Lecture · 16 min · advanced

Vectors

Linear combinations and the span of a set

The four fundamental subspaces · linear combinations

A vector is not an arrow with a pretty head. It is an object you can add and scale. In $R^{n}$ that object is an $n$ -tuple, and the whole subject of linear algebra is: which lists of vectors are enough to build every other vector by addition and scaling?

On the board today

—Form linear combinations $c_{1} v_{1} + \dots + c_{k} v_{k}$ .
—Describe $span {v_{1}, \dots, v_{k}}$ as a subspace.
—Detect linear dependence by a nontrivial relation.

Span

span {v_{1}, \dots, v_{k}} = {c_{1} v_{1} + \dots + c_{k} v_{k} : c_{i} \in R}

Every vector you can build from the list, using addition and scaling only.

Two nonzero vectors in $R^{2}$ span the plane if they do not lie on the same line. Three vectors in $R^{2}$ are always dependent: there is a nontrivial linear relation $c_{1} v_{1} + c_{2} v_{2} + c_{3} v_{3} = 0$ . Independence means the only way to get $0$ is all coefficients zero — no vector in the list is redundant.

Worked example · A dependent triple

Are $v_{1} = (1, 0)$ , $v_{2} = (0, 1)$ , $v_{3} = (1, 1)$ independent in $R^{2}$ ?

1. $v_{3} = v_{1} + v_{2}$ , so $1 \cdot v_{1} + 1 \cdot v_{2} + (- 1) \cdot v_{3} = 0$ .
2.A nontrivial relation exists, so the triple is dependent.
3.Any two of them already span $R^{2}$ .

Dependent. The third vector lives in the span of the first two.

Pause the lecture

Can four vectors in $R^{3}$ be independent?

Common mistake. Calling a single zero vector “independent.” The set ${0}$ is dependent: $1 \cdot 0 = 0$ is a nontrivial relation (the coefficient is not zero).

Practice this lecture

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