Theorem SLSLC tells us that a solution to a homogeneous system of equations is a linear combination of the columns of the coefficient matrix that equals the zero vector. We used just this situation to our advantage (twice!) in Example SCAD where we reduced the set of vectors used in a span construction from four down to two, by declaring certain vectors as surplus. The next two definitions will allow us to formalize this situation.
Given a set of vectors $S=\set{\vectorlist{u}{n}}$, a true statement of the form \begin{equation*} \lincombo{\alpha}{u}{n}=\zerovector \end{equation*} is a relation of linear dependence on $S$. If this statement is formed in a trivial fashion, i.e. $\alpha_i=0$, $1\leq i\leq n$, then we say it is the trivial relation of linear dependence on $S$.
The set of vectors $S=\set{\vectorlist{u}{n}}$ is linearly dependent if there is a relation of linear dependence on $S$ that is not trivial. In the case where the only relation of linear dependence on $S$ is the trivial one, then $S$ is a linearly independent set of vectors.
Notice that a relation of linear dependence is an equation. Though most of it is a linear combination, it is not a linear combination (that would be a vector). Linear independence is a property of a set of vectors. It is easy to take a set of vectors, and an equal number of scalars, all zero, and form a linear combination that equals the zero vector. When the easy way is the only way, then we say the set is linearly independent. Here are a couple of examples.
Example LDS and Example LIS relied on solving a homogeneous system of equations to determine linear independence. We can codify this process in a time-saving theorem.
Suppose that $S=\set{\vectorlist{v}{n}}\subseteq\complex{m}$ is a set of vectors and $A$ is the $m\times n$ matrix whose columns are the vectors in $S$. Then $S$ is a linearly independent set if and only if the homogeneous system $\homosystem{A}$ has a unique solution.
Since Theorem LIVHS is an equivalence, we can use it to determine the linear independence or dependence of any set of column vectors, just by creating a matrix and analyzing the row-reduced form. Let us illustrate this with two more examples.
As an equivalence, Theorem LIVHS gives us a straightforward way to determine if a set of vectors is linearly independent or dependent.
Review Example LIHS and Example LDHS. They are very similar, differing only in the last two slots of the third vector. This resulted in slightly different matrices when row-reduced, and slightly different values of $r$, the number of nonzero rows. Notice, too, that we are less interested in the actual solution set, and more interested in its form or size. These observations allow us to make a slight improvement in Theorem LIVHS.
Suppose that $S=\set{\vectorlist{v}{n}}\subseteq\complex{m}$ is a set of vectors and $A$ is the $m\times n$ matrix whose columns are the vectors in $S$. Let $B$ be a matrix in reduced row-echelon form that is row-equivalent to $A$ and let $r$ denote the number of pivot columns in $B$. Then $S$ is linearly independent if and only if $n=r$.
So now here is an example of the most straightforward way to determine if a set of column vectors is linearly independent or linearly dependent. While this method can be quick and easy, do not forget the logical progression from the definition of linear independence through homogeneous system of equations which makes it possible.
The situation in Example LLDS is slick enough to warrant formulating as a theorem.
Suppose that $S=\set{\vectorlist{u}{n}}\subseteq\complex{m}$ and $n>m$. Then $S$ is a linearly dependent set.