Let V0 ⊂ R
3 be the subset of all real solutions (x,
y,
z)
of the system of equations
$3x − 2y + 4z = a$
$−x + y − 2z = b $,
where $a, b ∈ R.$
(a) Prove that V0 is a subspace of $R^3$
if and only if $a = b = 0.$
(b) Find a basis of V0.
If you are willing, could you please walk me through this? Thank you.
For a)
I try to prove it is closed under vector addition:
$\bar{u} =(x_1, x_2, x_3), \bar{v} = (y_1, y_2, y_3)$
$3x_1 -2x_2 +4x_3=a$
$-x_1+x_2-2x_3 = b$
and,
$3y_1 -2y_2 +4y_3=a$
$-y_1+y_2-2y_3 = b$
So, $\bar{u}+\bar{v} = (2x_1+2y_1, -x_2-y_2, 2x_3+2y_3)$
So I input this vector into one of the systems of equations above I think.
$3(2x_1+2y_1) -2(-x_2-y_2) +4(2x_3+2y_3)=a$
$-(2x_1+2y_1)+(-x_2-y_2)-2(2x_3+2y_3) = b \implies$
$6x_1+6y_1 +2x_2+2y_2 +8x_3+8y_3=a$
$-2x_1-2y_1-x_2-y_2-4x_3-4y_3 = b \implies$
$4x_1-4y_1+ x_2+ y_2 +4x_3+4y_3 = a+b$
Best Answer
For (a):
If $a = b = 0$ then adding $2$ times the second equation from the first gives $x = 0$. The solution space is then $\{(x,y,z)\in\mathbb{R}^3:x = 0\}$ (can you justify this statement?), which is a two-dimensional subspace of $\mathbb{R}^3$.
If we don't have $a=b=0$, this means either $a=b\neq 0$, or $a\neq b$. In the first case, the same calculation as above gives you $x = a+2b$. But for any $C\neq 0$, the set $$\{(x,y,z)\in\mathbb{R}^3:x=C\}$$ is not a subspace, since it does not contain the point $(0,0,0)$. So the solution space does not form a subspace.
On the other hand, if $a\neq b$, then the same calculation gives $x = a+2b$ again. In this case, either $x = 0$, in which case substituting this into the original set of equations gives something inconsistent (why?), or $x\neq 0$, in which case the solution space is not a subspace (why?).
For (b):
You now know that the solution space is the subspace $\{(x,y,z)\in\mathbb{R}^3:x = 0\}$. What are two vectors that span this space that are not multiples of each other?