Equivalent Statements for Vector Subspace Dimension One Less

Theorem
Let $$K$$ be a field.

Let $$M$$ be a subspace of the $n$-dimensional vector space $K^n$.

The following statements are equivalent:


 * 1) $$\dim \left({M}\right) = n - 1$$;
 * 2) $$M$$ is the kernel of a nonzero linear form;
 * 3) There exists a sequence $$\left \langle {\alpha_n} \right \rangle$$ of scalars, not all of which are zero, such that $$M = \left\{{\left({\lambda_1, \ldots, \lambda_n}\right) \in K^n: \alpha_1 \lambda_1 + \cdots + \alpha_n \lambda_n = 0}\right\}$$.

Also, suppose the above hold.

Let $$\left \langle {\beta_n} \right \rangle$$ be a sequence of scalars such that $$M = \left\{{\left({\lambda_1, \ldots, \lambda_n}\right) \in K^n: \beta_1 \lambda_1 + \cdots + \beta_n \lambda_n = 0}\right\}$$.

Then there is a non-zero scalar $$\gamma$$ such that $$\forall k \in \left[{1 \,. \, . \, n}\right]: \beta_k = \gamma \alpha_k$$.

Proof

 * Let $$M^\circ$$ be the annihilator of $$M$$.

Let $$N = M^{\circ}$$.

By Results Concerning Annihilator of Vector Subspace, $$N$$ is one-dimensional and $$M = J^{-1} \left({N^\circ}\right)$$.

Let $$\phi \in N: \phi \ne 0$$.

Then $$N$$ is the set of all scalar multiples of $$\phi$$.

Because $$J^{-1} \left({N^\circ}\right) = \left\{{x \in K^n: \forall \psi \in N: \psi \left({x}\right) = 0}\right\}$$

it follows that $$J^{-1} \left({N^\circ}\right)$$ is simply the kernel of $$\phi$$.

Hence (1) implies (2).

By Sum of Nullity and Rank of Linear Transformation, (2) also implies (1).


 * Suppose $$\left \langle {\alpha_n} \right \rangle$$ is any sequence of scalars.

Let $$\left \langle {e'_n} \right \rangle$$ be the ordered basis of $$\left({K^n}\right)^*$$ dual to the standard ordered basis of $$K^n$$.

Let $$\phi = \sum_{k=1}^n \alpha_k e'_k$$.

Then, by simple calculation, $$\ker \left({\phi}\right) = \left\{{\left({\lambda_1, \ldots, \lambda_n}\right): \alpha_1 \lambda_1 + \cdots + \alpha_n \lambda_n = 0}\right\}$$.

It follows that $$\phi \ne 0 \iff \exists k \in \left[{1 \,. \, . \, n}\right]: \alpha_k \ne 0$$.

Thus (2) and (3) are equivalent.


 * Suppose $$M = \ker \left({\psi}\right)$$, where $$\psi = \sum_{k=1}^n \beta_k e'_k$$.

Then $$\psi = M^\circ$$.

As $$M^\circ$$ is one-dimensional and since $$\psi \ne 0$$, it follows that $$\exists \gamma \ne 0: \psi = \gamma \phi$$

Therefore $$\forall k \in \left[{1 \,. \, . \, n}\right]: \beta_k = \gamma \alpha_k$$.