# Dimension of the sum of two vector subspaces

$\dim(U_1+U_2) = \dim U_1 +\dim U_2 – \dim(U_1\cap U_2).$

I want to make sure that my intuition is correct. Suppose we have two planes $U_1,U_2$ though the origin in $\mathbb{R^3}$. Since the planes meet at the origin, they also intersect, which in this case is a one-dimensional line in $\mathbb{R^3}$. To obtain the dimension of $U_1$ and $U_2$, we add the dimensions of the planes (4), and the subtract the dimensions of the line (1), which results in (3).

Can we generalize this notion to $\mathbb{F^{n}}$?

Suppose we have an additional case where $U_1$ and $U_2$ are planes in $\mathbb{R^3}$, but $U_1 \subseteq U_2$. In this instance, $dim(U_1 + U_2) < 3$, because the first two-dimensional plane is contained in the second and as a result, the dimensions of the subspaces when summed cannot exceed two. Since both subspaces $U_1,U_2$ are two dimensional and $U_1 \subseteq U_2$, then their intersection is also two-dimensional, concluding $dim(U_1+U_2)=2+2-2 = 2$.

Is this proper intuition?

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In the latter case, they are actually the same plane, so their sum is again the same plane (as they are closed under addition).

Here is another (analogous) way to think about it. Let’s start with a basis $B_0$ for $U_1\cap U_2.$ We can extend $B_0$ to a basis $B_1$ for $U_1$ and a basis $B_2$ for $U_2$. Then $B_1\cup B_2$ is a basis for $U_1+U_2,$ and $B_1\cap B_2=B_0,$ so \begin{align}\dim(U_1+U_2) &= |B_1\cup B_2|\\ &= |B_1|+|B_2|-|B_1\cap B_2|\\ &= |B_1|+|B_2|-|B_0|\\ &= \dim(U_1)+\dim(U_2)-\dim(U_1\cap U_2).\end{align} This generalizes nicely to $\Bbb F^n$, and allows us to avoid geometric arguments that may be less sensible for an arbitrary field $\Bbb F$.

Also, it will never be the case that the intersection of two planes in space is precisely $\{0\}.$ If there were two such planes $U_1$ and $U_2,$ then we would have $$\dim(U_1+U_2)=\dim(U_1)+\dim(U_2)-\dim(U_1\cap U_2)=2+2-0=4>3=\dim(\Bbb R^3),$$ which is not possible, since $U_1+U_2$ is a subspace of $\Bbb R^3$.