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:

$(1): \quad \dim \left({M}\right) = n - 1$;

$(2): \quad M$ is the kernel of a nonzero linear form;

$(3): \quad$ 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:

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

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965): $\S 28$: Theorem $28.11$