Span.
Let \(S=\{v_1,v_2,\ldots,v_k\}\) be a subset of a vector space \(V\text{.}\) Define
\begin{equation*}
\operatorname{span}(S)=\{x_{1}v_{1}+x_{2}v_{2}+\ldots+x_{k}v_{k}|x_{i}\in \mathbb{R}\}
\end{equation*}
Section 4.1 Vector Space \(\mathbb{R}^n\)
The set
\begin{equation*}
\mathbb{R}^n=\left\{\left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{array} \right]\middle| x_i\in \mathbb{R} \right\}
\end{equation*}
is a
vector space with operations
\begin{equation*}
\left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{array} \right]+\left[ \begin{array}{c} y_1 \\ y_2 \\ \vdots \\ y_n \\ \end{array} \right]=\left[ \begin{array}{c} x_1+y_1 \\ x_2+y_2 \\ \vdots \\ x_n+y_n \\ \end{array} \right]
\qquad \text{and} \qquad
c\left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{array} \right]=\left[ \begin{array}{c} cx_1 \\ cx_2 \\ \vdots \\ cx_n \\ \end{array} \right].
\end{equation*}
Subsection 4.1.1 Linear Combinations
An important type of problem in linear algebra involves writing one vector \(\mathbf{x}\) as the sum of scalar multiples of other vectors \(\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots\text{,}\) and \(\mathbf{v}_{n}\text{.}\) That is, for scalars \(c_{1}\text{,}\) \(c_{2}, \ldots, c_{n}\)
\begin{equation*}
\mathbf{x}=c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{n} \mathbf{v}_{n}
\end{equation*}
The vector \(\mathbf{x}\) is called a linear combination of the vectors \(\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{n}\) with weight \(c_1,c_2,\ldots,c_n\text{.}\)
Subsection 4.1.2 Span
Example 4.1.1.
Let \(S=\left\{\left[\begin{array}{r}
1 \\
1 \\
\end{array}\right],\left[\begin{array}{r}
0 \\
1
\end{array}\right]\right\}\text{.}\)
\begin{equation*}
\operatorname{span}(S)=\{x_{1}v_{1}+x_{2}v_{2}|x_{i}\in \mathbb{R}\}=\mathbb{R}^{2}
\end{equation*}
A diagram showing two vectors v1 and v2 spanning R^2, with their linear combination illustrated
Activity 4.1.1.
Show that the set \(S\) below spans \(\mathbb{R}^{4}\text{.}\)
\begin{equation*}
S=\{v_1,v_2,v_3,v_4,v_5\}=\left\{\left[\begin{array}{r}
1 \\
0 \\
-1 \\
0
\end{array}\right],\left[\begin{array}{r}
1 \\
1 \\
0 \\
2
\end{array}\right],\left[\begin{array}{r}
0 \\
3 \\
1 \\
-2
\end{array}\right],\left[\begin{array}{r}
0 \\
1 \\
-1 \\
2
\end{array}\right], \left[\begin{array}{r}
1 \\
1 \\
1 \\
1
\end{array}\right]\right\}.
\end{equation*}
That is, for any \(v=\left[\begin{array}{r}
a \\
b \\
c \\
d
\end{array}\right]\text{,}\) the vector equation
\begin{equation*}
v = x_{1}v_{1}+x_{2}v_{2}+x_{3}v_{3}+x_{4}v_{4}+x_{5}v_{5}
\end{equation*}
is consistent(has a solution).
Activity 4.1.2.
Let
\begin{equation*}
S=\{v_1,v_2,v_3,v_4\}=\left\{\left[\begin{array}{r}
1 \\
0 \\
-1 \\
0
\end{array}\right],\left[\begin{array}{r}
1 \\
1 \\
0 \\
2
\end{array}\right],\left[\begin{array}{r}
0 \\
3 \\
1 \\
-2
\end{array}\right],\left[\begin{array}{r}
0 \\
1 \\
-1 \\
2
\end{array}\right]\right\}.
\end{equation*}
Suppose that \(v=\left[\begin{array}{r}
a \\
b \\
c \\
d
\end{array}\right]\) is in \(\operatorname{span}(S)\text{.}\) Find a relation of \(a,b,c,d\text{.}\)
Activity 4.1.3.
Let
\begin{equation*}
S=\{v_1,v_2,v_3,v_4\}=\left\{\left[\begin{array}{r}
1 \\
0 \\
-1 \\
0
\end{array}\right],\left[\begin{array}{r}
1 \\
1 \\
0 \\
2
\end{array}\right],\left[\begin{array}{r}
0 \\
3 \\
1 \\
2
\end{array}\right],\left[\begin{array}{r}
0 \\
1 \\
-1 \\
-2
\end{array}\right]\right\}.
\end{equation*}
Suppose that \(v=\left[\begin{array}{r}
a \\
b \\
c \\
d
\end{array}\right]\) is in \(\operatorname{span}(S)\text{.}\) Find a relation of \(a,b,c,d\text{.}\)
Hint: Form the augmented matrix and its
row echelon form:
\begin{equation*}
\left(\begin{array}{rrrrr}
1 & 1 & 0 & 0 & a \\
0 & 1 & 3 & 1 & b \\
-1 & 0 & 1 & -1 & c \\
0 & 2 & 2 & -2 & d
\end{array}\right) \sim \left(\begin{array}{rrrrr}
1 & 1 & 0 & 0 & a \\
0 & 1 & 3 & 1 & b \\
0 & 0 & 1 & 1 & -\frac{a - b + c}{2} \\
0 & 0 & 0 & 0 & -2a - 2c + d
\end{array}\right)
\end{equation*}
Subsection 4.1.3 Linear Independence
Linear Independent Set.
A set of vectors \(\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n\}\) is called linear independent if the linear combination
\begin{equation*}
x_{1} \mathbf{v}_{1}+x_{2} \mathbf{v}_{2}+\cdots+x_{n} \mathbf{v}_{n}=\mathbf{0}
\end{equation*}
has only trivial solution, that is, \(x_1=x_2=\ldots =x_n=0\) is the only solution. Otherwise, the set is called linear dependent.
Activity 4.1.4.
Let
\begin{equation*}
S=\{v_1,v_2,v_3,v_4\}=\left\{\left[\begin{array}{r}
1 \\
0 \\
-1 \\
0
\end{array}\right],\left[\begin{array}{r}
1 \\
1 \\
0 \\
2
\end{array}\right],\left[\begin{array}{r}
0 \\
3 \\
1 \\
-2
\end{array}\right],\left[\begin{array}{r}
0 \\
1 \\
-1 \\
2
\end{array}\right]\right\}.
\end{equation*}
Show that \(S\) is linear independent.
Activity 4.1.5.
Let
\begin{equation*}
S=\{v_1,v_2,v_3,v_4,v_5\}=\left\{\left[\begin{array}{r}
1 \\
0 \\
-1 \\
0
\end{array}\right],\left[\begin{array}{r}
1 \\
1 \\
0 \\
2
\end{array}\right],\left[\begin{array}{r}
0 \\
3 \\
1 \\
-2
\end{array}\right],\left[\begin{array}{r}
0 \\
1 \\
-1 \\
2
\end{array}\right], \left[\begin{array}{r}
1 \\
1 \\
1 \\
1
\end{array}\right]\right\}.
\end{equation*}
Show that \(S\) is linear dependent.
Activity 4.1.6.
Find a maximum linear independent subset \(W\) of \(S\text{.}\) That is, \(W\) is linear independent, but for any \(W\subsetneq U\)(proper included), \(U\) is linear dependent.
Activity 4.1.7.
Write the fifth vector \(v_5\text{,}\) as in Activity 4.1.5, a linear combination of the other four vectors \(v_1, v_2,v_3\) and \(v_4\text{.}\)
Activity 4.1.8.
Find a maximal linear independent subset of the set
\begin{equation*}
S=\{v_1,v_2,v_3,v_4,v_5\}=\left\{\left[\begin{array}{r}
1 \\
0 \\
-1 \\
0
\end{array}\right],\left[\begin{array}{r}
1 \\
1 \\
0 \\
2
\end{array}\right],\left[\begin{array}{r}
0 \\
3 \\
1 \\
2
\end{array}\right],\left[\begin{array}{r}
0 \\
1 \\
-1 \\
-2
\end{array}\right], \left[\begin{array}{r}
1 \\
1 \\
1 \\
1
\end{array}\right]\right\}.
\end{equation*}
Subsection 4.1.4 Basis
Definition 4.1.3. Basis.
A set \(\mathcal{B} = \{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\}\) of vectors in a vector space \(V\) is called a basis for \(V\) if:
-
\(\mathcal{B}\) is linearly independent
-
\(\mathcal{B}\) spans \(V\text{,}\) that is, \(\operatorname{span}(\mathcal{B}) = V\)
A basis provides the most efficient way to describe a vector space: it gives us the minimum number of vectors needed to span the entire space, with no redundancy (since they are linearly independent).
Theorem 4.1.4. Basis Test via Determinant.
Let \(\mathcal{B} = \{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\}\) be a set of \(n\) vectors in \(\mathbb{R}^n\text{,}\) and let \(B\) be the \(n \times n\) matrix with columns \(\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\text{.}\) Then \(\mathcal{B}\) is a basis for \(\mathbb{R}^n\) if and only if \(\det(B) \neq 0\text{.}\)
Proof.
Direction 1: If \(\mathcal{B}\) is a basis, then \(\mathcal{B}\) is linearly independent. This means the homogeneous system \(B\mathbf{x} = \mathbf{0}\) has only the trivial solution, which occurs if and only if \(\det(B) \neq 0\text{.}\)
Direction 2: If \(\det(B) \neq 0\text{,}\) then \(B\) is invertible, so the columns of \(B\) are linearly independent. Since we have \(n\) linearly independent vectors in \(\mathbb{R}^n\text{,}\) they automatically span \(\mathbb{R}^n\text{,}\) making \(\mathcal{B}\) a basis.
Example 4.1.5. Testing if a Set is a Basis.
Consider the first four vectors from activity 4.1.1:
\begin{equation*}
\mathcal{B} = \left\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\right\} = \left\{\begin{bmatrix}1\\0\\-1\\0\end{bmatrix}, \begin{bmatrix}1\\1\\0\\2\end{bmatrix}, \begin{bmatrix}0\\3\\1\\-2\end{bmatrix}, \begin{bmatrix}0\\1\\-1\\2\end{bmatrix}\right\}
\end{equation*}
Determine whether \(\mathcal{B}\) is a basis for \(\mathbb{R}^4\text{.}\)
Solution.
To test if \(\mathcal{B}\) is a basis for \(\mathbb{R}^4\text{,}\) we form the matrix \(B\) with these vectors as columns and compute its determinant.
From the computation above, we can see that \(\det(B) \neq 0\text{,}\) which confirms that \(\mathcal{B} = \{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}\) is indeed a basis for \(\mathbb{R}^4\text{.}\)
Key Insights:
-
The determinant test provides a quick computational method to verify if a set of vectors forms a basis.
-
For \(n\) vectors in \(\mathbb{R}^n\text{,}\) linear independence automatically implies spanning the entire space.
-
This result will be useful later when we work with coordinate vectors relative to this basis.
Activity 4.1.9.
Practice: Use the determinant test to verify whether each of the following sets forms a basis for \(\mathbb{R}^3\text{:}\)
-
\(\displaystyle \left\{\begin{bmatrix}1\\0\\1\end{bmatrix}, \begin{bmatrix}0\\1\\1\end{bmatrix}, \begin{bmatrix}1\\1\\0\end{bmatrix}\right\}\)
-
\(\displaystyle \left\{\begin{bmatrix}1\\2\\3\end{bmatrix}, \begin{bmatrix}2\\4\\6\end{bmatrix}, \begin{bmatrix}1\\0\\1\end{bmatrix}\right\}\)
