Spanning Set of a Subspace in $\mathbb R^3$

Question

Find a spanning set for the subspace of $\mathbb R^3$ denoted by $U=${$(x_1,x_2,x_3) \in \mathbb R^3 | 3x_1 - x_2 -2x_3 = 0)$}.

My answer is span{$(1,1,1),(\frac{1}{3},3,-1)$}. Am I correct in assuming any two vectors which solve the equation would be a sufficient spanning set, and that the spanning set will always have a size of $2$? Is there any other way of solving this, e.g. using gaussian elimination / matrices?

Thanks!

Answer 1

• You are correct, by the rank-nullity theorem applied to the linear map $f:\Bbb R^3→\Bbb R,(x,y,z)↦3x−y−2z .$ Since $f$ is onto, the dimension of $U=\ker f$ is $3−1=2.$
  Well... almost correct: not "any two vectors" but "any two linearly independent vectors", of course.

• My favorite method is elimination. Here, e.g.: $U=\{(x,3x-2z,z)\mid x,z\in\Bbb R\}=\operatorname{span}((1,3,0),(0,-2,1))$ directly gives you a spanning set. With this method, you do not have to check that these two vectors are linearly independent.
  What I did here was to choose to eliminate $y$ by expressing it in terms of $x$ and $z$ from the equation (I could as well have chosen to eliminate $x$ or $z$ but I prefered $y$ because its coefficient is $-1$, so I will not get fractions): $3x-y-2z=0\iff y=3x-2z$, so the elements of $U$ are exactly the vectors of the form $$(x,3x-2z,z)=x(1,3,0)+z(0,-2,1),\quad x,z\in\Bbb R.$$

• Oh but wait! you are just asked for a spanning set, not for a basis. So, your second claim is not quite correct: a spanning set will always have a size $≥2,$ not necessarily $=2.$
  And you can even fool your teacher by answering: "$U$ is a spanning set of itself". I would greatly appreciate such an answer (taking it as a proof that the student has well understood the various concepts, and as an advice to write my questions more carefully), but I am not sure (s)he would.

Answer 2

To see that a spanning set of $U$ must consist of two vectors, you can apply the rank-nullity theorem to $f: \Bbb R^3\to\Bbb R$; $(x,y,z)\mapsto 3x-y-2z$, as suggested by @Anne Bauval in the comments. Note that $\ker f = U$.

No, any two vectors which solve the equation do not necessarily constitute a spanning set; it is important to ensure that these two vectors are linearly independent.

One way to do this is as follows. Consider the equation $3x-y-2z = 0$. First, fix $x = 0$ and obtain a solution, i.e., solve $y + 2z = 0$. Observe that $y = 2$ and $z = -1$ work. So, $v_1 = (0,2,-1) \in U$. Next, fix $y = 0$ and obtain a solution, i.e., solve $3x - 2z = 0$. Observe that $x=2$ and $z =3$ work. So, $v_2 = (2,0,3) \in U$.

Then, $$S = \{v_1,v_2\} = \{(0,2,-1), (2,0,3)\}$$ is a spanning set of $U$ over $\Bbb R$.