Let S = \{v_1, v_2, \ldots, v_n\} be a set of vectors in a vector space V.
Linear Independence
Let \(S = \{v_1, v_2, \ldots,
v_n\}\) be a set of vectors in a vector space \(V\).
If there exist scalars \(\alpha_1, \alpha_2 \ldots \alpha_n \in F\) such that not all of the \(\alpha_i\)'s are zero and
\[\begin{aligned}\end{aligned}\],
then \(S\) is said to be linearly dependent.
If the set \(S\) is not linearly dependent,
then it is said to be linearly independent.
More specifically, \(S\) is a linearly independent set if
\[\begin{aligned}\end{aligned}\]
implies that
\[\begin{aligned}\end{aligned}\]
for any set of scalars \(\{ \alpha_1, \alpha_2 \ldots \alpha_n \}\).
The definition of linear dependence makes more sense if we consider the following proposition.
The following proposition is a consequence of the fact that any system of homogeneous linear equations with more unknowns than equations will have a nontrivial solution.
We leave the details of the proof for the end-of-chapter exercises.
A set \(\{ e_1, e_2, \ldots,
e_n \}\) of vectors in a vector space \(V\) is called a
basis for \(V\) if
\(\{ e_1, e_2, \ldots,
e_n \}\) is a linearly independent set that spans \(V\).
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A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group. That is where the idea came from.
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A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.