# Examples of dim(V+W)+dim(V∩W)=dim(V)+dim(W).

I saw a few posts that ask to prove this identity, but I'm still trying to understand what it means intuitively. Can anyone show a few examples of it?

• Take $W=0$ and $V=\Bbb R^n$, for example. Find an example yourself with $V$ and $W$ subspaces of $\Bbb R^2$. More intuition is given here. Jul 4 '19 at 18:40

Think about the $$x-y$$ and $$y-z$$ coordinate planes in space. Each has dimension $$2$$. Their intersection is the $$1$$ dimensional $$x$$ axis. Their sum is the full $$3$$ dimensional space.

• thanks! any more examples ? :-) Jul 4 '19 at 18:49
• Ok, you could probably also use 2 lines intersecting (e.g. x-axis, and y-axis) each has dimension of 1, and their intersection has a dimension of 0, so dim(U+V) = dim(U) + dim(V) = 1 + 1 = 2. Am I close? Jul 4 '19 at 18:54
• @DavidRefaeli Yes, that's an example too. I think mine is the simplest one where none of the four dimensions is $0$ and neither $V$ nor $W$ is a subspace of the other, and essentially the only such that fits in three space. You can use coordinate subspaces of different sizes in more dimensions to create as much overlap in the intersection as you like. Jul 4 '19 at 19:02
• yeah, and it can be expanded to 4 dimensions, say with U = span of (1,0,0,0), (0,1,0,0), (0,0,1,0), V = span of (0,1,0,0), (0,0,1,0), (0,0,0,1): dim(U) = dim(V) = 3. dim(U+V) = dim(U) + dim(V) - dim(U∩V) = 3 + 3 - 2 = 4. Jul 4 '19 at 19:06
• You've got it. When you learn about bases of vector spaces and subspaces you will see that those are actually all the examples. Jul 4 '19 at 19:07

Let $$V$$ be a finite dimensional $$k$$-vector space and let $$U_1,U_2 \subset V$$ be subspaces. Then one can consider the vector space $$U = U_1 + U_2$$. Now a natural question is which dimension $$U$$ has and how that dimension depends on the dimensions of $$U_1$$ and $$U_2$$. The observation is, that in general $$\text{dim}(U) \neq \text{dim}(U_1) + \text{dim}(U_2)$$. Consider for example:

The dimension of a subspace is always bounded by the dimension of the parent space, such that we have $$\text{dim}(U) \leq 3$$ in the case of $$V = \mathbb{R}^3$$. Therefore the dimension of the two planes cannot add up. One has to take the intersection into account and substract the dimension of the intersection.

In the case that the intersection is trivial, we call $$U$$ the direct sum of $$U_1$$ and $$U_2$$, which basically means that $$U$$ has a decomposition into $$U_1$$ and $$U_2$$. An equivalent way of stating that is that each vector $$u \in U$$ has a unique representation as $$u = u_1 + u_2$$, where $$u_i \in U_i$$. In that way the notion of a direct sum can be generalized to arbitrary finite sums of vector spaces $$U_1 + \dots + U_n$$.