dim E \cdot \dim \mathbb{F} = \dim E$ Jan 24, 2022 at 23:00
• In finite dimension no need to mention the continuity, all linear transformation are continuous. Jan 24, 2022 at 23:00
• This looks good! I especially like the usage of the lemma (which is personal favourite of mine) to show spanning. Unlike Essaidi, I think you should be mentioning continuity, but you should reference the fact that linear transformations on finite-dimensional normed spaces are automatically continuous (rather than, perhaps, just saying "clearly"). Jan 24, 2022 at 23:03
• This is in fact true for any base field. Jan 24, 2022 at 23:12

The basis you constructed is the dual basis of $$e_1,\ldots,e_n$$. The linear independence can be obtained in a marginally easier way if you just evaluate on each $$e_k$$, since $$\Big(\sum_j\alpha_jf_j\Big)(e_k)=\alpha_k$$.
Like Theo I love the Lemma, but it is a bit overkill here. You could just note directly, by evaluating on an $$x$$, that $$f=\sum_jf(e_j)\,f_j.$$