← Back to Math 010

4.3 Spanning Sets: Class Notes

Definition: (generalizes linear combination from Section 3.1)

#6

In each part, express the vector as a linear combination of ${𝐩}_1=2+x+4x^2$, ${𝐩}_2=1-x+3x^2$, and ${𝐩}_3=3+2x+5x^2$.

a. $-9-7x-15x^2$

b. $6+11x+6x^2$

c. $0$

d. $7+8x+9x^2$

Solution

We note that $c_1{\overrightarrow{p}}_1+c_2{\overrightarrow{p}}_2+c_3{\overrightarrow{p}}_3$ becomes

$$2c_1+c_{1}x+4c_1{x}^2+c_2-c_{2}x+3c_2{x}^2+3c_3+2c_{3}x+5c_3{x}^2$$

Then $c_1{\overrightarrow{p}}_1+c_2{\overrightarrow{p}}_2+c_3{\overrightarrow{p}}_3=\overrightarrow{p}$, where $\overrightarrow{p}=a_0+a_{1}x+a_2{x}^2$, yields (matching the constant terms, the $x$-terms, and the $x^2$-terms)

$$\begin{array}{cc}\hfill {\displaystyle 2c_1+c_2+3c_3}& {\displaystyle =a_0}\hfill \\ \hfill {\displaystyle c_1-c_2+2c_3}& {\displaystyle =a_1}\hfill \\ \hfill {\displaystyle 4c_1+3c_2+5c_3}& {\displaystyle =a_2}\hfill \end{array}$$

In matrix form, each column holds the coefficients of one polynomial:

$$\left[\begin{array}{ccc}2& 1& 3\\ 1& -1& 2\\ 4& 3& 5\end{array}\right]\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\end{array}\right]=\left[\begin{array}{c}{a}_0\\ a_1\\ a_2\end{array}\right]$$

We can solve (a)–(d) using the augmented matrix

$$\left[\begin{array}{cccc}\hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle a_0}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle a_1}\hfill \\ \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 5}\hfill & \hfill {\displaystyle a_2}\hfill \end{array}\right]$$

and substituting for each $a_i$. Row reduction gives

$${\displaystyle \left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -\frac{11}{2}{a}_0+2a_1+\frac{5}{2}{a}_2}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle \frac{3}{2}{a}_0-a_1-\frac{1}{2}{a}_2}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle \frac{7}{2}{a}_0-a_1-\frac{3}{2}{a}_2}\hfill \end{array}\right]}$$

a. If $a_0=-9$, $a_1=-7$, $a_2=-15$, then

$${\displaystyle c_1=-\frac{11}{2}(-9)+2(-7)+\frac{5}{2}(-15)=-2.}$$

Likewise, $c_2=1$ and $c_3=-2$.

Check:

$$\begin{array}{cc}\hfill {\displaystyle }& {\displaystyle -4-2x-8x^2+1-x+3x^2-6-4x-10x^2}\hfill \\ \hfill {\displaystyle }& {\displaystyle =-9-7x-15x^2}\hfill \end{array}$$

b.

c.

d.

Note: we can always find a solution, since there are no restrictions on $a_0$, $a_1$, $a_2$.

Definition

#9

Determine whether the following polynomials span $P_2$.

${𝐩}_1=1-x+2x^2$, ${𝐩}_2=3+x$, ${𝐩}_3=5-x+4x^2$, ${𝐩}_4=-2-2x+2x^2$

Solution

That is, can we build every polynomial of degree $\le 2$ using ${\overrightarrow{p}}_1,{\overrightarrow{p}}_2,{\overrightarrow{p}}_3,{\overrightarrow{p}}_4$? That is, does

$$a{\overrightarrow{p}}_1+b{\overrightarrow{p}}_2+c{\overrightarrow{p}}_3+d{\overrightarrow{p}}_4=\overrightarrow{p}$$

always have a solution, where $\overrightarrow{p}=b_1+b_{2}x+b_3{x}^2$? Writing each polynomial as its column of coefficients,

$${\overrightarrow{p}}_1=\left[\begin{array}{c}1\\ -1\\ 2\end{array}\right],{\overrightarrow{p}}_2=\left[\begin{array}{c}3\\ 1\\ 0\end{array}\right],{\overrightarrow{p}}_3=\left[\begin{array}{c}5\\ -1\\ 4\end{array}\right],{\overrightarrow{p}}_4=\left[\begin{array}{c}-2\\ -2\\ 2\end{array}\right]$$

Row reduce:

$${\displaystyle \left[\begin{array}{ccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 5}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle b_1}\hfill \\ \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle b_2}\hfill \\ \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle b_3}\hfill \end{array}\right]⟶\left[\begin{array}{ccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle \frac{1}{4}{b}_1-\frac{3}{4}{b}_2}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle \frac{1}{4}{b}_1+\frac{1}{4}{b}_2}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle b_1-3b_2-2b_3}\hfill \end{array}\right]}$$

This has a solution if and only if $b_1-3b_2-2b_3=0$. So if $b_1-3b_2-2b_3\ne 0$, there is no such linear combination.

#12

Let $T_A:R^2\to R^2$ be multiplication by $A$. Determine whether the vector $𝐮=(1,2)$ is in the span of $\{T_A({𝐞}_1),T_A({𝐞}_2)\}$.

a. $A=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$

b. $A=\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$

Solution

a.

$$T_A({\overrightarrow{e}}_1)=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right],T_A({\overrightarrow{e}}_2)=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}2\\ 1\end{array}\right]$$

Tip: Note that these are the columns of $A$.

$$\overrightarrow{u}=-3T_A({\overrightarrow{e}}_1)+2T_A({\overrightarrow{e}}_2)$$

b.

$$T_A({\overrightarrow{e}}_1)=\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 1\end{array}\right],T_A({\overrightarrow{e}}_2)=\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 1\end{array}\right]$$

We cannot express $\left[\begin{array}{c}1\\ 2\end{array}\right]$ as a linear combination of $\left[\begin{array}{c}1\\ 1\end{array}\right]$ and $\left[\begin{array}{c}1\\ 1\end{array}\right]$.

#16

Let $W$ be the solution space to the system $A𝐱=\mathrm{𝟎}$. Determine whether the set $\{𝐮,𝐯\}$ spans $W$.

$A=\left[\begin{array}{cccc}0& 1& -1& 1\\ 0& 2& -2& 2\\ 0& 3& -3& 3\end{array}\right]$

a. $𝐮=(1,1,1,0),𝐯=(0,-1,0,1)$

b. $𝐮=(0,1,1,0),𝐯=(1,0,1,1)$

Solution

Can an arbitrary vector in $W$ be expressed as a linear combination of $\overrightarrow{u}$ and $\overrightarrow{v}$? First find $W$:

$$\left[\begin{array}{ccccc}\hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]⟶\left[\begin{array}{ccccc}\hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]$$

$x_1$ is free: $x_1=t$. The first row says $x_2-x_3+x_4=0\Rightarrow x_2=x_3-x_4$. Let $x_3=r$ and $x_4=s$. Then

$$\overrightarrow{x}=\left[\begin{array}{c}t\\ 0\\ 0\\ 0\end{array}\right]+\left[\begin{array}{c}0\\ r-s\\ r\\ s\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]t+\left[\begin{array}{c}0\\ 1\\ 1\\ 0\end{array}\right]r+\left[\begin{array}{c}0\\ -1\\ 0\\ 1\end{array}\right]s$$

Thus the solution space is spanned by

$$\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 1\\ 0\end{array}\right],\text{and}\left[\begin{array}{c}0\\ -1\\ 0\\ 1\end{array}\right].$$

In both (a) and (b), $\overrightarrow{u}$ and $\overrightarrow{v}$ do not span this set, so they do not span the solution space.

Theorem 4.3.1

Proof of (a)

Let $\overrightarrow{u},\overrightarrow{v}\in \mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}(S)$. Then

$$\overrightarrow{u}=\sum _{i=1}^r{c}_i{\overrightarrow{w}}_i,\overrightarrow{v}=\sum _{i=1}^r{k}_i{\overrightarrow{w}}_i.$$

$$\overrightarrow{u}+\overrightarrow{v}=\sum _{i=1}^r(c_i{\overrightarrow{w}}_i+k_i{\overrightarrow{w}}_i)=\sum _{i=1}^r(c_i+k_i){\overrightarrow{w}}_i,$$

which has the form $\sum _{i=1}^r{d}_i{\overrightarrow{w}}_i$, a linear combination of the ${\overrightarrow{w}}_i$. So $\overrightarrow{u}+\overrightarrow{v}\in \mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}(S)$.

Let $k\in ℝ$. Then

$$k\overrightarrow{u}=k\sum _{i=1}^r{c}_i{\overrightarrow{w}}_i=\sum _{i=1}^r(k{c}_i){\overrightarrow{w}}_i,$$

so $k\overrightarrow{u}\in \mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}(S)$.

Theorem 4.3.2

#20

Let ${𝐯}_1=(1,6,4)$, ${𝐯}_2=(2,4,-1)$, ${𝐯}_3=(-1,2,5)$, and ${𝐰}_1=(1,-2,-5)$, ${𝐰}_2=(0,8,9)$. Use Theorem 4.3.2 to show that $\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\{{𝐯}_1,{𝐯}_2,{𝐯}_3\}=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}\{{𝐰}_1,{𝐰}_2\}$.

Solution

Can I use the ${\overrightarrow{v}}_i$ to make each ${\overrightarrow{w}}_i$? Put ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3$ in the coefficient columns and ${\overrightarrow{w}}_1,{\overrightarrow{w}}_2$ in the augmented columns, then row reduce:

$$\left[\begin{array}{ccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 6}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 8}\hfill \\ \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 5}\hfill & \hfill {\displaystyle -5}\hfill & \hfill {\displaystyle 9}\hfill \end{array}\right]⟶\left[\begin{array}{ccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 2}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]$$

Both systems are consistent. Reading the first augmented column (with $c_3=0$) and the second augmented column:

$${\overrightarrow{w}}_1=-{\overrightarrow{v}}_1+{\overrightarrow{v}}_2$$

$${\overrightarrow{w}}_2=2{\overrightarrow{v}}_1-{\overrightarrow{v}}_2$$

Now can I use the ${\overrightarrow{w}}_i$ to make each ${\overrightarrow{v}}_i$?

$$\left[\begin{array}{ccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -1}\hfill \\ \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 8}\hfill & \hfill {\displaystyle 6}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 2}\hfill \\ \hfill {\displaystyle -5}\hfill & \hfill {\displaystyle 9}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 5}\hfill \end{array}\right]⟶\left[\begin{array}{ccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]$$

Yes: all three systems are consistent (${\overrightarrow{v}}_1={\overrightarrow{w}}_1+{\overrightarrow{w}}_2$, ${\overrightarrow{v}}_2=2{\overrightarrow{w}}_1+{\overrightarrow{w}}_2$, ${\overrightarrow{v}}_3=-{\overrightarrow{w}}_1$). By Theorem 4.3.2,