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Coordinated Linear Algebra

Anna Davis, Paul Zachlin, Allan Donsig, Levi Heath, Mikkel Munkholm

Section 3.1 Linear Combinations of Vectors and their Span

When studying vectors, the two main operations we have learned about are vector addition and scalar multiplication. Both are involved in the important concept of a linear combination of vectors.

Definition 3.1.1.

A vector \(\mathbf{v}\) is said to be a linear combination of vectors \(\mathbf{v}_1, \mathbf{v}_2,\ldots, \mathbf{v}_n\) if
\begin{equation*} \mathbf{v}=a_1\mathbf{v}_1+ a_2\mathbf{v}_2+\ldots + a_n\mathbf{v}_n \end{equation*}
for some scalars \(a_1, a_2, \ldots ,a_n\text{.}\)
For example, \(\begin{bmatrix} -4\\9\\-10\\-1\end{bmatrix}\) is a linear combination of \(\begin{bmatrix} -1\\3\\-3\\0\end{bmatrix}\text{,}\) \(\begin{bmatrix} 2\\0\\1\\4\end{bmatrix}\) and \(\begin{bmatrix} 0\\1\\-1\\1\end{bmatrix}\) because
\begin{equation*} \begin{bmatrix} -4\\9\\-10\\-1\end{bmatrix}=2\begin{bmatrix} -1\\3\\-3\\0\end{bmatrix}+(-1)\begin{bmatrix} 2\\0\\1\\4\end{bmatrix}+3\begin{bmatrix} 0\\1\\-1\\1\end{bmatrix} \end{equation*}
In this section we will focus on vectors in \(\R^2\) and \(\R^3\text{.}\)

Remark 3.1.2.

Moving forward, vectors will also be written in horizontal notation instead of vertical. This is mainly for notational reasons. For instance, the vector
\begin{equation*} \begin{bmatrix} -4\\9\\-10\\-1\end{bmatrix} \end{equation*}
would instead in midtext be denoted by \([-4,9,10,-1]\text{.}\) This is justified, as any point in \((a,b,c,d)\) in \(\R^4 \) can be thought of as the vector starting from the orign \((0,0,0,0) \) and with direction \([a,b,c,d] \text{.}\)

Subsection 3.1.1 Visualizing Linear Combinations in \(\R^2\) and \(\R^3\)

Exploration 3.1.1.

Answer the questions below using the GeoGebra interactive. To use the interactive, you can
  1. change vectors \(\mathbf{v}\) and \(\mathbf{w}\) by dragging the tips of these vectors.
  2. change the coefficients \(k_1\) and \(k_2\) of the linear combination by using sliders.
Figure 3.1.3.
Problem 3.1.4.
Let \(\mathbf{w}=[1,2]\) and \(\mathbf{v}=[1,-1]\text{.}\) Find \(k_1\) and \(k_2\) such that
\begin{equation*} k_1\mathbf{w}+k_2\mathbf{v}=\begin{bmatrix}4\\-1\end{bmatrix}. \end{equation*}
Answer.
\(k_1=1,\quad k_2=3.\)
Problem 3.1.5.
Let \(\mathbf{w}=[1,2]\) and \(\mathbf{v}=[-2,0]\text{.}\) Find \(k_1\) and \(k_2\) such that
\begin{equation*} k_1\mathbf{w}+k_2\mathbf{v}=[3,-2]. \end{equation*}
Answer.
\(k_1=-1,\quad k_2=-2 \)
Problem 3.1.6.
Use the same vectors \(\mathbf{w}\) and \(\mathbf{v}\) as in the previous part. Do you think it is possible to express any vector in \(\R^2\) as a linear combination of \(\mathbf{w}\) and \(\mathbf{v}\text{?}\)
Answer.
\begin{equation*} \text{Yes} \end{equation*}
Problem 3.1.7.
Let \(\mathbf{w}=[4, 2]\) and \(\mathbf{v}=[-2,-1]\text{.}\) Do you think it is possible to express any vector in \(\R^2\) as a linear combination of \(\mathbf{w}\) and \(\mathbf{v}\text{.}\) Argue why/why not and write?
Answer.
\begin{equation*} \text{No} \end{equation*}
Visualizing linear combinations of vectors in \(\R^3\) is more difficult than doing so in \(\R^2\text{.}\) The following GeoGebra interactive will help you do this.

Exploration 3.1.2.

To use the interactive, define vectors \(\mathbf{u}\text{,}\) \(\mathbf{v}\) and \(\mathbf{w}\text{.}\) Use sliders to change the coefficients \(k_1, k_2\) and \(k_3\) of the linear combination. The linear combination \(k_1\mathbf{u}+k_2\mathbf{v}+k_3\mathbf{w}\) is shown as the pink vector along the diagonal of the parallelepiped. RIGHT-CLICK and DRAG the left panel to rotate the image.
Figure 3.1.8.

Subsection 3.1.2 Geometry of Linear Combinations

We now explore how to geometrically examine linear combinations, starting with an example.

Example 3.1.9.

Use geometry to express \([3,-4]\) as a linear combination of \([-1,3]\) and \([-4,2]\text{.}\)
Answer.
We are looking for \([3,-4]\) to be the diagonal of a parallelogram determined by scalar multiples of \([-1,3]\) and \([-4,2]\text{.}\)
The three vectors drawn in the real plane.
Because a scalar multiple of a vector can point in the same direction as the vector or in the opposite direction, we will start by drawing straight lines determined by the two vectors.
Scaling the vectors from before
The two lines that we drew will contain the sides of the parallelogram we are looking for. To find the other two sides we will draw lines parallel to \([-1,3]\) and \([-4,2]\) through the head of vector \([3,-4]\text{.}\)
Paralell lines added to second step
Now the parallelogram is clearly visible.
Parallelogram drawn
The last remaining task is to identify the sides of the parallelogram as scalar multiples of \([-1,3]\) and \([-4,2]\text{.}\) We do this by identifying vectors \([1,-3]\) and \([2,-1]\) as the vectors that determine the parallelogram.
Two sidepoints added
Observe that vector \([2,-1]\) is half the length of \([-4,2]\) and points in the opposite direction, while the vector \([1,-3]\) is the same length as \([-1,3]\) and also points in the opposite direction. Now we can write \([3,-4]\) as a linear combination of \([-1,3]\) and \([-4,2]\) as follows
\begin{equation*} \begin{bmatrix}3\\-4\end{bmatrix}=\begin{bmatrix}1\\-3\end{bmatrix}+\begin{bmatrix}2\\-1\end{bmatrix}=(-1)\begin{bmatrix}-1\\3\end{bmatrix}+\left(-\frac{1}{2}\right)\begin{bmatrix}-4\\2\end{bmatrix}. \end{equation*}
The method we used in Example 3.1.9 to express the given vector as a linear combination of two other vectors is sufficiently useful that we summarize the steps.

Exploration 3.1.3.

This GeoGebra interactive will allow you to go through the steps given in Algorithm 3.1.10 for a combination of vectors of your choice. To use the interactive
  1. Enter components of vectors \(\mathbf{u}\) and \(\mathbf{v}\text{.}\)
  2. Enter components of vector \(\mathbf{w}\) that you want to express as a linear combination of \(\mathbf{u}\) and \(\mathbf{v}\text{.}\)
  3. Use the navigation bar to go through the steps of Algorithm 3.1.10
Figure 3.1.11.
One of the stipulations in Algorithm 3.1.10 is that vectors \(\mathbf{u}\) and \(\mathbf{v}\) should be non-collinear. You can use the interactive in Exploration 3.1.3 to investigate what happens when \(\mathbf{u}\) and \(\mathbf{v}\) are collinear. The following example examines what happens from a geometric as well as an algebraic standpoint.

Example 3.1.12.

Can the vector \([3,2]\) be written as a linear combination of vectors \([-3,1]\) and \([6,-2]\text{?}\)
Answer.
We will start with a geometric approach.
Three vectors drawn
Observe that \([-3,1]\) and \([6,-2]\) are scalar multiples of each other and lie on the same line. A linear combination of \([-3,1]\) and \([6,-2]\) has the form:
\begin{equation*} a\begin{bmatrix}6\\-2\end{bmatrix}+b\begin{bmatrix}-3\\1\end{bmatrix}=a(-2)\begin{bmatrix}-3\\1\end{bmatrix}+b\begin{bmatrix}-3\\1\end{bmatrix}=(-2a+b)\begin{bmatrix}-3\\1\end{bmatrix}. \end{equation*}
This shows that all linear combinations of \([-3,1]\) and \([6,-2]\) will be scalar multiples of \([-3,1]\text{,}\) and therefore lie on the same line as \([-3,1]\text{.}\) Since \([3,2]\) does not lie on the line determined by \([-3,1]\) it cannot be expressed as a linear combination of \([-3,1]\) and \([6,-2]\text{.}\) We can also address this question algebraically. To express \([3,2]\) as a linear combination of \([-3,1]\) and \([6,-2]\text{,}\) we need to solve the equation.
\begin{equation*} a\begin{bmatrix}-3\\1\end{bmatrix}+b\begin{bmatrix}6\\-2\end{bmatrix}=\begin{bmatrix}3\\2\end{bmatrix} \end{equation*}
This gives us a system of equations
\begin{align*} -3a+6b\amp =3 \\ a-2b\amp =2 \end{align*}
When you try to solve this system, you will find that the system is inconsistent. Thus, \([3,2]\) cannot be written as a linear combination of \([-3,1]\) and \([6,-2]\text{.}\) We know that there is no way to express \([3,2]\) as a linear combination of vectors \([-3,1]\) and \([6,-2]\text{.}\) What would happen if we tried to apply Algorithm 3.1.10 to these vectors? You can use the GeoGebra interactive in Exploration 3.1.3 to find out.

Example 3.1.13.

Express \([2,4]\) as a linear combination of \([2,1]\) and \([2,-2]\text{.}\) Interpret your results geometrically.
Answer.
We need to find scalars \(a\) and \(b\) such that
\begin{equation*} \begin{bmatrix}2\\4\end{bmatrix}=a\begin{bmatrix}2\\1\end{bmatrix}+b\begin{bmatrix}2\\-2\end{bmatrix}. \end{equation*}
This amounts to solving a system of linear equations
\begin{align*} 2a+2b\amp =2 \\ a-2b\amp =4 \end{align*}
Use your favorite method to solve this system. (Hint: adding the second equation to the first will work well for this system.) You will find that \(a=2\) and \(b=-1\text{.}\) Now we can write \([2,4]\) as a linear combination of \([2,1]\) and \([2,-2]\) as follows:
\begin{equation*} \begin{bmatrix}2\\4\end{bmatrix}=2\begin{bmatrix}2\\1\end{bmatrix}+(-1)\begin{bmatrix}2\\-2\end{bmatrix}. \end{equation*}
Geometrically speaking, this means that the \([2,4]\) is the diagonal of the parallelogram determined by \(2[2,1]\) and \((-1) [2,-2]\text{.}\)
The original vectors \([2,1]\) and \([2,-2]\) are shown below together with the parallelogram and its diagonal.
Paralellogram with diagonal drawn

Example 3.1.14.

If possible, express \([7, 4, -5]\) as a linear combination of \([1,-2,1]\) and \([3, 0, -1]\text{.}\)
Answer.
We are looking for coefficients \(a\) and \(b\) such that
\begin{equation*} a\begin{bmatrix}1\\-2\\1\end{bmatrix}+b\begin{bmatrix}3\\0\\-1\end{bmatrix}=\begin{bmatrix}7\\4\\-5\end{bmatrix}. \end{equation*}
This translates into a system of equations
\begin{align*} a+3b\amp =7 \\ -2a\amp =4 \\ a-b\amp =-5 \end{align*}
Solve this system for \(a\) and \(b\text{,}\) and enter your answers below:
\begin{equation*} a=-2\quad\text{and}\quad b=3. \end{equation*}
We conclude that \([7,4,-5]\) is a linear combination of \([1,-2,1]\) and \([3,0,-1]\text{,}\) and write:
\begin{equation*} \begin{bmatrix}7\\4\\-5\end{bmatrix}=-2\begin{bmatrix}1\\-2\\1\end{bmatrix}+3\begin{bmatrix}3\\0\\-1\end{bmatrix}. \end{equation*}

Example 3.1.15.

Set up a system of equations that can be used to express \([2,-1,3,0]\) as a linear combination of \([1,0,4,-2]\text{,}\) \([-2,-1,1,-1]\text{,}\) \([0,4,-3,1]\) and \([1,1,-1,4]\text{,}\) or to determine that such a combination does not exist.
\(\textbf{Do not solve the system}. \)
Answer.
We are looking for \(x_1\text{,}\) \(x_2\text{,}\) \(x_3\) and \(x_4\) such that
\begin{equation*} x_1\begin{bmatrix}1\\0\\4\\-2\end{bmatrix}+x_2\begin{bmatrix}-2\\-1\\1\\-1\end{bmatrix}+x_3\begin{bmatrix}0\\4\\-3\\1\end{bmatrix}+x_4\begin{bmatrix}1\\1\\-1\\4\end{bmatrix}=\begin{bmatrix}2\\-1\\3\\0\end{bmatrix}. \end{equation*}
This translates into the following system of equations:
\begin{equation*} \begin{array}{ccccccccc} x_1 \amp -\amp 2x_2 \amp \amp \amp + \amp x_4\amp = \amp 2 \\ \amp \amp -x_2 \amp + \amp 4x_3 \amp +\amp x_4\amp = \amp -1 \\ 4x_1 \amp + \amp x_2 \amp -\amp 3x_3 \amp -\amp x_4\amp = \amp 3\\ -2x_1\amp -\amp x_2 \amp +\amp x_3\amp +\amp 4x_4\amp =\amp 0 \end{array}. \end{equation*}

Subsection 3.1.3 The Linear Span

Recall that a vector \(\mathbf{v}\) is said to be a linear combination of vectors \(\mathbf{v}_1, \mathbf{v}_2,\ldots, \mathbf{v}_n\) if one has
\begin{equation*} \mathbf{v}=a_1\mathbf{v}_1+ a_2\mathbf{v}_2+\ldots + a_n\mathbf{v}_n \end{equation*}
for some scalars \(a_1, a_2, \ldots ,a_n\text{.}\)
We provide an explicit with explicit, concrete, numbers to demystify the above.

Example 3.1.16.

If possible, express the given vector as a linear combination of
\begin{equation*} \mathbf{v}_1=\begin{bmatrix}-2\\-3\\4\end{bmatrix},\quad\mathbf{v}_2=\begin{bmatrix}2\\3\\2\end{bmatrix}. \end{equation*}
Interpret your results geometrically.
  1. \begin{equation*} \mathbf{u}=\begin{bmatrix}2\\3\\5\end{bmatrix}. \end{equation*}
  2. \begin{equation*} \mathbf{w}=\begin{bmatrix}5\\5\\1\end{bmatrix}. \end{equation*}
Answer.
For Item 1, we need to find coefficients \(a_1\) and \(a_2\) such that \(\mathbf{u}=a_1\mathbf{v}_1+a_2\mathbf{v}_2\text{.}\) To do this we need to solve the vector equation:
\begin{equation*} a_1\begin{bmatrix}-2\\-3\\4\end{bmatrix}+a_2\begin{bmatrix}2\\3\\2\end{bmatrix}=\begin{bmatrix}2\\3\\5\end{bmatrix}. \end{equation*}
This equation translates into the following system:
\begin{equation*} \begin{array}{ccccc} -2a_1 \amp +\amp 2a_2\amp = \amp 2 \\ -3a_1\amp +\amp 3a_2\amp = \amp 3 \\ 4a_1 \amp + \amp 2a_2\amp = \amp 5\\ \end{array}. \end{equation*}
We write the system in augmented matrix form and apply elementary row operations to bring it to reduced row-echelon form.
\begin{equation*} \left[\begin{array}{cc|c} -2\amp 2\amp 2\\-3\amp 3\amp 3\\4\amp 2\amp 5 \end{array}\right]\rightsquigarrow\left[\begin{array}{cc|c} 1\amp 0\amp 1/2\\0\amp 1\amp 3/2\\0\amp 0\amp 0 \end{array}\right]. \end{equation*}
This shows that \(a_1=\frac{1}{2}\) and \(a_2=\frac{3}{2}\text{,}\) and we can express \(\mathbf{u}\) as a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\) as follows:
\begin{equation*} \frac{1}{2}\begin{bmatrix}-2\\-3\\4\end{bmatrix}+\frac{3}{2}\begin{bmatrix}2\\3\\2\end{bmatrix}=\begin{bmatrix}2\\3\\5\end{bmatrix}. \end{equation*}
Observe that because vector \(\mathbf{u}\) is a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{,}\) \(\mathbf{u}\) is the diagonal of a parallelogram whose sides are scalar multiples of \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{.}\) As such, \(\mathbf{u}\) lies in the same plane as \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{,}\) as illustrated below.
Span of two vectors graphed
For Item 2, we need to solve the following vector equation:
\begin{equation*} a_1\begin{bmatrix}-2\\-3\\4\end{bmatrix}+a_2\begin{bmatrix}2\\3\\2\end{bmatrix}=\begin{bmatrix}5\\5\\1\end{bmatrix}. \end{equation*}
This equation corresponds to the system:
\begin{equation*} \begin{array}{ccccc} -2a_1 \amp +\amp 2a_2\amp = \amp 5 \\ -3a_1\amp +\amp 3a_2\amp = \amp 5 \\ 4a_1 \amp + \amp 2a_2\amp = \amp 1\\ \end{array}. \end{equation*}
Writing the system in augmented matrix form and applying elementary row operations gives us the following reduced row-echelon form:
\begin{equation*} \left[\begin{array}{cc|c} -2\amp 2\amp 5\\-3\amp 3\amp 5\\4\amp 2\amp 1 \end{array}\right]\rightsquigarrow\left[\begin{array}{cc|c} 1\amp 0\amp 0\\0\amp 1\amp 0\\0\amp 0\amp 1 \end{array}\right]. \end{equation*}
We conclude that there are no solutions, and \(\mathbf{w}\) is not a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{.}\)
Geometrically, this means that \(\mathbf{w}\) is not the diagonal of any parallelogram whose sides are scalar multiples of \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{.}\) Thus, \(\mathbf{w}\) does not lie in the plane determined by \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{.}\)
Three vectors graphed without span
In Item 1 of Example 3.1.16 we expressed \(\mathbf{u}\) as a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{,}\) and concluded that \(\mathbf{u}\) lies in the plane determined by \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{.}\) We say that \(\mathbf{u}\) is in the span of \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{.}\) In fact, every vector in the plane determined by \(\mathbf{v}_1\) and \(\mathbf{v}_2\) is in the span of \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{.}\) We say that \(\mathbf{v}_1\) and \(\mathbf{v}_2\) span the plane.
In contrast, vector \(\mathbf{w}\) of Item 2 of Example 3.1.16 is not a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{.}\) We say that \(\mathbf{w}\) is not in the span of \(\mathbf{v}_1\) and \(\mathbf{v}_2\text{.}\)
The following video takes another look at Example 3.1.16 using our new vocabulary.

Definition 3.1.17.

Let \(\mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_p\) be vectors in \(\R^n\text{.}\) The set \(S\) of all linear combinations of \(\mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_p\) is called the span of \(\mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_p\text{.}\) We write
\begin{equation*} S=\text{span}(\mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_p), \end{equation*}
and we say that vectors \(\mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_p\) span \(S\text{.}\) Any vector in \(S\) is said to be in the span of \(\mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_p\text{.}\) The set \(\{\mathbf{v}_1, \mathbf{v}_2,\ldots ,\mathbf{v}_p\}\) is called a spanning set for the space \(S\text{.}\)
The definition is rather formal even with all the preceding examples and geometric intuition. Running through the examples below in detail is highly recommended.

Example 3.1.18.

Describe
\begin{equation*} \text{span}\left(\begin{bmatrix}-3\\1\end{bmatrix}\right). \end{equation*}
Answer.
The span of \([-3,1]\) is the set of all linear combinations of \([-3,1]\text{.}\) Since we are looking for linear combinations of only one vector, we are really looking for all of its scalar multiples. So, the span will be the set of all vectors of the form \(\mathbf{v}=a[-3,1]\text{.}\) All such vectors lie on the line determined by \([-3,1]\text{.}\)
Vector and its span graphed

Example 3.1.19.

Describe
\begin{equation*} \text{span}\left(\begin{bmatrix}2\\2\end{bmatrix}, \begin{bmatrix}-1\\0\end{bmatrix}\right). \end{equation*}
Answer.
First, observe that \([2,2]\) and \([-1,0]\) are not scalar multiples of each other.
Two vectors graphed
Geometrically, we can use Algorithm 3.1.10 to express any vector of \(\R^2\) as a linear combination of \([2,2]\) and \([-1,0]\text{,}\) indicating that the two vectors span all of \(\R^2\text{.}\)
To verify this claim algebraically we will show that an arbitrary vector \([s,t]\) of \(\R^2\) can be written as a linear combination of \([2,2]\) and \([-1,0]\text{.}\)
Consider the vector equation:
\begin{equation*} a_1\begin{bmatrix}2\\2\end{bmatrix}+a_2\begin{bmatrix}-1\\0\end{bmatrix}=\begin{bmatrix}s\\t\end{bmatrix}. \end{equation*}
This corresponds to the system:
\begin{equation*} \begin{array}{ccccc} 2a_1 \amp -\amp a_2\amp = \amp s \\ 2a_1\amp \amp \amp = \amp t \\ \end{array}. \end{equation*}
Writing the system in augmented matrix form and applying elementary row operations gives us the following reduced row-echelon form:
\begin{equation*} \left[\begin{array}{cc|c} 2\amp -1\amp s\\2\amp 0\amp t \end{array}\right]\rightsquigarrow\left[\begin{array}{cc|c} 1\amp 0\amp t/2\\0\amp 1\amp t-s \end{array}\right]. \end{equation*}
This shows that every vector of \(\R^2\) can be written as a linear combination of \([2,2]\) and \([-1,0]\text{:}\)
\begin{equation*} (t/2)\begin{bmatrix}2\\2\end{bmatrix}+(t-s)\begin{bmatrix}-1\\0\end{bmatrix}=\begin{bmatrix}s\\t\end{bmatrix}. \end{equation*}
We conclude that
\begin{equation*} \text{span}\left(\begin{bmatrix}2\\2\end{bmatrix}, \begin{bmatrix}-1\\0\end{bmatrix}\right)=\R^2. \end{equation*}

Example 3.1.20.

Describe
\begin{equation*} \text{span}\left(\begin{bmatrix}5\\0\\4\end{bmatrix}, \begin{bmatrix}0\\4\\2\end{bmatrix}\right). \end{equation*}
Answer.
First, observe that \([5,0,4], [0,4,2]\) are not scalar multiples of each other.
Two relevant vectors drawn
The span of \([5,0,4]\) and \([0,4,2]\) consists of elements of the form
\begin{equation*} a_1\begin{bmatrix}5\\0\\4\end{bmatrix}+a_2\begin{bmatrix}0\\4\\2\end{bmatrix}. \end{equation*}
Geometrically, we can interpret all such linear combinations as diagonals of parallelograms determined by scalar multiples of \([5,0,4]\) and \([0,4,2]\text{.}\) All such diagonals will lie in the plane determined by \([5,0,4]\) and \([0,4,2]\text{.}\) Let this plane be called \(p\text{.}\) A portion of \(p\) is shown below.
Span of previous two vectors drawn
Because Algorithm 3.1.10 can be applied to vectors that lie in \(p\) just as easily as it can be applied to vectors of \(\R^2\text{,}\) we conclude that every vector in \(p\) can be expressed as a linear combination of \([5,0,4]\) and \([0,4,2]\text{.}\) Thus,
\begin{equation*} \text{span}\left(\begin{bmatrix}5\\0\\4\end{bmatrix}, \begin{bmatrix}0\\4\\2\end{bmatrix}\right)=p. \end{equation*}

Exercises 3.1.4 Exercises

1.

Solve a system of linear equations to express \([-1, 7]\) as a linear combination of \([1,2]\) and \([-1,1]\text{.}\)
Answer.
System of linear equations:
\begin{equation*} \begin{array}{ccccc} 1a \amp +\amp -1b\amp = \amp -1 \\ 2a\amp +\amp 1b\amp =\amp 7 \end{array}. \end{equation*}
Values of \(a\) and \(b\text{:}\)
\begin{equation*} a=2\quad\text{and}\quad b=3 \end{equation*}
Linear Combination:
\begin{equation*} \begin{bmatrix}-1\\7\end{bmatrix}=2\begin{bmatrix}1\\2\end{bmatrix}+3\begin{bmatrix}-1\\1\end{bmatrix}. \end{equation*}

2.

Use Algorithm 3.1.10 to express \([-3, 0]\) as a linear combination of \([2,4]\) and \([-1, 1]\text{.}\)
Answer.
Linear Combination:
\begin{equation*} \begin{bmatrix}-3\\0\end{bmatrix}=-0.5\begin{bmatrix}2\\4\end{bmatrix}+2\begin{bmatrix}-1\\1\end{bmatrix}. \end{equation*}

3.

Use two different approaches (algebraic and geometric) to explain why the vector \([5,1]\) cannot be expressed as a linear combination of vectors \([2,-1]\) and \([-4,2]\text{.}\)

4.

We have seen Algorithm 3.1.10 applied to vectors in \(\R^2\text{.}\) The same process can, in certain cases be applied to vectors in \(\R^3\text{.}\) In both parts of this problem you will be asked to follow the steps in Algorithm 3.1.10 to express one vector as a linear combination of two given vectors. Then you will be asked to identify the condition which makes it possible to do so.
The following GeoGebra interactive shows vectors \(u=[3,0,-1]\text{,}\) \(v=[1,-2,1]\) and \(w=[7,4,-5]\text{.}\) RIGHT-CLICK and DRAG to rotate the image.
Figure 3.1.21.

5.

    Can \(\mathbf{w}\) be expressed as a linear combination of \(\mathbf{u}\) and \(\mathbf{v}\text{?}\)
  • No, because \(\mathbf{w}\) is not between \(\mathbf{u}\) and \(\mathbf{v}\)
    .
  • Yes, because all three vectors are in the same plane.
  • Yes, because all three vectors are in the same plane, AND \(\mathbf{u} \) and \(\mathbf{v}\) are not collinear.

6.

Use the navigation bar at the bottom of the interactive window to view construction steps of Algorithm 3.1.10 applied to vectors \(\mathbf{u}\text{,}\) \(\mathbf{v}\) and \(\mathbf{w}\) (right-click and drag to rotate the image). Use the final image to express \(w\) as a linear combination of \(v\) (blue) and \(u\) (red).
Answer.
\begin{equation*} w=-2v+3u \end{equation*}

7.

The following GeoGebra interactive shows vectors \(v=[1,-2,1]\text{,}\) \(u=[3,0,-1]\text{,}\) and \(w=[7,4,0]\text{.}\) RIGHT-CLICK and DRAG to rotate the image. Use geometry to explain why \(w\) cannot be expressed as a linear combination of \(v\) and \(u\text{.}\)
Figure 3.1.22.
Answer.
We can also show that \(w\) is not a linear combination of \(v\) and \(u\) algebraically by attempting to solve a system of equations corresponding to
\begin{equation*} x_1 v + x_2 u=w. \end{equation*}
Set up the system of equations
\begin{equation*} \begin{array}{ccccccc} 1x_1 \amp +\amp 3x_2\amp = \amp 7 \\ -2x_1\amp +\amp 0x_2\amp =\amp 4\\ 1x_1\amp +\amp -1x_2\amp =\amp 0 \end{array}. \end{equation*}
Find the reduced row echelon form.
\begin{equation*} \left[\begin{array}{cc|c} 1\amp 0\amp 0\\0\amp 1\amp 0\\0\amp 0\amp 1 \end{array}\right]. \end{equation*}

Exercise Group.

Choose the best description for each set below.
8.
    \begin{equation*} \text{span}\left(\begin{bmatrix}1\\1\\-2\end{bmatrix}, \begin{bmatrix}2\\2\\-4\end{bmatrix}\right) \end{equation*}
  • Plane in \(\R^3\)
  • Line in \(\R^2\)
  • Line in \(\R^3\)
  • \(\R^3\)
  • \(\R^2\)
9.
    \begin{equation*} \text{span}\left(\begin{bmatrix}1\\-2\end{bmatrix}, \begin{bmatrix}1\\0\end{bmatrix}\right) \end{equation*}
  • Plane in \(\R^3\)
  • Line in \(\R^2\)
  • Line in \(\R^3\)
  • \(\R^3\)
  • \(\R^2\)
10.
    \begin{equation*} \text{span}\left(\begin{bmatrix}-3\\1\end{bmatrix}, \begin{bmatrix}6\\-2\end{bmatrix}, \begin{bmatrix}3\\-1\end{bmatrix}\right) \end{equation*}
  • Plane in \(\R^3\)
  • Line in \(\R^2\)
  • Line in \(\R^3\)
  • \(\R^3\)
  • \(\R^2\)

11.

    Which of the following pairs of sets are equal?
  • \(V=\text{span}\left(\begin{bmatrix}5\\0\\0\end{bmatrix},\begin{bmatrix}10\\0\\0\end{bmatrix},\begin{bmatrix}0\\0\\-4\end{bmatrix}\right) \quad\text{and}\quad W=\text{span}\left(\begin{bmatrix}-2\\0\\0\end{bmatrix},\begin{bmatrix}0\\0\\1\end{bmatrix}\right)\)
  • \(V=\text{span}\left(\begin{bmatrix} 5\\ 3\end{bmatrix},\begin{bmatrix} 10\\ -1 \end{bmatrix},\begin{bmatrix} 0\\ 2 \end{bmatrix}\right) \quad\text{and}\quad W=\text{span}\left(\begin{bmatrix} -2\\ 0 \end{bmatrix},\begin{bmatrix} 0\\ 1\end{bmatrix}\right)\)
  • \(V=\text{span}\left(\begin{bmatrix}1\\-2\\4\end{bmatrix}\right)\quad\text{and}\quad W=\text{span}\left(\begin{bmatrix}-1\\2\\-4\end{bmatrix},\begin{bmatrix}0\\0\\1\end{bmatrix}\right)\)
  • \(V=\text{span}\left(\begin{bmatrix}5\\0\end{bmatrix}\right)\quad\text{and}\quad W=\text{span}\left(\begin{bmatrix}-2\\0\end{bmatrix},\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}-4\\0\end{bmatrix},\begin{bmatrix}0\\0\end{bmatrix}\right)\)

12.

Let \(\mathbf{v}=[3,4,5]\text{.}\) Give an example of at least one vector \(\mathbf{w}\) such that \(\mathbf{v}\text{,}\) \(\mathbf{w}\) do NOT span a plane in \(\R^3\text{.}\) Describe \(\text{span}(\mathbf{v}, \mathbf{w})\text{.}\)

13.

Prove or disprove. The zero vector of \(\R^n\) is contained in the span of any collection of vectors of \(\R^n\text{.}\)
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