2.3 Linear Independence

1. Determine if the set of vectors is linearly independent.

a. $\left[\begin{array}{l}1\\ 2\end{array}\right],\left[\begin{array}{c}3\\ -1\end{array}\right]$
These are linear independent because the vectors are not scalar multiples of one another.

b. $\left[\begin{array}{c}1\\ -1\end{array}\right],\left[\begin{array}{l}2\\ 2\end{array}\right],\left[\begin{array}{l}3\\ 5\end{array}\right]$
This set cannot be linearly independent because it has 3 vectors in $R^2$ space.

c. $\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right],\left[\begin{array}{c}2\\ -1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ -1\\ 2\end{array}\right]$
This set is linearly dependent because the third vector can be produced with a linear combination of the first two.

2. Determine if the columns of the given matrix are linearly independent.

$$A=\left[\begin{array}{ccc}1& 2& -5\\ -3& 6& 0\\ 0& 1& 4\end{array}\right]$$

The columns are linearly independent. (Found by RREF)

3. Find the general solution and the solutions to the associated homogeneous system for:

$$\begin{array}{rl}& x_1+3x_2+3x_4-5x_5=5\\ & x_3-6x_4+10x_5=-7\\ & 3x_4-4x_5=6\end{array}$$ $$A=\left[\begin{array}{ccccc}1& 3& 0& 3& -5\\ 0& 0& 1& -6& 10\\ 0& 0& 0& 3& -4\end{array}\right]$$

Specific Solution:

$$X=\left[\begin{array}{c}-1-3s_2+s_5\\ s_2\\ 5-2s_5\\ 2+\frac{4}{3}{s}_5\\ s_5\end{array}\right]$$

Homogeneous Solution:

$$X=\left[\begin{array}{c}-3s_2+s_5\\ s_2\\ -2s_5\\ \frac{4}{3}{s}_5\\ s_5\end{array}\right]$$

General Solution:

$$X=\left[\begin{array}{c}-1-6s_2+2s_5\\ 2s_2\\ 5-3s_5\\ 2+\frac{8}{3}{s}_5\\ 2s_5\end{array}\right]$$

4. Show that the columns of $A$ are linearly independent.

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

Because there’s a zero in the top position of the center column, the only potential linear combination of the first two to produce the third is $-2(v_1)+c(v_2)$. Because the center element of the right column is a 2, this means $c=4$, which would make the resulting bottom value also a 2, and $2\ne 3$.

5. Prove that two nonzero vectors $\mathbf{u}$ and $\mathbf{v}$ are linearly dependent if and only if $\mathbf{u}=c\mathbf{v}$ for some $c\in \mathbb{R}$.

If $u$ and $v$ are linearly dependent, then by the definitions of linear dependence it must be true that $au+bv=0$, where $a$ and $b$ are not both zero.

$$au+bv=0$$ $$u=\frac{-b}{a}v$$ $$c=\frac{-b}{a}\in R$$