Exercise 1899Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

Hyperplanes and linear forms

#hyperplane#linear form#kernel#dimension

Statement

Let $E$ be a finite-dimensional vector space of dimension $n\ge1$ . Show that a subspace $H$ is a hyperplane (i.e. $\dim H=n-1$ ) if and only if there exists a nonzero linear form $\varphi\in E^*$ with $H=\ker\varphi$ .

Hint

For the forward direction, take a basis of $H$ , extend it by one vector, and define $\varphi$ to vanish on $H$ and be $1$ on the extra vector.

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Solution

If $H=\ker\varphi$ with $\varphi\neq0$ . Then $\varphi$ is surjective (nonzero linear form onto $\mathbb{R}$ ), so by rank-nullity, $\dim\ker\varphi=n-1$ . So $H$ is a hyperplane.

Conversely, assume $\dim H=n-1$ . Choose a basis $(e_1,\ldots,e_{n-1})$ of $H$ , and complete it with $e_n$ to a basis of $E$ . Define $\varphi$ to be the linear form with $\varphi(e_i)=0$ for $i\le n-1$ and $\varphi(e_n)=1$ . Then $\varphi$ is a nonzero linear form with kernel exactly $H$ (vectors with zero $e_n$ -coordinate).

Conclusion. Hyperplanes are exactly the kernels of nonzero linear forms.

Hyperplanes and linear forms

Statement

Solution

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