find the basis of this subspace of $\Bbb{R}^3$

Question

How would I go about finding the basis of the subspace of $\Bbb{R}^3$ consisting of all $(x,y,z)$ such that $x+y+z= 0$?

I understand that even though its three dimensional, the span could be of less dimensions. If this is true, then is there a way to calculate the basis? my stratergy up until now has just been to look then what is and try to work it out. For this one, I thought the basis would be three-dimensional.

Answer 1

Hint:

Rewrite the equation as, say, $z=-x-y$, and find a linear map from $\mathbf R^2$ to $\mathbf R^3$ with image the subspace.

Answer 2

You can write $x = -y -z$. So you need values of $y$ and $z$ to determine values of x. Let $y=t,z=r$ so $x=-t-r$. So your subspace has form $(-t-r,t,r)$. So you have 2 free variables which are t and r. So dimension=2. Now to find basis you need 2 linearly independent vectors( because dimension is 2). You can find those by setting $t=0$ and $r=1$ and then $t=1$ and $r=0$.

Answer 3

by using the Hint given above, and that if $g$ is a linear map from $ E$ to $F$ ($E, F$ vector spaces over the same field), then the image of any generator system $S$ of $ E$ by $g$ is a generator system $g(S)$ of $g(F)$.

So we take $S=\{(1,0);(0,1)\}$ canonical basis of $\Bbb{R}$, and $g:(x,y)\mapsto (x,y,-x-z)$ then $g(S)=\{(1,0,-1);(0,1,-1)\}$ is a generator system of the desired subspace , also a basis of the kernel of the linear forms proposed.