Question

$W = {(x, y, z) ∈R3|3x −2y + 4z = 0}.$

Answer 1

$$3x−2y+4z=0$$

Any vector $\vec w=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ in $W$ satisfies $3x-2y+4z=0$ or equivalently $x=\dfrac{2}{3}y-\dfrac{4}{3}z.$

Thus we have

$$\vec w=\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} \dfrac{2}{3}y-\dfrac{4}{3}z \\ y \\ z \end{bmatrix}=y\begin{bmatrix} 2/3 \\ 1 \\ 0 \end{bmatrix}+z\begin{bmatrix} -4/3 \\ 0 \\ 1 \end{bmatrix}$$

Let

$$\vec u_1=\begin{bmatrix} 2/3 \\ 1 \\ 0 \end{bmatrix}, \vec u_2=\begin{bmatrix} -4/3 \\ 0\\ 1 \end{bmatrix}$$

The above computation shows that any vector $\vec w$  in $W$ can be written as a linear combination of the vectors $\vec u_1, \vec u_2.$

Hence the set $\{\vec u_1, \vec u_2\}$ is a spanning set for the subspace $W.$

We claim that $\{\vec u_1, \vec u_2\}$ is a linearly independent set.

Consider

$$a_1\vec u_1+a_2\vec u_2=\vec 0$$

The equation can be written as

$$\begin{bmatrix} (2/3)a_1-(4/3)a_2 \\ a_1 \\ a_2 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

Comparing entries, we obtain $a_1=a_2=0.$

Thus the equation $a_1\vec u_1+a_2\vec u_2=\vec 0$ has only the zero solution and hence the vectors $\vec u_1, \vec u_2$

are linearly independent.

We found a basis

$$\bigg\{\vec u_1=\begin{bmatrix} 2/3 \\ 1 \\ 0 \end{bmatrix}, \vec u_2=\begin{bmatrix} -4/3 \\ 0\\ 1 \end{bmatrix}\bigg\}.$$