The union of a linearly independent set with a vector outside of its span is also linearly independent

up:: Linear Independence

Let $(V,+,\cdot )$ be a Vector Space, and $S\subset V$ a Linearly Independent set.

Let $v\notin [S]$ outside of the Span of $S$. Then $S\cup \{v\}$ is linearly independent as well:

$$\sum _i{\alpha }_i{v}_i+\beta v=0⟹\beta =0$$

because, if it’s not zero, then we’d contradict our hypothesis ($v$ would be inside $S$‘s span), since

$$v=\sum _i\frac{{\alpha }_i}{\beta }{v}_i$$

Thus, since $S$ is linearly independent, we must have that ${\alpha }_i=0\wedge \beta =0$.

Thus, $S\cup \{v\}$ is linearly independent.

Corollaries

• A set is linearly dependent iff there is a vector which is a linear combination of previous vectors, which means constructing a Linearly Dependent set comes from including a vector which is a Linear Combination of the previous ones
• By contrapositive of the above, it also follows that A set is linearly independent iff all its finite subsets are also L.I.

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References

• Notas sobre Álgebra Linear