## Section2.2Linear Combinations (VS2)

### Subsection2.2.1Class Activities

#### Definition2.2.1.

A linear combination of a set of vectors $$\{\vec v_1,\vec v_2,\dots,\vec v_m\}$$ is given by $$c_1\vec v_1+c_2\vec v_2+\dots+c_m\vec v_m$$ for any choice of scalar multiples $$c_1,c_2,\dots,c_m\text{.}$$

For example, we can say $$\left[\begin{array}{c}3 \\0 \\ 5\end{array}\right]$$ is a linear combination of the vectors $$\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right]$$ and $$\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right]$$ since

\begin{equation*} \left[\begin{array}{c} 3 \\ 0 \\ 5 \end{array}\right] = 2 \left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right] + 1\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right]\text{.} \end{equation*}

#### Definition2.2.2.

The span of a set of vectors is the collection of all linear combinations of that set:

\begin{equation*} \vspan\{\vec v_1,\vec v_2,\dots,\vec v_m\} = \setBuilder{c_1\vec v_1+c_2\vec v_2+\dots+c_m\vec v_m}{ c_i\in\IR}\text{.} \end{equation*}

For example:

\begin{equation*} \vspan\setList { \left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right], \left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] } = \setBuilder { a\left[\begin{array}{c} 1 \\ -1 \\ 2 \end{array}\right]+ b\left[\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right] }{ a,b\in\IR }\text{.} \end{equation*}

#### Activity2.2.3.

Consider $$\vspan\left\{\left[\begin{array}{c}1\\2\end{array}\right]\right\}\text{.}$$

##### (a)

Sketch $$1\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}1\\2\end{array}\right]\text{,}$$ $$3\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}3\\6\end{array}\right]\text{,}$$ $$0\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}0\\0\end{array}\right]\text{,}$$ and $$-2\left[\begin{array}{c}1\\2\end{array}\right]=\left[\begin{array}{c}-2\\-4\end{array}\right]$$ in the $$xy$$ plane.

##### (b)

Sketch a representation of all the vectors belonging to $$\vspan\setList{\left[\begin{array}{c}1\\2\end{array}\right]} = \setBuilder{a\left[\begin{array}{c}1\\2\end{array}\right]}{a\in\IR}$$ in the $$xy$$ plane.

#### Activity2.2.4.

Consider $$\vspan\left\{\left[\begin{array}{c}1\\2\end{array}\right], \left[\begin{array}{c}-1\\1\end{array}\right]\right\}\text{.}$$

##### (a)

Sketch the following linear combinations in the $$xy$$ plane.

\begin{equation*} 1\left[\begin{array}{c}1\\2\end{array}\right]+ 0\left[\begin{array}{c}-1\\1\end{array}\right]\hspace{3em} 0\left[\begin{array}{c}1\\2\end{array}\right]+ 1\left[\begin{array}{c}-1\\1\end{array}\right]\hspace{3em} 1\left[\begin{array}{c}1\\2\end{array}\right]+ 1\left[\begin{array}{c}-1\\1\end{array}\right] \end{equation*}
\begin{equation*} -2\left[\begin{array}{c}1\\2\end{array}\right]+ 1\left[\begin{array}{c}-1\\1\end{array}\right]\hspace{3em} -1\left[\begin{array}{c}1\\2\end{array}\right]+ -2\left[\begin{array}{c}-1\\1\end{array}\right] \end{equation*}
##### (b)

Sketch a representation of all the vectors belonging to $$\vspan\left\{\left[\begin{array}{c}1\\2\end{array}\right], \left[\begin{array}{c}-1\\1\end{array}\right]\right\}=\setBuilder{a\left[\begin{array}{c}1\\2\end{array}\right]+b\left[\begin{array}{c}-1\\1\end{array}\right]}{a, b \in \IR}$$ in the $$xy$$ plane.

#### Activity2.2.5.

Sketch a representation of all the vectors belonging to $$\vspan\left\{\left[\begin{array}{c}6\\-4\end{array}\right], \left[\begin{array}{c}-3\\2\end{array}\right]\right\}$$ in the $$xy$$ plane.

#### Activity2.2.6.

The vector $$\left[\begin{array}{c}-1\\-6\\1\end{array}\right]$$ belongs to $$\vspan\left\{\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}-1\\-3\\2\end{array}\right]\right\}$$ exactly when there exists a solution to the vector equation $$x_1\left[\begin{array}{c}1\\0\\-3\end{array}\right]+ x_2\left[\begin{array}{c}-1\\-3\\2\end{array}\right] =\left[\begin{array}{c}-1\\-6\\1\end{array}\right]\text{.}$$

##### (a)

Reinterpret this vector equation as a system of linear equations.

##### (b)

Find its solution set, using technology to find $$\RREF$$ of its corresponding augmented matrix.

##### (c)

Given this solution set, does $$\left[\begin{array}{c}-1\\-6\\1\end{array}\right]$$ belong to $$\vspan\left\{\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}-1\\-3\\2\end{array}\right]\right\}\text{?}$$

#### Observation2.2.8.

The following are all equivalent statements:

• The vector $$\vec{b}$$ belongs to $$\vspan\{\vec v_1,\dots,\vec v_n\}\text{.}$$

• The vector equation $$x_1 \vec{v}_1+\cdots+x_n \vec{v}_n=\vec{b}$$ is consistent.

• The linear system corresponding to $$\left[\vec v_1\,\dots\,\vec v_n \,|\, \vec b\right]$$ is consistent.

• $$\RREF\left[\vec v_1\,\dots\,\vec v_n \,|\, \vec b\right]$$ doesn't have a row $$[0\,\cdots\,0\,|\,1]$$ representing the contradiction $$0=1\text{.}$$

#### Activity2.2.9.

Determine if $$\left[\begin{array}{c}3\\-2\\1 \\ 5\end{array}\right]$$ belongs to $$\vspan\left\{\left[\begin{array}{c}1\\0\\-3 \\ 2\end{array}\right], \left[\begin{array}{c}-1\\-3\\2 \\ 2\end{array}\right]\right\}$$ by solving an appropriate vector equation.

#### Activity2.2.10.

Determine if $$\left[\begin{array}{c}-1\\-9\\0\end{array}\right]$$ belongs to $$\vspan\left\{\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}-1\\-3\\2\end{array}\right]\right\}$$ by solving an appropriate vector equation.

#### Activity2.2.11.

Does the third-degree polynomial $$3y^3-2y^2+y+5$$ in $$\P_3$$ belong to $$\vspan\{y^3-3y+2,-y^3-3y^2+2y+2\}\text{?}$$

##### (a)

Reinterpret this question as a question about the solution(s) of a polynomial equation:

\begin{equation*} x_1(\cdots\unknown\cdots)+x_2(\cdots\unknown\cdots)= (\cdots\unknown\cdots) \end{equation*}
##### (b)

Write a Euclidean vector equation that has the same solution set:

\begin{equation*} x_1\left[\begin{array}{c}\unknown\\\unknown\\\unknown\\\unknown\end{array}\right]+ x_2\left[\begin{array}{c}\unknown\\\unknown\\\unknown\\\unknown\end{array}\right]= \left[\begin{array}{c}\unknown\\\unknown\\\unknown\\\unknown\end{array}\right] \end{equation*}
##### (c)

Answer this equivalent question, and use its solution to answer the original question.

#### Activity2.2.12.

Does the polynomial $$x^2+x+1$$ belong to $$\vspan\{x^2-x,x+1, x^2-1\}\text{?}$$

#### Activity2.2.13.

Does the matrix $$\left[\begin{array}{cc}3&-2\\1&5\end{array}\right]$$ belong to $$\vspan\left\{\left[\begin{array}{cc}1&0\\-3&2\end{array}\right], \left[\begin{array}{cc}-1&-3\\2&2\end{array}\right]\right\}\text{?}$$

##### (a)

Reinterpret this question as a question about the solution(s) of a matrix equation.

##### (b)

Answer this equivalent question, and use its solution to answer the original question.

### Subsection2.2.3Slideshow

Slideshow of activities available at https://teambasedinquirylearning.github.io/linear-algebra/2022/VS2.slides.html.

### Subsection2.2.5Mathematical Writing Explorations

#### Exploration2.2.14.

Suppose $$S = \{\vec{v_1},\ldots, \vec{v_n}\}$$ is a set of vectors. Show that $$\vec{v_0}$$ is a linear combination of members of $$S\text{,}$$ if an only if there are a set of scalars $$\{c_0,c_1,\ldots, c_n\}$$ such that $$\vec{z} = c_0\vec{v_0} + \cdots + c_n\vec{v_n}.$$ We can do this in a few parts. I've used bullets here to indicate all that needs to be done. This is an "if and only if" proof, so it needs two parts.

• First, assume that $$\vec{0} = c_0\vec{v_0} + \cdots + c_n\vec{v_n}$$ has a solution, with $$c_0 \neq 0\text{.}$$ Show that $$\vec{v_0}$$ is a linear combination of elements of $$S\text{.}$$

• Next, assume that $$\vec{v_0}$$ is a linear combination of elements of $$S\text{.}$$ Can you find the appropriate $$\{c_0,c_1,\ldots, c_n\}$$ to make the equation $$\vec{z} = c_0\vec{v_0} + \cdots + c_n\vec{v_n}$$ true?

• In either of your proofs above, does the case when $$\vec{v_0} = \vec{z}$$ change your thinking? Explain why or why not.

### Subsection2.2.6Sample Problem and Solution

Sample problem Example B.1.6.