Learning Outcomes
- Explain why a set of Euclidean vectors is or is not a basis of \(\IR^n\text{.}\)
Section 2.6 Identifying a Basis (VS6)
Subsection 2.6.1 Class Activities
Activity 2.6.1.
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)
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*}
\left[\begin{array}{c} 5 \\ 2 \\ 0 \\ 1 \end{array} \right] =
\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]\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 \(S\text{.}\)
(c)
How many ways can the vector \(\left[\begin{array}{c} 8 \\ 6 \\ 7 \\ 5 \end{array} \right]\) be written as a linear combination of the vectors in \(S\text{?}\)
Observation 2.6.2.
We saw that some vectors could be expressed as a linear combination of the vectors in \(S\) in lots of (infinitely many) ways, while others could not be expressed at all as a linear combination.
This motivates us to look for sets of vectors where every other vector can be constructed from them, and there is only one way to do so.
Definition 2.6.3.
A basis of a vector space \(V\) is a set of vectors \(S\) for which
- Every vector in \(V\) can be expressed as a linear combination of the vectors in \(S\)
- For each vector \(\vec{v} \in V\text{,}\) 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 succintly as the statement "Every vector in \(V\) can be expressed uniquely as a linear combination of the vectors in \(S\)".
Or, in terms of a vector equation, a set \(S=\left\{\vec{v}_1,\ldots,\vec{v}_n\right\}\) is a basis of \(V\) 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 \(\vec{w} \in V\text{.}\)
Definition 2.6.4.
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 2.6.5.
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.
Remark 2.6.6.
Saying that the vector equation
\begin{equation*}
x_1 \vec{v_1}+\cdots+x_n\vec{v_n}=\vec{w}
\end{equation*}
has a solution for every vector \(\vec{w} \in V\) is saying that the set \(\left\{\vec{v}_1,\ldots,\vec{v}_n\right\}\) spans the vector space \(V\text{.}\)
However, to be a basis, we need to also check that the solution is always unique
In other words, it is necessary for a set to span \(V\) in order to be a basis of \(V\text{,}\) but this is not sufficient.
Activity 2.6.7.
Recall that in Activity 2.6.1 we found two different ways to write the vector \(\left[\begin{array}{c} 5 \\ 2 \\ 0 \\ 1 \end{array} \right]\) as a linear combination of the vectors in the set \(S=\left\{
\left[\begin{array}{c} 3 \\ -2 \\ -1 \\ 0 \end{array} \right],
\left[\begin{array}{c} 2 \\ 4 \\ 1 \\ 1 \end{array} \right],
\left[\begin{array}{c} 0 \\ -16 \\ -5 \\ -3 \end{array} \right],
\left[\begin{array}{c} 1 \\ 2 \\ 3 \\ 0 \end{array} \right],
\left[\begin{array}{c} 3 \\ 3 \\ 0 \\ 1 \end{array} \right] \right\}
\text{.}\)
For example, we had
\begin{align*}
\left[\begin{array}{c} 5 \\ 2 \\ 0 \\ 1 \end{array} \right]& =
\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} 5 \\ 2 \\ 0 \\ 1 \end{array} \right]& =
3\left[\begin{array}{c} 3 \\ -2 \\ -1 \\ 0 \end{array} \right]
-2\left[\begin{array}{c} 2 \\ 4 \\ 1 \\ 1 \end{array} \right]
-\left[\begin{array}{c} 0 \\ -16 \\ -5 \\ -3 \end{array} \right]\text{.}
\end{align*}
Manipulate these two equations to show how to build the zero vector out of the vectors in \(S\text{.}\)
Observation 2.6.8.
Whenever a vector can be expressed as two different linear combinations of vectors in a set \(S\text{,}\) the set \(S\) must be linearly dependent.
Fact 2.6.9.
A set of vectors \(S\) is a basis of a vector space \(V\) if \(S\) spans \(V\) and is linearly independent.
If either of these fails, then \(S\) cannot be a basis of \(V\text{.}\)
Activity 2.6.10.
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 2.6.11.
If \(\{\vec v_1,\vec v_2,\vec v_3,\vec v_4\}\) is a basis for \(\IR^4\text{,}\) that means \(\RREF[\vec v_1\,\vec v_2\,\vec v_3\,\vec v_4]\) 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 2.6.12.
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.)
Subsection 2.6.2 Videos
Subsection 2.6.3 Slideshow
Slideshow of activities available at
https://teambasedinquirylearning.github.io/linear-algebra/2023e/VS6.slides.html.Exercises 2.6.4 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2023e/exercises/#/bank/VS6/.Subsection 2.6.5 Mathematical Writing Explorations
Exploration 2.6.13.
- 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 Eulidean space.
Exploration 2.6.14.
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 2.6.15.
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 2.6.6 Sample Problem and Solution
Sample problem Example B.1.10.
