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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 $\dim (V_1+V_2)= \dim V_1+\dim V_2- \dim(V_1 \cap V_2)$

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  • $\begingroup$ 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. $\endgroup$ – Paul Sinclair Feb 5 '17 at 19:48
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The dimension of a vector space is the number of linearly independent vectors in a basis for that space.

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 $V_1 + V_2$.

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  • $\begingroup$ I was just writing up a much wordier response of the same thing. $\endgroup$ – Paul Sinclair Feb 5 '17 at 19:46
  • $\begingroup$ Hmm I already tried doing that but I'm can't understand why necesserily the span of (V1 ∪ V2) necessarily equals V. I can inutiatively see it, but not mathematically. $\endgroup$ – Play4u Feb 5 '17 at 22:15
  • $\begingroup$ Ohh and 1 more thing.. I have some troubles proving that the vektors a1,...a(n),b(n+1), ...b(k), c(n+1),...,c(l) are linearly independant. (The things in bracelets are just lower case indexes, sorry I'm new to the formatting in this forum) dim(V1∩V2)=n dimV1=k dimV2=l a1...a(n) is the basis of V1∩V2 b(n+1), ...b(k) basis V1 c(n+1),...,c(l) basis V2 $\endgroup$ – Play4u Feb 5 '17 at 22:18
  • $\begingroup$ $V_1 + V_2$ does not have to be all of $V$. It is only a subspace. $\endgroup$ – wckronholm Feb 5 '17 at 22:44
  • $\begingroup$ 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. $\endgroup$ – wckronholm Feb 5 '17 at 22:48

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