(The *degenerate* case) If $\mathbf a \in K^{n}$, and if $\mathbf a \cdot \mathbf x=0$ for all $\mathbf x \in K^{n}$ then $\mathbf a=\mathbf 0$ $\mathbf a \in K^{n}:\ \forall\mathbf x \in K^{n},\ \mathbf a \cdot \mathbf x=0 \implies\mathbf a=\mathbf 0$ (The *non-degenerate* case) If a vector $\mathbf a$ is multiplied with at least one element to get a non-zero result, then $\mathbf a \in K^{n}:\ \exists\mathbf x \in K^{n},\ \mathbf a \cdot \mathbf x\neq0 \implies\mathbf a\neq\mathbf 0$ In other words, the [[zero vector]] is the only suitable [[absorbing element]] for the [[dot product]] This is particularly obvious with vectors over $\mathbb{R}^{n}$ since all non-zero vectors will have a positive dot product with themselves $\mathbf b\neq 0 \in \mathbb{R}^n \iff \mathbf b^2 = \mathbf b \cdot \mathbf b > 0 \quad(\mathbf b\in\mathbb{R}^n)$