Sum and intersection of 2 vectors spaces?

Question

Okey, I have the following vector spaces :

F1 = {(x, y, z) $\in$ $R^3$, x+y=0}

F2 = {(x, y, z) $\in$ $R^3$, x+z=y}

I know that the F1 contains all the vectors of form: (a, -a, b) where a and b $\in$ $R^3$ and also F2 contains all the vectors of form: (t, u, -t+v) where t, v $\in$ $R^3$

I am just getting started with vectors spaces and some concepts I can't understand like summing 2 and the intersection of them, so that I am stuck right now with this one the F1 + F2 and F1 $\cap$ F2?

Any help will be greatly appreciated.

Answer 1

$F_1 \cap F_2$ are vectors that satisfy both conditions (and you can start with the original conditions), so a vector lies in $F_1 \cap F_2$ iff $x=-y$ and $x+z = y$. The first one is equivalent to the condition $x=-y$ and you can use this fact in the second equation, so $-y+z=y$, which is equivalent to $z=2y$. Such vectors are then in the form $(a,-b,2b)$ for $a,b\in\mathbb{R}$. (Note that this reasoning works for general sets as well, you don't need vector spaces.)

The sum would be easier could you handle with bases and so on, but still we make it:

Let us claim, that the sum is the whole space; let's see that we can write any vector $v=(x,y,z)$ as a sum of vectors from $F_1$ and $F_2$ by step-by-step approximations. Take the vector $v_1=(x,-x, z)$, which lies in $F_1$, then $v-v_1 = (0,y-x,0)$, so it remains to decompose this 'remainder'. From $F_2$ the best one can take is $v_2=(0,y-x,y-x)$, so now we have to approximate $v-v_1-v_2 = (0,0,x-y) =: v_3$ which, however, lies in $F_1$. So $v=v_1+v_2+v_3$, $v_1+v_3\in F_1$, $v_2\in F_2$ and we're done.