# Linear Combinations and Span of Vectors

   

## Definition of Linear Combinations of Vectors

Vector $\textbf{v}$ is a linear combinations of the vectors $\textbf{u}_1, \textbf{u}_2, ... \textbf{u}_n$ if it can be written in the form $\textbf{v} = r_1 \textbf{u}_1 + r_2 \textbf{u}_2 + ... + r_n \textbf{u}_n$ where $r_1, r_2, ... , r_n$ are scalars.

## Examples with Solutions

Note solve the examples in the order that they are presented in order to fully understand them.

Example 1
Express vector $\textbf {v} = \begin{bmatrix} 2 \\ 6 \\ 12 \\ \end{bmatrix}$ as a linear combination of the vectors $\textbf{u}_1 = \begin{bmatrix} 1 \\ 3 \\ 0 \\ \end{bmatrix}$ and $\textbf{u}_2 = \begin{bmatrix} 0 \\ 0 \\ 4 \\ \end{bmatrix}$

Solution to Example 1
For vector $\textbf{v}$ to be a linear combination of the vectors $\textbf{u}_1$ and $\textbf{u}_2$, we need to find a scalars $r_1$ and $r_2$ such that (see definition above)
$\textbf{v} = r_1 \textbf{u}_1 + r_2 \textbf{u}_2$
Substitute $\textbf{v}, \textbf{u}_1 , \textbf{u}_2$ by their components to obtain the equation
$\begin{bmatrix} 2 \\ 6 \\ 12 \\ \end{bmatrix} = r_1 \begin{bmatrix} 1 \\ 3 \\ 0 \\ \end{bmatrix} + r_2 \begin{bmatrix} 0 \\ 0 \\ 4 \\ \end{bmatrix}$
Use scalar mutliplication on the right side to write
$\begin{bmatrix} 2 \\ 6 \\ 12 \\ \end{bmatrix} = \begin{bmatrix} r_1 \\ 3 r_1 \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 4 r_2 \\ \end{bmatrix}$
Add the vectors on the right
$\begin{bmatrix} 2 \\ 6 \\ 12 \\ \end{bmatrix} = \begin{bmatrix} r_1 \\ 3 r_1 \\ 4 r_2 \\ \end{bmatrix}$
Equality of the components of the vectors above gives the system of equations
$2 = r_1 , 6 = 3 r_1 , 12 = 4 r_2$
The solutions are
$r_1 = 2$ and $r_2 = 3$
Hence
$\begin{bmatrix} 2 \\ 6 \\ 12 \\ \end{bmatrix} = 2 \begin{bmatrix} 1 \\ 3 \\ 0 \\ \end{bmatrix} + 3 \begin{bmatrix} 0 \\ 0 \\ 4 \\ \end{bmatrix}$

Example 2
Show that the vector $\textbf {v} = \begin{bmatrix} 1 \\ 3 \\ \end{bmatrix}$ cannot be expressed as a linear combination of the vectors $\textbf{u}_1 = \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}$ and $\textbf{u}_2 = \begin{bmatrix} 2 \\ 4 \\ \end{bmatrix}$
Solution to Example 2
We need to show that we cannot find $r_1$ and $r_2$ such that
$\begin{bmatrix} 1 \\ 3 \\ \end{bmatrix} = r_1 \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix} + r_2 \begin{bmatrix} 2 \\ 4 \\ \end{bmatrix}$
Scalar multiply and add the vectors on the right side of the above equation
$\begin{bmatrix} 1 \\ 3 \\ \end{bmatrix} = \begin{bmatrix} r_1 + 2 r_2 \\ 2 r_1 + 4 r_2\\ \end{bmatrix}$
Solve the above as a system of equations
$\begin{array}{lcl} r_1 + 2 r_2 & = & 1 \\ 2 r_1 + 4 r_2 & = & 3 \end{array}$
Solve the above system using any method. We use the method of elimination.
Multiply the top equation by $- 2$
$\begin{array}{lcl} -2 r_1 - 4 r_2 & = & -2 \\ 2 r_1 + 4 r_2 & = & 3 \end{array}$
Add the two equations to obtain
$0 = 1$
The system has no solution; therefore vector $\textbf {v}$ cannot be expressed as a linear combinations of the given vectors $\textbf {u}_1$ and $\textbf {u}_2$.

## Space Spanned by Vectors

If vectors $\textbf{u}_1, \textbf{u}_2, ... \textbf{u}_n$ are in a vector space $V$, then the set $W$ of of these vectors is a subspace of $V$ and is called the span of the vectors $\textbf{u}_1, \textbf{u}_2, ... \textbf{u}_n$.
We also say that the vectors $\textbf{u}_1, \textbf{u}_2, ... \textbf{u}_n$ span $W$ which may be written as
$W = \text{span} \{ \textbf{u}_1, \textbf{u}_2, ... \textbf{u}_n \}$

Example 3
Show that the vectors $\textbf{u}_1 = \begin{bmatrix} 1 \\ 5\\ \end{bmatrix}$ and $\textbf{u}_2 = \begin{bmatrix} -1 \\ 2\\ \end{bmatrix}$ span the vector space $\mathbb{R}^2$
Solution to Example 3
A vector in the space $\mathbb{R}^2$ is of the form $\begin{bmatrix} a \\ b \\ \end{bmatrix}$ where $a$ and $b$ can take any value in the set of real numbers $\mathbb{R}$.
To show that the vectors $\textbf{u}_1$ and $\textbf{u}_2$ span $\mathbb{R}^2$, we need to show that any vector $\begin{bmatrix} a \\ b \\ \end{bmatrix}$ in $\mathbb{R}^2$ is a linear combination of the vectors $\textbf{u}_1$ and $\textbf{u}_2$. We therefore need to show that we can find the scalars $r_1$ and $r_2$ such that
$\begin{bmatrix} a \\ b \\ \end{bmatrix} = r_1 \begin{bmatrix} 1 \\ 5\\ \end{bmatrix} + r_2 \begin{bmatrix} -1 \\ 2\\ \end{bmatrix}$
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 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 4
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 4
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 5
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 5
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 6
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 6
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 7
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 7
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 $\mathbb{R}^3$