Vtyjkyuk

I am reading this text:

and I'm confused as to how/why they made y and z free variables? Also, why is this set linearly independent?

Is it linearly independent because:

$$c_1 beginbmatrix -2 & 1 & 0 endbmatrix + c_2 beginbmatrix -3 & 0 & 1 endbmatrix = beginbmatrix 0 & 0 & 0 endbmatrix$$

produces:
$$-2 * c_1 + (-3) * c_2 = 0$$
$$c_1 = 0$$
$$c_2 = 0$$

so the only solution is trivial. Is that why this is linearly independent?

Lastly, what does 2-dimensional mean? Is it because it has two vectors? Why is this not surprising? Do basis for $mathbbR^3$ have 3 vectors?

We are looking for the set of all points $(x,y,z)$ which satisfy the following equation

$$x+2y+3z=0$$

and we can easily show, by the definition, that the set is a subspace of $mathbbR^3$.

To find a basis of the subspace observe that we have 3 variable and only one equation therefore we can set any 2 variables free and determine the third as for example

$$y=s quad z=t implies x=-2s-3t$$

therefore the subspace can be written as

$$beginbmatrixx\y\zendbmatrix=sbeginbmatrix-2\1\0endbmatrix+tbeginbmatrix-3\0\1endbmatrix$$

therefore since the subspace is spanned by two linearly independent vectors, by definition, we have that the dimension is $2$, the given equation indeed represents a plane.

Therefore with reference to you questions

• yes the way you follows show that the two vectors are linearly
  independent, that is by definition $c_1v_1+c_2v_2=0 iff c_1=c_2=0$;
  in that case we don't need to check that directly since these two vectors have
  been obtained in such way to be linearly independent; indeed they are
  in the form $(a,1,0)$ and $(b,0,1)$ and therefore by linear
  combination we can't never reduce to $(0,0,0)$;


• $2$ dimensional means that the subspace is spanned by $2$ linearly independent vectors (revise the definition of basis and dimension) and since $mathbbR^3$ is spanned by the three linearly independent vectors $e_1=(1,0,0)$, $e_2=(0,1,0)$ and $e_3=(0,0,1)$ its dimension is $3$ of course; the same argument applies to $mathbbR^n$ which has dimension $n$.