Skip to main content
Logo image

Linear Algebra for Team-Based Inquiry Learning 2025 Edition

Steven Clontz, Drew Lewis

Section 2.5 Identifying a Basis (EV5)

Subsection 2.5.1 Warm Up

Remark 2.5.1.

Recall that in Activity 2.2.1 we used the words vector, linear combination, and span to make an analogy with recipes, ingredients, and meals. In this analogy, a recipe was defined to be a list of amounts of each ingredient to build a particular meal.
🔗
🔗

Activity 2.5.2.

Consider the following set of ingredients:
\begin{equation*} S=\left\{\textrm{tomato}, \textrm{olive oil}, \textrm{dough}, \textrm{cheese}, \textrm{pizza sauce}, \textrm{garlic}\right\} \end{equation*}
🔗
(a)
Does "pizza" live inside of \(\vspan(S)\text{?}\)
🔗
🔗
(b)
Identify which ingredients in \(S\) make the set linearly dependent.
🔗
🔗
(c)
Can you think of a subset \(S'\) of \(S\) that is linearly independent and for which "pizza" is still in \(\vspan{S'}\text{?}\)
🔗
🔗
🔗
🔗

Subsection 2.5.2 Class Activities

Activity 2.5.3.

Consider the set of vectors
\begin{equation*} S=\left\{ \left[\begin{array}{c} 3 \\ -2 \\ -1 \\ 0 \end{array} \right], \left[\begin{array}{c} 2 \\ 4 \\ 1 \\ 1 \end{array} \right], \left[\begin{array}{c} 0 \\ -16 \\ -5 \\ -3 \end{array} \right], \left[\begin{array}{c} 1 \\ 2 \\ 3 \\ 0 \end{array} \right], \left[\begin{array}{c} 3 \\ 3 \\ 0 \\ 1 \end{array} \right] \right\}\text{.} \end{equation*}
🔗
(a)
Given
\begin{equation*} \left[\begin{array}{ccccc|c} 3 & 2 & 0 & 1 & 3 & 5\\ -2 & 4 & -16 & 2 & 3 & 2\\ -1 & 1 & -5 & 3 & 0 & 0\\ 0 & 1 & -3 & 0 & 1 & 1 \end{array}\right] \sim \left[\begin{array}{ccccc|c} 1 & 0 & 2 & 0 & 0 & 1\\ 0 & 1 & -3 & 0 & 0 & 1\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right] \end{equation*}
Express the vector \(\left[\begin{array}{c} 5 \\ 2 \\ 0 \\ 1 \end{array} \right]\) as a linear combination of the vectors in \(S\text{,}\) i.e. find scalars such that
\begin{equation*} \unknown \left[\begin{array}{c} 3 \\ -2 \\ -1 \\ 0 \end{array} \right] + \unknown \left[\begin{array}{c} 2 \\ 4 \\ 1 \\ 1 \end{array} \right] + \unknown \left[\begin{array}{c} 0 \\ -16 \\ -5 \\ -3 \end{array} \right] + \unknown \left[\begin{array}{c} 1 \\ 2 \\ 3 \\ 0 \end{array} \right] + \unknown \left[\begin{array}{c} 3 \\ 3 \\ 0 \\ 1 \end{array} \right] = \left[\begin{array}{c} 5 \\ 2 \\ 0 \\ 1 \end{array} \right]\text{.} \end{equation*}
🔗
🔗
(b)
Find a different way to express the vector \(\left[\begin{array}{c} 5 \\ 2 \\ 0 \\ 1 \end{array} \right]\) as a linear combination of the vectors in \(S\text{:}\)
\begin{equation*} \unknown \left[\begin{array}{c} 3 \\ -2 \\ -1 \\ 0 \end{array} \right] + \unknown \left[\begin{array}{c} 2 \\ 4 \\ 1 \\ 1 \end{array} \right] + \unknown \left[\begin{array}{c} 0 \\ -16 \\ -5 \\ -3 \end{array} \right] + \unknown \left[\begin{array}{c} 1 \\ 2 \\ 3 \\ 0 \end{array} \right] + \unknown \left[\begin{array}{c} 3 \\ 3 \\ 0 \\ 1 \end{array} \right] = \left[\begin{array}{c} 5 \\ 2 \\ 0 \\ 1 \end{array} \right]\text{.} \end{equation*}
🔗
🔗
(c)
Consider another vector \(\left[\begin{array}{c} 8 \\ 6 \\ 7 \\ 5 \end{array} \right]\text{.}\) Without computing the RREF of another matrix, do we already know how many ways can this vector be written as a linear combination of the vectors in \(S\text{?}\)
  1. Yes, zero.
    🔗
    🔗
  2. Yes, one.
    🔗
    🔗
  3. Yes, infinitely-many.
    🔗
    🔗
  4. No, computing a new matrix RREF is necessary.
    🔗
    🔗
🔗
🔗
🔗

Activity 2.5.4.

Let’s review some of the terminology we’ve been dealing with...
🔗
(a)
If every vector in a vector space can be constructed as one or more linear combinations of vectors in a set \(S\text{,}\) we can say...
  1. the set \(S\) spans the vector space.
    🔗
    🔗
  2. the set \(S\) fails to span the vector space.
    🔗
    🔗
  3. the set \(S\) is linearly independent.
    🔗
    🔗
  4. the set \(S\) is linearly dependent.
    🔗
    🔗
🔗
🔗
(b)
If the zero vector \(\vec 0\) can be constructed as a unique linear combination of vectors in a set \(S\) (the combination multiplying every vector by the scalar value \(0\)), we can say...
  1. the set \(S\) spans the vector space.
    🔗
    🔗
  2. the set \(S\) fails to span the vector space.
    🔗
    🔗
  3. the set \(S\) is linearly independent.
    🔗
    🔗
  4. the set \(S\) is linearly dependent.
    🔗
    🔗
🔗
🔗
(c)
If every vector of a vector space can either be constructed as a unique linear combination of vectors in a set \(S\text{,}\) or not at all, we can say...
  1. the set \(S\) spans the vector space.
    🔗
    🔗
  2. the set \(S\) fails to span the vector space.
    🔗
    🔗
  3. the set \(S\) is linearly independent.
    🔗
    🔗
  4. the set \(S\) is linearly dependent.
    🔗
    🔗
🔗
🔗
🔗

Definition 2.5.5.

A basis of a vector space \(V\) is a set of vectors \(S\) contained in \(V\) for which
  1. Every vector in the vector space can be expressed as a linear combination of the vectors in \(S\text{.}\)
    🔗
    🔗
  2. For each vector \(\vec{v}\) in the vector space, there is only one way to write it as a linear combination of the vectors in \(S\text{.}\)
    🔗
    🔗
These two properties may be expressed more succinctly as the statement "Every vector in \(V\) can be expressed uniquely as a linear combination of the vectors in \(S\)".
🔗
🔗

Observation 2.5.6.

In terms of a vector equation, a set \(S=\left\{\vec{v}_1,\ldots,\vec{v}_n\right\}\) is a basis of a vector space if the vector equation
\begin{equation*} x_1 \vec{v_1}+\cdots+x_n\vec{v_n}=\vec{w} \end{equation*}
has a unique solution for every vector \(\vec{w}\) in the vector space.
🔗
Put another way, a basis may be thought of as a minimal set of “building blocks” that can be used to construct any other vector of the vector space.
🔗
🔗

Activity 2.5.7.

Let \(S\) be a basis (Definition 2.5.5) for a vector space. Then...
  1. the set \(S\) must both span the vector space and be linearly independent.
    🔗
    🔗
  2. the set \(S\) must span the vector space but could be linearly dependent.
    🔗
    🔗
  3. the set \(S\) must be linearly independent but could fail to span the vector space.
    🔗
    🔗
  4. the set \(S\) could fail to span the vector space and could be linearly dependent.
    🔗
    🔗
🔗
🔗

Activity 2.5.8.

The vectors
\begin{align*} \hat i &= (1,0,0)=\left[\begin{array}{c}1 \\ 0 \\ 0 \\ \end{array}\right] & \hat j &= (0,1,0)=\left[\begin{array}{c}0 \\ 1 \\ 0 \end{array}\right] & \hat k &=(0,0,1)= \left[\begin{array}{c}0 \\ 0 \\ 1 \end{array}\right] \end{align*}
form a basis \(\{\hat i,\hat j,\hat k\}\) used frequently in multivariable calculus.
🔗
Find the unique linear combination of these vectors
\begin{equation*} \unknown\hat i+\unknown\hat j+\unknown\hat k \end{equation*}
that equals the vector
\begin{equation*} (3,-2,4)=\left[\begin{array}{c}3 \\ -2 \\ 4\end{array}\right] \end{equation*}
in \(xyz\) space.
🔗
🔗

Definition 2.5.9.

The standard basis of \(\IR^n\) is the set \(\{\vec{e}_1, \ldots, \vec{e}_n\}\) where
\begin{align*} \vec{e}_1 &= \left[\begin{array}{c}1 \\ 0 \\ 0 \\ \vdots \\ 0 \\ 0 \end{array}\right] & \vec{e}_2 &= \left[\begin{array}{c}0 \\ 1 \\ 0 \\ \vdots \\ 0 \\ 0 \end{array}\right] & \cdots & & \vec{e}_n = \left[\begin{array}{c}0 \\ 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{array}\right]\text{.} \end{align*}
🔗
In particular, the standard basis for \(\mathbb R^3\) is \(\{\vec e_1,\vec e_2,\vec e_3\}=\{\hat i,\hat j,\hat k\}\text{.}\)
🔗
🔗

Activity 2.5.10.

Use technology to find the RREF of an appropriate matrix and determine if each of the following sets is a basis for \(\IR^4\text{.}\) (Don’t forget to use format rat to nicely format Octave’s output.)
🔗
(a)
\begin{equation*} \left\{ \left[\begin{array}{c}1\\0\\0\\0\end{array}\right], \left[\begin{array}{c}0\\1\\0\\0\end{array}\right], \left[\begin{array}{c}0\\0\\1\\0\end{array}\right], \left[\begin{array}{c}0\\0\\0\\1\end{array}\right] \right\} \end{equation*}
  1. A basis, because it both spans \(\IR^4\) and is linearly independent.
    🔗
    🔗
  2. Not a basis, because while it spans \(\IR^4\text{,}\) it is linearly dependent.
    🔗
    🔗
  3. Not a basis, because while it is linearly independent, it fails to span \(\IR^4\text{.}\)
    🔗
    🔗
  4. Not a basis, because not only does it fail to span \(\IR^4\text{,}\) it’s also linearly dependent.
    🔗
    🔗
🔗
🔗
(b)
\begin{equation*} \left\{ \left[\begin{array}{c}2\\3\\0\\-1\end{array}\right], \left[\begin{array}{c}2\\0\\0\\3\end{array}\right], \left[\begin{array}{c}4\\3\\0\\2\end{array}\right], \left[\begin{array}{c}-3\\0\\1\\3\end{array}\right] \right\} \end{equation*}
  1. A basis, because it both spans \(\IR^4\) and is linearly independent.
    🔗
    🔗
  2. Not a basis, because while it spans \(\IR^4\text{,}\) it is linearly dependent.
    🔗
    🔗
  3. Not a basis, because while it is linearly independent, it fails to span \(\IR^4\text{.}\)
    🔗
    🔗
  4. Not a basis, because not only does it fail to span \(\IR^4\text{,}\) it’s also linearly dependent.
    🔗
    🔗
🔗
🔗
(c)
\begin{equation*} \left\{ \left[\begin{array}{c}2\\3\\0\\-1\end{array}\right], \left[\begin{array}{c}2\\0\\0\\3\end{array}\right], \left[\begin{array}{c}3\\13\\7\\16\end{array}\right], \left[\begin{array}{c}-1\\10\\7\\14\end{array}\right], \left[\begin{array}{c}4\\3\\0\\2\end{array}\right] \right\} \end{equation*}
  1. A basis, because it both spans \(\IR^4\) and is linearly independent.
    🔗
    🔗
  2. Not a basis, because while it spans \(\IR^4\text{,}\) it is linearly dependent.
    🔗
    🔗
  3. Not a basis, because while it is linearly independent, it fails to span \(\IR^4\text{.}\)
    🔗
    🔗
  4. Not a basis, because not only does it fail to span \(\IR^4\text{,}\) it’s also linearly dependent.
    🔗
    🔗
🔗
🔗
(d)
\begin{equation*} \left\{ \left[\begin{array}{c}2\\3\\0\\-1\end{array}\right], \left[\begin{array}{c}4\\3\\0\\2\end{array}\right], \left[\begin{array}{c}-3\\0\\1\\3\end{array}\right], \left[\begin{array}{c}3\\6\\1\\5\end{array}\right] \right\} \end{equation*}
  1. A basis, because it both spans \(\IR^4\) and is linearly independent.
    🔗
    🔗
  2. Not a basis, because while it spans \(\IR^4\text{,}\) it is linearly dependent.
    🔗
    🔗
  3. Not a basis, because while it is linearly independent, it fails to span \(\IR^4\text{.}\)
    🔗
    🔗
  4. Not a basis, because not only does it fail to span \(\IR^4\text{,}\) it’s also linearly dependent.
    🔗
    🔗
🔗
🔗
(e)
\begin{equation*} \left\{ \left[\begin{array}{c}5\\3\\0\\-1\end{array}\right], \left[\begin{array}{c}-2\\1\\0\\3\end{array}\right], \left[\begin{array}{c}4\\5\\1\\3\end{array}\right] \right\} \end{equation*}
  1. A basis, because it both spans \(\IR^4\) and is linearly independent.
    🔗
    🔗
  2. Not a basis, because while it spans \(\IR^4\text{,}\) it is linearly dependent.
    🔗
    🔗
  3. Not a basis, because while it is linearly independent, it fails to span \(\IR^4\text{.}\)
    🔗
    🔗
  4. Not a basis, because not only does it fail to span \(\IR^4\text{,}\) it’s also linearly dependent.
    🔗
    🔗
🔗
🔗
🔗

Activity 2.5.11.

If \(\{\vec v_1,\vec v_2,\vec v_3,\vec v_4\}\) is a basis for \(\IR^4\text{,}\) that means \(\RREF[\vec v_1\,\vec v_2\,\vec v_3\,\vec v_4]\) has a pivot in every row (because it spans), and has a pivot in every column (because it’s linearly independent).
🔗
What is \(\RREF[\vec v_1\,\vec v_2\,\vec v_3\,\vec v_4]\text{?}\)
\begin{equation*} \RREF[\vec v_1\,\vec v_2\,\vec v_3\,\vec v_4] = \left[\begin{array}{cccc} \unknown & \unknown & \unknown & \unknown \\ \unknown & \unknown & \unknown & \unknown \\ \unknown & \unknown & \unknown & \unknown \\ \unknown & \unknown & \unknown & \unknown \\ \end{array}\right] \end{equation*}
🔗
🔗
🔗

Subsection 2.5.3 Individual Practice

Activity 2.5.13.

Let \(S\) denote a set of vectors in \(\IR^n\text{.}\) Without referring to your Activity Book, write down:
🔗
(a)
The definition of what it means for \(S\) to be linearly independent.
🔗
🔗
(b)
The definition of what it means for \(S\) to span \(\IR^n\text{.}\)
🔗
🔗
(c)
The definition of what it means for \(S\) to be a basis for \(\IR^n\text{.}\)
🔗
🔗
🔗

Activity 2.5.14.

You are going on a trip and need to pack. Let \(S\) denote the set of items that you are packing in your suitcase.
🔗
(a)
Give an example of such a set of items \(S\) that you would say "spans" everything you need, but is linearly dependent.
🔗
🔗
(b)
Give an example of such a set of items \(S\) that is linearly independent, but does not "span" everything you need.
🔗
🔗
(c)
Give an example of such a set \(S\) that you might reasonably consider to be a "basis" for what you need?
🔗
🔗
🔗
🔗

Subsection 2.5.4 Videos

Figure 16. Video: Verifying that a set of vectors is a basis of a vector space
🔗
🔗

Subsection 2.5.5 Exercises

🔗

Subsection 2.5.6 Mathematical Writing Explorations

Exploration 2.5.15.

  • What is a basis for \(M_{2,2}\text{?}\)
    🔗
  • What about \(M_{3,3}\text{?}\)
    🔗
  • Could we write each of these in a way that looks like the standard basis vectors in \(\mathbb{R}^m\) for some \(m\text{?}\) Make a conjecture about the relationship between these spaces of matrices and standard Euclidean space.
    🔗
🔗
🔗

Exploration 2.5.16.

Recall our earlier definition of symmetric matrices. Find a basis for each of the following:
  • The space of \(2 \times 2\) symmetric matrices.
    🔗
  • The space of \(3 \times 3\) symmetric matrices.
    🔗
  • The space of \(n \times n\) symmetric matrices.
    🔗
🔗
🔗

Exploration 2.5.17.

Must a basis for the space \(P_2\text{,}\) the space of all quadratic polynomials, contain a polynomial of each degree less than or equal to 2? Generalize your result to polynomials of arbitrary degree.
🔗
🔗
🔗

Subsection 2.5.7 Sample Problem and Solution

Sample problem Example B.1.9.
🔗
🔗
🔗
PrevTopNext