Let $N$ be a $p$-dimensional subspace of $E^*$ and $G := N^\perp$. Then $\operatorname{codim} G = p$

Question

Two Hamel bases of the same vector space have the same cardinality, so we define the dimension of a vector space as the cardinality of one of its Hamel basis. I'm trying to prove that

Let $N$ be a $p$-dimensional subspace of $E^*$ and $$ G := N^\perp := \{x \in E : \langle f, x \rangle =0 \text{ for all } f \in N\}. $$ Then $p = \operatorname{codim} G := \dim(E/G)$.

Could you have a check on my attempt?

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Let $\{f_1, \ldots, f_p\}$ be a basis of $N$. We consider the function $$ \Phi:E \to \mathbb R^p, x \mapsto (\langle f_1, x \rangle, \ldots, \langle f_p, x \rangle). $$

Then $\Phi$ is a continuous linear map with $\ker \Phi = G$. Let's prove that $\Phi$ is surjective. Suppose not; then there is $\alpha =(\alpha_1, \ldots, \alpha_p) \in \mathbb R^p \setminus \operatorname{im} \Phi$. Because $\dim (\operatorname{im} \Phi) <\infty$, we get $\operatorname{im} \Phi$ is closed. By Hahn–Banach theorem, there are real numbers $\lambda_1 < \lambda_2$ and a vector $\beta =(\beta_1, \ldots, \beta_p) \in \mathbb R^p \setminus \{0\}$ such that $$ \sum_{i=1}^p \beta_i \langle f_i, x \rangle < \lambda_1 < \lambda_2 < \sum_{i=1}^p \beta_i \alpha_i, $$ which is a contradiction. Hence $\Phi$ is surjective. By first isomorphism theorem for groups, $$ \operatorname{im} \Phi \cong E / \ker \Phi. $$

It follows that $\mathbb R^p \cong E/G$. The claim then follows.

Answer 1

As commented 4 hours ago by @DavidGao, you should probably clarify what exactly is the contradiction, and you don’t really need to say $Φ$ is continuous, or really use Hahn-Banach.

Here is an improved version of your proof, valid for a vector space over an arbitrary field of scalars $K$:

Let $(f_1,\dots,f_p)$ be a basis of $N$. We consider the function $$\Phi:E\to K^p,\;x\mapsto\left(f_1(x),\dots,f_p(x)\right).$$ By the first isomorphism theorem for modules, the two vector spaces $E/G=E/\ker\Phi$ and $\operatorname{im}\Phi$ are isomorphic. Let us prove by contradiction that the latter is of dimension $p$. If it were not, it would be a strict subspace of $K^p$, hence contained in some hyperplane $H$, defined by an equation $\beta_1y_1+\dots+\beta_py_p=0$ for some non-zero $(\beta_1,\dots,\beta_p)\in K^p$. But $\operatorname{im}\Phi\subseteq H$ is equivalent to $\sum_{i=1}^p\beta_if_i=0$, which is impossible since $f_1,\dots,f_p$ are linearly independent.