Span of Vectors

The definitions of the span of vectors are presented including with examples and their solutions


Space Spanned by Vectors

If vectors Vectors u1, u2, ..., un 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 Vectors u1, u2, ..., un.
We also say that the vectors Vectors u1, u2, ..., un span W which may be written as
Span W .

Examples with Solutions

Example 1
Show that the vectors Vectors u1=[1 , 5] and Vectors u2=[-1 , 2] span the vector space 2.
Solution to Example 1
A vector in the space 2 is of the form Vectors [a , b] where a and b can take any value in the set of real numbers .
To show that the vectors u1 and u2 span 2, we need to show that any vector Vectors [a , b] in 2 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 \( \) \( \) \( \) \( \)
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 \( \mathbb{R}^3\)


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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