Find a base in linear algebra

Question

So I have this problem:

Let $V_1 = \{(x_1,x_2,x_3) \in \mathbb R^3 : x_1+x_2-x_3 = 0\} $ and $V_2 = \{(x_1,x_2,x_3) \in \mathbb R^3 : x_1-2x_3 = 0\} $. Find a base in $V_1∩V_2$ and show that $V_1+V_2 =\mathbb R^3$.

I have no idea how to solve this, I literally busted my brains in an attempt to solve it but I don't even now how to begin in solving it. I would really appreciate if you guys explained how to solve this problems and this kind of problems in general.

Answer 1

For $V_1∩V_2$: \begin{cases} x_1+x_2-x_3 = 0\\ x_1-2x_3 = 0 \end{cases} Solving it in the parameter $x_3=\alpha$, we have: $x_1=2\alpha$, $x_2=-\alpha$. Whence: $(x_1,x_2,x_3)=(2\alpha,-\alpha,\alpha)=\alpha(2,-1,1)$. Therefore the basis is builded by $(2,-1,1)$.

For the second question, we note that, since $$x_1=-x_2+x_3$$ a basis for $V_1$ is $$(x_1,x_2,x_3)=(-x_2+x_3,x_2,x_3)=x_2(-1,1,0)+x_3(1,0,1).$$

For $V_2$, since $x_1-2x_3=0$, we have: $$(x_1,x_2,x_3)=(2x_3,x_2,x_3)=x_2(0,1,0)+x_3(2,0,1)$$ The four vectors thus obtained can generated the entire space $\mathbb R^3$; only two carriers would not be sufficient, but four, three of which are linearly independent, as in this case, yes.

Answer 2

You might want to check out the Zassenhaus Algorithm. It is quite easy to calculate a basis for the given subspaces $V_1$ and $V_2$ and this algorithm will do the rest.

Answer 3

OK. I'm going to lead you through this. The first question is "Can you find a vector in $V_1 $? Hint: pick just about any values you like for $x_1$ and $x_2$, and see of you can figure out what $x_3$ has to be. Hint 2: If you make $x_1$ and $x_2$ simply numbers like 0 and 1, it's even easier. Call the resulting vector $\mathbf a$. Then do it again and find a vector $\mathbf b$ that's in $V_2$.

Second step: None of the vectors you just found lie in both $V_1$ and $V_2$, so they're pretty useless for solving the first part of the problem. If a vector $(x_1, x_2, x_3)$ lies in both $V_1$ and $V_2$, what two equations must it satisfy?

Third step: You could find a solution to that pair of equations by row-elimination or some similar approach, but you can also, generally, just set one of the variables to $1$ and then solve for the other two. (If setting $x_1 = 1$ doesn't work, then you try $x_2$...and then $x_3$.) So go ahead and do that, and find the intersection vector, which I'll call $\mathbf z$.

Now you've got a vector in the intersection. And you've got a second vector in each of $V_1$ and $V_2$ from part 1. They're very likely to be linearly independent -- it's easy to check: two vectors are dependent only if one of them is a multiple of the other. Assuming you hit it lucky, and they really are independent, you have a basis for $V_1$, namely $\mathbf a$ and
\mathbf z$, and a basis for $V_2$, namely $\mathbf b$ and $\mathbf z$.

That means that every vector in $V_1$ is a combination of the first two,a nd every vector in $V_2$ is a combination of the second two. So every SUM of a vector in $V_1$ and $V_2$ can be written as a combination of all three. If you can show that every vector in 3-space can be written as a combination of all three, then you'll have shown that $V_1 + V_2 = \mathbb R^3$.