Let $S = \{(x,y,z) \in \mathbb{R^3} | x+y+z = 0\}$ be a subspace of vector space. Find a basis and dimension for the subspace.
I'm not sure if I am approaching this correctly.
I started off by rewriting the plane as:
$x = -y-z$
$y = -x-z$
$z = -x-y$
Which gives me the vector $(-y-z, -x-z, -x-y)$ which can be broken down into $x(0,-1,-1) + y(-1,0,-1) + z(-1,-1,0)$.
This means that $S = sp\{(0,-1,-1), (-1,0,-1), (-1,-1,0)\}$
You can express this as a matrix and row reduce to see that you get a rank of 3. What can I conclude from this? I get pivots along the diagonal, and it is a 3×3 matrix, so it is safe to say thsoe vectors are linearly independent, and so they do form a basis. Furthermore, since we have three basis vectors, then the dimension of the subspace is 3.
But I am not sure if this approach is correct.
Best Answer
$$x+y+z=0$$
Let $y=s$, let $z=t$, then $x = -s-t$
Hence we can write $\begin{bmatrix} x\\ y \\ z \end{bmatrix} = s\begin{bmatrix} -1\\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} -1\\ 0 \\ 1 \end{bmatrix} $
Try to see if you can read off the basis from here.
Remark about your approach: It doesn't seems that $(0,-1,-1)$ is a solution to your equation.