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4.4 Linear Independence: Class Notes

Definition

For example, $\{\widehat{ı},\widehat{ȷ}\}$ is a linearly independent set in $R^2$. The set $\{\widehat{ı},\widehat{ȷ},(3,2)\}$ is not, since $-3\widehat{ı}-2\widehat{ȷ}+1(3,2)=\overrightarrow{0}$.

Theorem 4.4.1

Proof

($\Rightarrow$) Let $S=\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,\dots ,{\overrightarrow{v}}_r\}$ be a linearly independent set. Suppose there exist $k_1,k_2,\dots ,k_r$, not all zero, such that

$$k_1{\overrightarrow{v}}_1+k_2{\overrightarrow{v}}_2+\cdots +k_i{\overrightarrow{v}}_i+\cdots +k_r{\overrightarrow{v}}_r=\overrightarrow{0}.$$

If $k_i\ne 0$, then $k_i{\overrightarrow{v}}_i=-k_1{\overrightarrow{v}}_1-k_2{\overrightarrow{v}}_2-\cdots -k_r{\overrightarrow{v}}_r$ (all terms except the $i$th on the right), so

$${\displaystyle \Rightarrow {\overrightarrow{v}}_i=-\frac{k_1}{k_i}{\overrightarrow{v}}_1-\frac{k_2}{k_i}{\overrightarrow{v}}_2-\cdots -\frac{k_r}{k_i}{\overrightarrow{v}}_r,}$$

a linear combination of the vectors ${\overrightarrow{v}}_j,j\ne i$, contradicting linear independence.

($\Leftarrow$) Suppose $\{{\overrightarrow{v}}_i\}$ is not a linearly independent set. Then one vector, say ${\overrightarrow{v}}_1$, is a linear combination of the others:

$${\overrightarrow{v}}_1=c_2{\overrightarrow{v}}_2+\cdots +c_r{\overrightarrow{v}}_r\Rightarrow {\overrightarrow{v}}_1-c_2{\overrightarrow{v}}_2-\cdots -c_r{\overrightarrow{v}}_r=\overrightarrow{0},$$

contradicting that all the coefficients are zero (the coefficient of ${\overrightarrow{v}}_1$ is 1).

The standard basis vectors for $R^n$ are a linearly independent set.

#3

In each part, determine whether the vectors are linearly independent or are linearly dependent in $R^4$.

a. $(3,8,7,-3)$, $(1,5,3,-1)$, $(2,-1,2,6)$, $(4,2,6,4)$

b. $(3,0,-3,6)$, $(0,2,3,1)$, $(0,-2,-2,0)$, $(-2,1,2,1)$

Solution

a. The set is independent if the only way $c_1{\overrightarrow{v}}_1+c_2{\overrightarrow{v}}_2+c_3{\overrightarrow{v}}_3+c_4{\overrightarrow{v}}_4=\overrightarrow{0}$ is if $c_i=0$ for all $i$. This is the homogeneous system with augmented matrix

$$\left[\begin{array}{ccccc}\hfill {\displaystyle {\overrightarrow{v}}_1}\hfill & \hfill {\displaystyle {\overrightarrow{v}}_2}\hfill & \hfill {\displaystyle {\overrightarrow{v}}_3}\hfill & \hfill {\displaystyle {\overrightarrow{v}}_4}\hfill & \hfill {\displaystyle \begin{array}{c}0\\ 0\\ 0\\ 0\end{array}}\hfill \end{array}\right]$$

The system has only the trivial solution if and only if

• $A=[{\overrightarrow{v}}_1{\overrightarrow{v}}_2{\overrightarrow{v}}_3{\overrightarrow{v}}_4]$ is invertible
• $det(A)\ne 0$
• $\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{f}(A)$ is $I$

Compute the determinant, with the vectors as columns. Use $R_1\to R_1+R_4$, then expand along the new first row with the signs + − + −:

$$\left|\begin{array}{cccc}3& 1& 2& 4\\ 8& 5& -1& 2\\ 7& 3& 2& 6\\ -3& -1& 6& 4\end{array}\right|=\left|\begin{array}{cccc}0& 0& 8& 8\\ 8& 5& -1& 2\\ 7& 3& 2& 6\\ -3& -1& 6& 4\end{array}\right|=0$$

It turns out to be 0. The system has nontrivial solutions, and so the set is not linearly independent. Row reduction yields

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

The leading 1s are in columns 1, 2, and 3, so $c_4$ is free. A general solution is

$$c_1=-t,c_2=t,c_3=-t,c_4=t.$$

With $t=1$, this means

$$-\left[\begin{array}{c}3\\ 8\\ 7\\ -3\end{array}\right]+\left[\begin{array}{c}1\\ 5\\ 3\\ -1\end{array}\right]-\left[\begin{array}{c}2\\ -1\\ 2\\ 6\end{array}\right]+\left[\begin{array}{c}4\\ 2\\ 6\\ 4\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right]$$

b. $det(A)\ne 0$, so the set is independent. (Verify when you make time to do so: here $det(A)=36$.)

#5

In each part, determine whether the matrices are linearly independent or dependent.

a. $\left[\begin{array}{cc}1& 0\\ 1& 2\end{array}\right],\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right],\left[\begin{array}{cc}0& 1\\ 2& 1\end{array}\right]$ in $M_{22}$

b. $\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\end{array}\right],\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\end{array}\right],\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\end{array}\right]$ in $M_{23}$

Solution

a.

$$c_1\left[\begin{array}{cc}1& 0\\ 1& 2\end{array}\right]+c_2\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]+c_3\left[\begin{array}{cc}0& 1\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$$

Matching the four entries gives the system

$$\Rightarrow \left[\begin{array}{ccc}1& 1& 0\\ 0& 2& 1\\ 1& 2& 2\\ 2& 1& 1\end{array}\right]\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\\ 0\end{array}\right]$$

Use $R_3\to R_3-R_1$ ($(1,2,2)-(1,1,0)=(0,1,2)$) and $R_4\to R_4-2R_1$ ($(2,1,1)-(2,2,0)=(0,-1,1)$):

$$\left[\begin{array}{ccc}1& 1& 0\\ 0& 2& 1\\ 0& 1& 2\\ 0& -1& 1\end{array}\right]$$

Then ${\displaystyle R_3\to R_3-\frac{1}{2}{R}_2}$ ($(0,1,2)-(0,1,\frac{1}{2})=(0,0,\frac{3}{2})$) and ${\displaystyle R_4\to R_4+\frac{1}{2}{R}_2}$ ($(0,-1,1)+(0,1,\frac{1}{2})=(0,0,\frac{3}{2})$), and then $R_4\to R_4-R_3$:

$${\displaystyle \left[\begin{array}{ccc}1& 1& 0\\ 0& 2& 1\\ 0& 0& \frac{3}{2}\\ 0& 0& 0\end{array}\right]}$$

Since each column has a pivot, the set is independent.

b. We can see the set is independent by inspection: each matrix has its single 1 in a position where the others have 0.

#9

a. Show that the three vectors ${𝐯}_1=(0,3,1,-1)$, ${𝐯}_2=(6,0,5,1)$, ${𝐯}_3=(4,-7,1,3)$ form a linearly dependent set in $R^4$.

b. Express each vector in part (a) as a linear combination of the other two.

Solution

Put ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3$ in the columns and row reduce:

$${\displaystyle \left[\begin{array}{ccc}0& 6& 4\\ 3& 0& -7\\ 1& 5& 1\\ -1& 1& 3\end{array}\right]⟶\left[\begin{array}{ccc}1& 5& 1\\ 0& 3& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]⟶\left[\begin{array}{ccc}1& 0& -\frac{7}{3}\\ 0& 1& \frac{2}{3}\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]}$$

So $c_1{\overrightarrow{v}}_1+c_2{\overrightarrow{v}}_2+c_3{\overrightarrow{v}}_3=\overrightarrow{0}$ if

$${\displaystyle c_1=\frac{7}{3}{c}_3,c_2=-\frac{2}{3}{c}_3.}$$

If $c_3=3$, then $c_1=7$ and $c_2=-2$, a nontrivial solution. Then

so the set is linearly dependent. Isolating each vector in turn finishes the problem.

#14

In each part, let $T_A:R^3\to R^3$ be multiplication by $A$ and let ${𝐮}_1=(1,0,0)$, ${𝐮}_2=(2,-1,1)$, and ${𝐮}_3=(0,1,1)$. Determine whether the set $\{T_A({𝐮}_1),T_A({𝐮}_2),T_A({𝐮}_3)\}$ is linearly independent in $R^3$.

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

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

Solution

a.

$$T_A({\overrightarrow{u}}_1)=\left[\begin{array}{c}1\\ 1\\ 2\end{array}\right],T_A({\overrightarrow{u}}_2)=\left[\begin{array}{ccc}1& 1& 2\\ 1& 0& -3\\ 2& 2& 0\end{array}\right]\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right]=\left[\begin{array}{c}3\\ -1\\ 2\end{array}\right]$$

$$T_A({\overrightarrow{u}}_3)=\left[\begin{array}{ccc}1& 1& 2\\ 1& 0& -3\\ 2& 2& 0\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}3\\ -3\\ 2\end{array}\right]$$

Take the determinant of the matrix with these columns, by the diagonal method: copy the first two columns to the right, add the products down the three diagonals going down to the right, and subtract the products down the three diagonals going up to the right.

$$|\begin{array}{ccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 3}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill \\ \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 2}\hfill \end{array}$$

$$\left|\begin{array}{ccc}1& 3& 3\\ 1& -1& -3\\ 2& 2& 2\end{array}\right|=(-2)+(-18)+6+6+6-6=-8\ne 0$$

b.

$$T_A({\overrightarrow{u}}_1)=\left[\begin{array}{c}1\\ 1\\ 2\end{array}\right],T_A({\overrightarrow{u}}_2)=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& -3\\ 2& 2& 0\end{array}\right]\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 2\end{array}\right]$$

$$T_A({\overrightarrow{u}}_3)=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& -3\\ 2& 2& 0\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 2\end{array}\right]$$

Since $T_A({\overrightarrow{u}}_2)$ and $T_A({\overrightarrow{u}}_3)$ match, the set is not linearly independent.

Note: for (a), $det(A)\ne 0$. For (b), $det(A)=0$. We'll revisit this in Ch. 8.

Theorem 4.4.2

Theorem 4.4.3

This is because a matrix equation for an associated homogeneous system has more columns than rows, and so contains free variables.

Definition

Theorem 4.4.4

#19

Use the Wronskian to show that the following sets of vectors are linearly independent.

a. $1,x,e^x$

b. $1,x,x^2$

Solution

In each part, expand along the first column, which has only one nonzero entry.

a.

$$W(1,x,e^x)=\left|\begin{array}{ccc}1& x& e^x\\ 0& 1& e^x\\ 0& 0& e^x\end{array}\right|=1\left|\begin{array}{cc}1& e^x\\ 0& e^x\end{array}\right|=e^x\ne 0$$

b.

$$W(1,x,x^2)=\left|\begin{array}{ccc}1& x& x^2\\ 0& 1& 2x\\ 0& 0& 2\end{array}\right|=1\left|\begin{array}{cc}1& 2x\\ 0& 2\end{array}\right|=2\ne 0$$