Linear Combinations and Span of Vectors

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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 all linear combinations 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 \( \mathbb{R}^3\)


More References and links

  1. Linear Algebra - Questions with Solutions
  2. Linear Algebra and its Applications - 5 th Edition - David C. Lay , Steven R. Lay , Judi J. McDonald
  3. Elementary Linear Algebra - 7 th Edition - Howard Anton and Chris Rorres

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