Learning Outcomes
- Explain why a set of Euclidean vectors is or is not a basis of \(\IR^n\text{.}\)
Section 2.5 Identifying a Basis (EV5)
Subsection 2.5.1 Class Activities
Activity 2.5.1.
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)
Consider another vector \(\left[\begin{array}{c} 8 \\ 6 \\ 7 \\ 5 \end{array} \right]\text{.}\) Without computing the RREF of another matrix, how many ways can this vector be written as a linear combination of the vectors in \(S\text{?}\)
- Zero.
- One.
- Infinitely-many.
- Computing a new matrix RREF is necessary.
Activity 2.5.2.
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 combination of vectors in a set \(S\text{,}\) we can say...
- the set \(S\) spans the vector space.
- the set \(S\) fails to span the vector space.
- the set \(S\) is linearly independent.
- 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...
- the set \(S\) spans the vector space.
- the set \(S\) fails to span the vector space.
- the set \(S\) is linearly independent.
- 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...
- the set \(S\) spans the vector space.
- the set \(S\) fails to span the vector space.
- the set \(S\) is linearly independent.
- the set \(S\) is linearly dependent.
Definition 2.5.3.
A basis of a vector space \(V\) is a set of vectors \(S\) contained in \(V\) for which
- Every vector in the vector space can be expressed as a linear combination of the vectors in \(S\text{.}\)
- 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 succintly as the statement "Every vector in \(V\) can be expressed uniquely as a linear combination of the vectors in \(S\)".
Observation 2.5.4.
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 2.5.5.
Let \(S\) be a basis (Definition 2.5.3) for a vector space. Then...
- the set \(S\) must both span the vector space and be linearly independent.
- the set \(S\) must span the vector space but could be linearly dependent.
- the set \(S\) must be linearly independent but could fail to span the vector space.
- the set \(S\) could fail to span the vector space and could be linearly dependent.
Activity 2.5.6.
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 2.5.7.
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 2.5.8.
Take the RREF of an appropriate matrix to determine if each of the following sets is a basis for \(\IR^4\text{.}\)
(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*}
- A basis, because it both spans \(\IR^4\) and is linearly independent.
- Not a basis, because while it spans \(\IR^4\text{,}\) it is linearly dependent.
- Not a basis, because while it is linearly independent, it fails to span \(\IR^4\text{.}\)
- 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*}
- A basis, because it both spans \(\IR^4\) and is linearly independent.
- Not a basis, because while it spans \(\IR^4\text{,}\) it is linearly dependent.
- Not a basis, because while it is linearly independent, it fails to span \(\IR^4\text{.}\)
- 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*}
- A basis, because it both spans \(\IR^4\) and is linearly independent.
- Not a basis, because while it spans \(\IR^4\text{,}\) it is linearly dependent.
- Not a basis, because while it is linearly independent, it fails to span \(\IR^4\text{.}\)
- 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*}
- A basis, because it both spans \(\IR^4\) and is linearly independent.
- Not a basis, because while it spans \(\IR^4\text{,}\) it is linearly dependent.
- Not a basis, because while it is linearly independent, it fails to span \(\IR^4\text{.}\)
- 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*}
- A basis, because it both spans \(\IR^4\) and is linearly independent.
- Not a basis, because while it spans \(\IR^4\text{,}\) it is linearly dependent.
- Not a basis, because while it is linearly independent, it fails to span \(\IR^4\text{.}\)
- Not a basis, because not only does it fail to span \(\IR^4\text{,}\) it’s also linearly dependent.
Activity 2.5.9.
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*}
Fact 2.5.10.
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 2.5.2 Videos
Exercises 2.5.3 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2024e/exercises/#/bank/EV5/.Subsection 2.5.4 Mathematical Writing Explorations
Exploration 2.5.11.
- 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 2.5.12.
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 2.5.13.
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 2.5.5 Sample Problem and Solution
Sample problem Example B.1.9.
