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A finite family of vectors in a vector space over a specified field $F$ is linearly dependent if it has a nontrivial linear relation: ${\sum }_i{a}_i{v}_i=0$ with $a_i\in F$ and at least one coefficient nonzero. One can then solve for a vector whose coefficient is nonzero as a linear combination of the others; removing that entry preserves the linear span. Repeated vectors in a list give linear dependence: for example $(v,v)$ has the linear relation $1v+(-1)v=0$. A list keeps both entries even though the set $\{v,v\}$ has only one element. Strang's notes illustrate redundant columns by expressing one as the sum of two others.^[ref]

Keep $v_1=(1,1)$ and $v_2=(1,-1)$, and add $v_3=(2,0)=v_1+v_2$. The three vectors still have linear span ${\mathbb{R}}^2$, but nowThe two coefficient triples $(2,1,0)$ and $(1,0,1)$ differ by the nontrivial linear relation $(-1,-1,1)$. In fact all linear combination representations of this target areTo check both completeness and correctness, use $c_1{v}_1+c_2{v}_2+c_3{v}_3=(c_1+c_3)v_1+(c_2+c_3)v_2$. Linear independence of $v_1,v_2$ forces $c_1+c_3=2$ and $c_2+c_3=1$, giving exactly the displayed family. Redundancy therefore preserves attainable targets while allowing their coefficients to move. For solving a linear system, these alternatives matter; for assigning unique coordinate vectors, one instead chooses a basis.

The following example shows why linear independence depends on the scalar field.