# Intersection of three dimensional objects. Linear algebra

I'm just starting out with linear algebra and I got to admit is quite an interesting topic. I've reached the point of subspaces and I got to this interesting formula: Let $V_1$ and $V_2$ be finite subspaces of the vector field $V$. Then $$\dim(V_1+V_2)=\dim V_1+\dim V_2-\dim(V_1\cap V_2)$$ First thing is I can't seem to think of a proof of the euqality by myself. All I can think of is an example: Let $V=\Bbb R^2, V_1=[a, 0], V_2=[0, b]$. Basically $X$ and $Y$ axis. They intersection is the point $(0,0)$. $\dim (V_1+V_2)=2$. So we have $2=1+1-0$. Which is in fact correct. However if I try to apply this for $\Bbb R^3$, all examples I can think of don't satisfy the equation. And I believe that it is because I take the intersection of two three-dimensional objects to be either another 3d object or a 2d plane.

Is that correct at all?

If not, what is the intersection of two three dimensional spaces?

How can I prove that $\dim (V_1+V_2)= \dim V_1+\dim V_2- \dim(V_1 \cap V_2)$

• Proper subspaces of $\Bbb R^3$ are either planes thru the origin, lines thru the origin, or the origin itself. The intersection of two distinct planes is a line. The interesection of a plane and a line not in it is a point. In all cases, the formula works. – Paul Sinclair Feb 5 '17 at 19:48

If $V_1$ and $V_2$ are subspaces of some vector space $V$, then $V_1 \cap V_2$ is also a vector space. Start with a basis for $V_1 \cap V_2$, and then extend it to a basis for $V_1$, and separately extend it to a basis for $V_2$. The union of these two bases forms a basis for $V_1 + V_2$.
• $V_1 + V_2$ does not have to be all of $V$. It is only a subspace. – wckronholm Feb 5 '17 at 22:44
• If $\sum \alpha_i a_i + \sum \beta_j b_j + \sum \gamma_k c_k = 0$, then $\sum \gamma_k c_k \in V_1 \cap V_2$. But this means that $\sum \gamma_k c_k$ is a linear combination of the $a_i$'s, since they are a basis for $V_1$. But the $c$'s and $a$'s are linearly independent, so all $\gamma$'s are zero. Take it from there. – wckronholm Feb 5 '17 at 22:48