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Linear Algebra for Team-Based Inquiry Learning

2022 Edition

Steven Clontz	Drew Lewis
University of South Alabama	University of South Alabama

August 2, 2022

Section 2.6: Identifying a Basis (VS6)

Observation 2.6.1

Suppose you are building a starship, which is for some reason in the shape of a cube. Due to some clever engineering, each part of the ship can be made out of a finite set of components. In fact, there are only 5 basic components. Assemble them in different ways, and you make every part of the cube! However, at the last minute, the design is changed from a cube to an octahedron. Would it make more sense to take all of the parts you were planning to build, build them anyway and modify them later, or to just modify the 5 basic components?

Activity 2.6.2

Start with three vectors

\begin{equation*} \vec e_1=\hat\imath=\left[\begin{array}{c}1 \\ 0 \\ 0\end{array}\right], \vec e_2=\hat\jmath=\left[\begin{array}{c}0 \\ 1 \\ 0\end{array}\right], \text{ and } \vec e_3=\hat k=\left[\begin{array}{c}0 \\ 0 \\ 1\end{array}\right]\text{.} \end{equation*}

Part 1.

Let $\vec{v}$ be an unspecified vector in $\IR\text{.}$ Show that $\vec{v}$ can be expressed as a linear combination of $\vec{e}_1, \vec{e}_2\text{,}$ and $\vec{e}_3\text{.}$

Activity 2.6.2

Start with three vectors

Part 2.

Let $\vec{w} = \left[\begin{array}{c}1 \\ 1 \\ 0\end{array}\right]\text{.}$ Show that $\vec{v}$ cannot be expressed as a linear combination of $\vec{e}_1, \vec{e}_2\text{,}$ and $\vec{w}\text{.}$

Activity 2.6.2

Start with three vectors

Part 3.

Does this imply that all vectors in $\IR^3$ can be written as a linear combination of $\vec{e}_1, \vec{e}_2\text{,}$ and $\vec{e}_3\text{?}$ If you think so, explain what makes these vectors special. If not, explain why not.

Definition 2.6.3

A basis is a linearly independent set that spans a vector space.

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*}

Observation 2.6.4

A basis may be thought of as a collection of building blocks for a vector space, since every vector in the space can be expressed as a unique linear combination of basis vectors.

For example, in many calculus courses, vectors in $\IR^3$ are often expressed in their component form

\begin{equation*} (3,-2,4)=\left[\begin{array}{c}3 \\ -2 \\ 4\end{array}\right] \end{equation*}

or in their standard basic vector form

\begin{equation*} 3\vec e_1-2\vec e_2+4\vec e_3 = 3\hat\imath-2\hat\jmath+4\hat k . \end{equation*}

Since every vector in $\IR^3$ can be uniquely described as a linear combination of the vectors in $\setList{\vec e_1,\vec e_2,\vec e_3}\text{,}$ this set is indeed a basis.

Activity 2.6.5 (~15 min)

Label each of the sets $A,B,C,D,E$ as

SPANS $\IR^4$ or DOES NOT SPAN $\IR^4$
LINEARLY INDEPENDENT or LINEARLY DEPENDENT
BASIS FOR $\IR^4$ or NOT A BASIS FOR $\IR^4$

by finding $\RREF$ for their corresponding matrices.

\begin{align*} A&=\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\} & B&=\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\}\\ C&=\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\} & D&=\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\}\\ E&=\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{align*}

Activity 2.6.6 (~10 min)

If $\{\vec v_1,\vec v_2,\vec v_3,\vec v_4\}$ is a basis for $\IR^4\text{,}$ that means $\RREF[\vec v_1\,\vec v_2\,\vec v_3\,\vec v_4]$ doesn't have a non-pivot column, and doesn't have a row of zeros. 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*}

Fact 2.6.7

The set $\{\vec v_1,\dots,\vec v_m\}$ is a basis for $\IR^n$ if and only if $m=n$ and $\RREF[\vec v_1\,\dots\,\vec v_n]= \left[\begin{array}{cccc} 1&0&\dots&0\\ 0&1&\dots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\dots&1 \end{array}\right] \text{.}$

That is, a basis for $\IR^n$ must have exactly $n$ vectors and its square matrix must row-reduce to the so-called identity matrix containing all zeros except for a downward diagonal of ones. (We will learn where the identity matrix gets its name in a later module.)