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:


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

Existence of Scalar for Vector Subspace Dimension One Less
Suppose the above hold.

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 $M = \map {J^{-1} } {N^\circ}$.

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

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

Because:
 * $\map {J^{-1} } {N^\circ} = \set {x \in K^n: \forall \psi \in N: \map \psi x = 0}$

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

Hence $(1)$ implies $(2)$.

By Rank Plus Nullity Theorem, $(2)$ also implies $(1)$.

Suppose $\sequence {\alpha_n}$ is any sequence of scalars.

Let $\sequence { {e_n}'}$ be the ordered basis of $\paren {K^n}^*$ dual to the standard ordered basis of $K^n$.

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

Then, by simple calculation:
 * $\map \ker \phi = \set {\tuple {\lambda_1, \ldots, \lambda_n}: \alpha_1 \lambda_1 + \cdots + \alpha_n \lambda_n = 0}$

It follows that:
 * $\phi \ne 0 \iff \exists k \in \closedint 1 n: \alpha_k \ne 0$

Thus $(2)$ and $(3)$ are equivalent.