### Vector spaces: Spans

### Spanning sets

For each set of vectors of a vector space, there is a smallest *linear subspace* that contains all the vectors of the set. The linear subspace can be defined as the set of all linear combinations of the vectors from the system and is called the linear span. Below we discuss the concepts of linear combination and span.

Linear combination

Let #n# be a natural number and let #\vec{a}_1 ,\ldots ,\vec{a}_n# be #n# vectors in a vector space #V#.

A vector #\vec{x}# of #V# is called a **linear** **combination** of #\vec{a}_1 ,\ldots ,\vec{a}_n# if there exist *scalars* #\lambda_1,\ldots ,\lambda_n# such that \[ \vec{x} =\lambda_1 \cdot\vec{a}_1 +\lambda_2 \cdot\vec{a}_2 +\cdots +\lambda_n\cdot \vec{a}_n \]

We then also say that the vector #\vec{x}# (**linearly**)** depends** on the vectors #\vec{a}_1 ,\ldots ,\vec{a}_n#.

We are particularly interested in the sets of all linear combinations of a given set of vectors.

Span

Let #n# be a natural number and let #\vec{a}_1 ,\ldots ,\vec{a}_n# be #n# vectors in a vector space.

The set of all linear combinations of #\vec{a}_1 ,\ldots ,\vec{a}_n# is called the **space spanned** by the vectors #\vec{a}_1 ,\ldots ,\vec{a}_n# or their (linear) **span**, and is denoted as

\[\linspan{ \vec{a}_1 , \ldots, \vec{a}_n}\]

We agree that the span of nothing (the empty set) is equal to #\linspan{}=\{0\}#.

The span of a set of vectors forms a constructive way of determining the smallest subspace containing those vectors:

Spans are linear subspaces

Let #\vec{a}_1 ,\ldots ,\vec{a}_n# be #n# vectors of a vector space #V#. The span #\linspan{\vec{a}_1 ,\ldots ,\vec{a}_n }# is a linear subspace of #V#.

The span is contained in all linear subspaces of #V# which contain the vectors #\vec{a}_1 ,\ldots ,\vec{a}_n#. In other words, the span is the smallest linear subspace containing #\vec{a}_1 ,\ldots ,\vec{a}_n#.

Write #W=\linspan{\vec{a},\vec{b},\vec{c}}# .

Every linear combination of #\vec{a}# and #\vec{c}# is also a linear combination of #\vec{a}, \vec{b}#, and #\vec{c}# (add #0\,\vec{b}# to it), so #\linspan{\vec{a},\vec{c}}# is contained in #W#.

The point is to show that every #\vec{x} \in W# also belongs to #\linspan{\vec{a},\vec{c}}#. Such a vector #\vec{x}# can be written as

\[ \vec{x} =x_1 \cdot\vec{a} +x_2\cdot \vec{b} +x_3\cdot \vec{c}\] Crucial for the proof is the observation \[\vec{b} =2\vec{c} -\vec{a}\] This means that

\[ \begin{array}{rcl}

\vec{x}&=& x_1 \cdot\vec{a} + x_2\cdot (2\vec{c} -\vec{a}) +x_3\cdot \vec{c} \\

& =&\left(x_1 -x_2\right )\cdot \vec{a} +\left(2x_2 +x_3\right )\cdot \vec{c}

\end{array} \]

Thus, each vector of #W# is a linear combination of #\vec{a}# and #\vec{c}#, so #W# is contained in #\linspan{\vec{a},\vec{c}}#. The conclusion is #W=\linspan{\vec{a},\vec{c}}#, which is what we had to prove.

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