are in a vector space V , then the set W of all linear combinations of these vectors is a subspace of V and is called the span of the vectors .
We also say that the vectors
span W which may be written as
.The definitions of the span of vectors are presented including with examples and their solutions
If vectors
are in a vector space V , then the set W of all linear combinations of these vectors is a subspace of V and is called the span of the vectors .
We also say that the vectors span W which may be written as
.
Example 1
Show that the vectors
![Vectors u1=[1 , 5]](https://www.analyzemath.com/linear-algebra/spaces/span/im-3.png)
and
![Vectors u2=[-1 , 2]](https://www.analyzemath.com/linear-algebra/spaces/span/im-4.png)
span the vector space ℝ2 .
Solution to Example 1
A vector in the space ℝ2 is of the form
![Vectors [a , b]](https://www.analyzemath.com/linear-algebra/spaces/span/im-5.png){kind=small}
where a and b can take any value in the set of real numbers ℝ.
To show that the vectors u1 and u2 span ℝ2 , we need to show that any vector
in ℝ2 is a linear combinations of the vectors u1 and u2. We therefore need to show that we can find the scalars r1 and r2 such that
![Vector [a , b] as a Linear Combination of u1 and u2](https://www.analyzemath.com/linear-algebra/spaces/span/im-6.png)
\( \) \( \) \( \) \( \)
for any values of \( a \) and \( b \).
Scalar multiply and add the vectors on the right side in the above equation
\( \begin{bmatrix}
a \\
b \\
\end{bmatrix}
= \begin{bmatrix}
r_1 - r_2\\
5 r_1 + 2 r_2 \\
\end{bmatrix}
\) (I)
Solve the above for \( r_1 \) and \( r_2 \) using any method. Here we use the method of elimination
Multiply all terms of the top equation by \( 2 \)
\( \begin{bmatrix}
2 a \\
b \\
\end{bmatrix}
= \begin{bmatrix}
2 r_1 - 2 r_2\\
5 r_1 + 2 r_2 \\
\end{bmatrix}
\)
Add the two equations to eliminate \( r_2 \) and place the result in place of the second equation to obtain the system
\( \begin{bmatrix}
2 a \\
b + 2 a \\
\end{bmatrix}
= \begin{bmatrix}
2 r_1 - 2 r_2\\
7 r_1 \\
\end{bmatrix}
\)
Solve for \( r_1\)
\( r_1 = \dfrac{b + 2 a}{7} \)
From the top equation in the system (I) above, we can write
\( r_2 = r_1 - a = \dfrac{b + 2 a}{7} - a = \dfrac{b-5a}{7} \)
We have proved that any vector \( \begin{bmatrix}
a \\
b \\
\end{bmatrix} \) in
\( \mathbb{R}^2\) may be expressed as a linear combination of \( \textbf{u}_1 \) and \( \textbf{u}_2 \) and therefore \( \textbf{u}_1 \) and \( \textbf{u}_2 \) span \( \mathbb{R}^2\).
Example 2
Show that the vectors \( \textbf{u}_1 = \begin{bmatrix}
1 \\
-2\\
\end{bmatrix} \) and \( \textbf{u}_2 = \begin{bmatrix}
2 \\
-4\\
\end{bmatrix} \) DO NOT span the vector space \( \mathbb{R}^2\)
Solution to Example 2
We need to show that we cannot find scalars \( r_1 \) and \( r_2 \) for any real numbers \( a \) and \( b \) such that
\( \begin{bmatrix}
a \\
b \\
\end{bmatrix}
= r_1 \begin{bmatrix}
1 \\
-2\\
\end{bmatrix} + r_2 \begin{bmatrix}
2 \\
-4\\
\end{bmatrix}
\)
Scalar multiply and add the terms on the right side
\( \begin{bmatrix}
a \\
b \\
\end{bmatrix}
= \begin{bmatrix}
r_1 + 2 r_2 \\
-2 r_1 - 4 r_2\\
\end{bmatrix}
\)
Multiply the top equation by \( 2 \)
\( \begin{bmatrix}
2 a \\
b \\
\end{bmatrix}
= \begin{bmatrix}
2 r_1 + 4 r_2 \\
-2 r_1 - 4 r_2\\
\end{bmatrix}
\)
Add the two equations and place the result in the bottom
\( \begin{bmatrix}
2 a \\
2a + b \\
\end{bmatrix}
= \begin{bmatrix}
2 r_1 + 4 r_2 \\
0\\
\end{bmatrix}
\)
The last equation shows the there are no solutions if \( a \) and \( b \) are such that \( 2a + b \ne 0 \). Hence the given vectors do not span \( \mathbb{R}^2\).
Example 3
Show that the vectors \( \textbf{u}_1 = \begin{bmatrix}
1 \\
0\\
0
\end{bmatrix} \) and \( \textbf{u}_2 = \begin{bmatrix}
0 \\
1\\
0
\end{bmatrix} \) and \( \textbf{u}_3 = \begin{bmatrix}
0 \\
0\\
1
\end{bmatrix} \) span the vector space \( \mathbb{R}^3\)
Solution to Example 3
To show that the vectors \( \textbf{u}_1 , \textbf{u}_2 \) and \( \textbf{u}_3 \) span \( \mathbb{R}^3\), we need to show that any vector \( \begin{bmatrix}
a \\
b \\
c
\end{bmatrix} \) in \( \mathbb{R}^3\) is a linear combination of the vectors \( \textbf{u}_1 \), \( \textbf{u}_2 \) and \( \textbf{u}_3 \). We therefore need to show that we can find the scalars \( r_1 \), \( r_2 \) and \( r_3 \) such that
\( \begin{bmatrix}
a \\
b \\
c
\end{bmatrix}
= r_1 \begin{bmatrix}
1 \\
0\\
0
\end{bmatrix} +
r_2 \begin{bmatrix}
0\\
1\\
0
\end{bmatrix}
+
r_3 \begin{bmatrix}
0\\
0\\
1
\end{bmatrix}
\)
for any values of \( a \), \( b \) and \( c \).
Scalar multiply and add the vectors on the right side in the above equation
\( \begin{bmatrix}
a \\
b \\
c
\end{bmatrix}
= \begin{bmatrix}
r_1\\
r_2 \\
r_3
\end{bmatrix}
\) (I)
Solve the above for \( r_1 \), \( r_2 \) and \( r_3 \).
\( r_1 = a \) , \( r_2 = b \) and \( r_3 = c \)
Any vector \( \begin{bmatrix}
a \\
b \\
c
\end{bmatrix} \) in \( \mathbb{R}^3\) may be expressed as a linear combination of \( \textbf{u}_1 \) , \( \textbf{u}_2 \) and \( \textbf{u}_3 \) and therefore these 3 vectors span \( \mathbb{R}^3\).
Example 4
Show that the vectors \( \textbf{u}_1 = \begin{bmatrix}
1 \\
2\\
0
\end{bmatrix} \) and \( \textbf{u}_2 = \begin{bmatrix}
0 \\
1\\
-1
\end{bmatrix} \) and \( \textbf{u}_3 = \begin{bmatrix}
0 \\
2\\
1
\end{bmatrix} \) span the vector space \( \mathbb{R}^3\)
Solution to Example 4
Vectors \( \textbf{u}_1 , \textbf{u}_2 \) and \( \textbf{u}_3 \) span \( \mathbb{R}^3\) if we can show that any vector \( \begin{bmatrix}
a \\
b \\
c
\end{bmatrix} \) in \( \mathbb{R}^3\) is a linear combination of the vectors \( \textbf{u}_1 \), \( \textbf{u}_2 \) and \( \textbf{u}_3 \). We therefore need to show that we can find the scalars \( r_1 \), \( r_2 \) and \( r_3 \) such that
\( \begin{bmatrix}
a \\
b \\
c
\end{bmatrix}
= r_1 \begin{bmatrix}
1 \\
2\\
0
\end{bmatrix} +
r_2 \begin{bmatrix}
0 \\
1\\
-1
\end{bmatrix}
+
r_3 \begin{bmatrix}
0\\
2\\
1
\end{bmatrix}
\)
for any values of \( a \) , \( b \) and \( c \).
Scalar multiply and add the vectors on the right side in the above equation
\( \begin{bmatrix}
a \\
b \\
c
\end{bmatrix}
= \begin{bmatrix}
r_1\\
2 r_1 + r_2 + 2 r_3\\
-r_2 + r_3
\end{bmatrix}
\) (I)
Use any method to solve the above for \( r_1 \), \( r_2 \) and \( r_3 \).
\( r_1 = a \) , \( r_2 = \dfrac{b-2c-2a}{3} \) and \( r_3 = \dfrac{b+c-2a}{3} \)
Any vector \( \begin{bmatrix}
a \\
b \\
c
\end{bmatrix} \) in \( \mathbb{R}^3\) may be expressed as a linear combination of \( \textbf{u}_1 \) , \( \textbf{u}_2 \) and \( \textbf{u}_3 \) and therefore these 3 vectors span \( \mathbb{R}^3\).
Example 5
Show that the vectors \( \textbf{u}_1 = \begin{bmatrix}
-1 \\
2\\
0
\end{bmatrix} \) and \( \textbf{u}_2 = \begin{bmatrix}
0 \\
2\\
-1
\end{bmatrix} \) and \( \textbf{u}_3 = \begin{bmatrix}
1 \\
-4\\
1
\end{bmatrix} \) DO NOT span the vector space \( \mathbb{R}^3\)
Solution to Example 5
Express vector \( \begin{bmatrix}
a \\
b \\
c
\end{bmatrix} \) in \( \mathbb{R}^3\) as a linear combination of \( \textbf{u}_1, \textbf{u}_2\) and \( \textbf{u}_1 \)
\( \begin{bmatrix}
a \\
b \\
c
\end{bmatrix}
= r_1 \begin{bmatrix}
-1 \\
2\\
0
\end{bmatrix} +
r_2 \begin{bmatrix}
0 \\
2\\
-1
\end{bmatrix}
+
r_3 \begin{bmatrix}
1 \\
-4\\
1
\end{bmatrix}
\)
Use matrices to write the above system of equation as
\(
\begin{bmatrix}
-1 & 0 & 1\\
2 & 2 & -4\\
0 & -1 & 1
\end{bmatrix}
\begin{bmatrix}
r_1 \\
r_2 \\
r_3
\end{bmatrix}
=
\begin{bmatrix}
a \\
b \\
c
\end{bmatrix}
\)
The augmented matrix of the above system of equations is
\(
\begin{bmatrix}
-1 & 0 & 1 &|& a\\
2 & 2 & -4 &|& b \\
0 & -1 & 1 & |& c
\end{bmatrix}
\)
Use elementary row operations
to reduce the above to row echelon form.
Interchange row (1) and row (2)
\(
\begin{bmatrix}
2 & 2 & -4 &|& b \\
-1 & 0 & 1 &|& a\\
0 & -1 & 1 & |& c
\end{bmatrix}
\)
Add row (1) to twice row (2) and put the result in row (2)
\(
\begin{bmatrix}
2 & 2 & -4 &|& b \\
0 & 2 & -2 &|& 2a + b\\
0 & -1 & 1 & |& c
\end{bmatrix}
\)
Add row (2) to twice row (3) and put the result in row (3)
\(
\begin{bmatrix}
2 & 2 & -4 &|& b \\
0 & 2 & -2 &|& 2a + b\\
0 & 0 & 0 & |& 2a+ b + 2c
\end{bmatrix}
\)
There is no need to go any further because the last row tells us that the system has no solution for \( 2a+ b + 2c \ne 0 \).
We conclude that since we cannot find \( r_1 \) , \(r_2 \) and \( r_3 \) for any vector in \( \mathbb{R}^3\) that satisfies the condition \( 2a+ b + 2c \ne 0 \), the given vectors do not span \( \mathbb{R}^3\)
