How to find a basis for a given subspace?

Question

Let $S=\{(x_1,x_2,x_3)\mid x_1+x_2+x_3=0\}$ be a subspace of $\mathbb R^3$. How to find a basis for S?

The dimension of $S$ is 3. I thought that $x_1+x_2+x_3=0$ is a linear combination of $(x_1,x_2,x_3)$. Then it spans $S$. But they are not linearly independent. Then how can I find another basis for $S$?. Then I thought the basis as $(x_1,x_2,0),(0,x_2,x_3),(x_1,0,x_3)$. Am I correct?

Then I need to know that how can I extend this basis to a basis $B$ in $\mathbb R^3$?

Answer 1

Its dimension is $2$. it is a plane.

to find the basis, write $S $ as

$$S=\{(x,y,z)\in \Bbb R^3 \;\;: z=-x-y\} $$

the elements of $S $ are of the form $$(x,y,-x-y)=$$ $$x (1,0,-1)+y (0,1,-1) =$$

$$x\vec {e_1}+y\vec {e_2} .$$

now prove that these two vectors are not dependent.

to complete them to a base of $\Bbb R^3$, take the normal vector to the plane $S : 1.x+1.y+1.z=0$, which is $$\vec {e_3}=(1,1,1) .$$

or

Put $$\vec {e_3}=\vec {e_1} \land \vec {e_2} $$

Answer 2

The linear map $f\colon \mathbb{R}^3 \to \mathbb{R}$ given by $$f(x_1,x_2,x_3) = x_1+x_2+x_3$$ is nonzero, so $\dim(\text{im}(f))=1$.

It follows that $\dim(\text{ker}(f))=3-1=2$, so any two linearly independent elements of $\text{ker}(f)$ will be a basis for $\text{ker}(f)$.

Thus, the two vectors $$(1,-1,0),\;\;(0,1,-1)$$ qualify as a basis.

If you adjoin any third vector not in the kernel, that will yield a basis for $R^3$.