Linear Independence and Spanning Sets

A vector space is defined as a set with two operations, meeting ten properties (). Just as the definition of span of a set of vectors only required knowing how to add vectors and how to multiply vectors by scalars, so it is with linear independence. A definition of a linearly independent set of vectors in an arbitrary vector space only requires knowing how to form linear combinations and equating these with the zero vector. Since every vector space must have a zero vector (), we always have a zero vector at our disposal.

In this section we will also put a twist on the notion of the span of a set of vectors. Rather than beginning with a set of vectors and creating a subspace that is the span, we will instead begin with a subspace and look for a set of vectors whose span equals the subspace.

The combination of linear independence and spanning will be very important going forward.

1. \set{ \begin{bmatrix} 1&3\\-2&4 \end{bmatrix},\, \begin{bmatrix} -2&3\\3&-5 \end{bmatrix},\, \begin{bmatrix} 0&9\\-1&3 \end{bmatrix} }
2. span
   1. $S$ is a subset of the vector space $V$ and the span of $S$ is a subspace of $V$.
   2. $W$ is a subspace of the vector space $Y$ and $T$ spans $W$.
3. S=\set{ \colvector{6\\2\\1},\, \colvector{4\\-3\\1},\, \colvector{5\\8\\2} }

Our previous definition of linear independence () employed a relation of linear dependence that was a linear combination on one side of an equality and a zero vector on the other side. As a linear combination in a vector space () depends only on vector addition and scalar multiplication, and every vector space must have a zero vector (), we can extend our definition of linear independence from the setting of $\complex{m}$ to the setting of a general vector space $V$ with almost no changes. Compare these next two definitions with and .

Notice the emphasis on the word only. This might remind you of the definition of a nonsingular matrix, where if the matrix is employed as the coefficient matrix of a homogeneous system then the only solution is the trivial one.

Condensed — the full section is in Beezer, A First Course in Linear Algebra.

In a vector space $V$, suppose we are given a set of vectors $S\subseteq V$. Then we can immediately construct a subspace, $\spn{S}$, using and then be assured by that the construction does provide a subspace. We now turn the situation upside-down. Suppose we are first given a subspace $W\subseteq V$. Can we find a set $S$ so that $\spn{S}=W$? Typically $W$ is infinite and we are searching for a finite set of vectors $S$ that we can combine in linear combinations and build all of $W$.

I like to think of $S$ as the raw materials that are sufficient for the construction of $W$. If you have nails, lumber, wire, copper pipe, drywall, plywood, carpet, shingles, paint (and a few other things), then you can combine them in many different ways to create a house (or infinitely many different houses for that matter). A fast-food restaurant may have beef, chicken, beans, cheese, tortillas, taco shells and hot sauce and from this small list of ingredients build a wide variety of items for sale. Or maybe a better analogy comes from Ben Cordes the additive primary colors (red, green and blue) can be combined to create many different colors by varying the intensity of each. The intensity is like a scalar multiple, and the combination of the three intensities is like vector addition. The three individual colors, red, green and blue, are the elements of the spanning set.

Because we will use terms like spanned by and spanning set, there is the potential for confusion with the span. Come back and reread the first paragraph of this subsection whenever you are uncertain about the difference. Here is the working definition.

The definition of a spanning set requires that two sets (subspaces actually) be equal. If $S$ is a subset of $V$, then $\spn{S}\subseteq V$, always. Thus it is usually only necessary to prove that $V\subseteq\spn{S}$. Now would be a good time to review .

Given a subspace and a set of vectors, as in it can take some work to determine that the set actually is a spanning set. An even harder problem is to be confronted with a subspace and required to construct a spanning set with no guidance. We will now work an example of this flavor, but some of the steps will be unmotivated. Fortunately, we will have some better tools for this type of problem later on.

Condensed — the full section is in Beezer, A First Course in Linear Algebra.

In we will take up the matter of representations fully, where will be critical for . We will now motivate and prove a critical theorem that tells us how to represent a vector. This theorem could wait, but working with it now will provide some extra insight into the nature of linearly independent spanning sets. First an example, then the theorem.

The converse of is true as well, but is not important enough to rise beyond an exercise (see ).

Condensed — the full section is in Beezer, A First Course in Linear Algebra.