# 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?

• The intuition is correct. – Cameron Williams Sep 21 '13 at 17:17

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 $\Bbb R^3$.

• Thanks for the answer. Could it be the case that a pair of two dimensional planes in $\mathbb{R^4}$ intersect at a point? – St Vincent Sep 21 '13 at 17:27
• Yes, indeed. Consider $$U_1=\{(w,x,y,z)\in\Bbb R^4:y=z=0\}$$ and $$U_2=\{(w,x,y,z)\in\Bbb R^4:w=x=0\}.$$ The upshot in $\Bbb R^3$ is that there "isn't enough room" for two planar subspaces to avoid each other that well. – Cameron Buie Sep 21 '13 at 17:38
• Why can we conclude that $B_{1} \cap B_{2} = B_{0}$? It seems perfectly reasonable that the extensions share vectors outside of $B_{0}$. – Jonathan Hebert Sep 14 '16 at 0:40
• @Jonathan: If there is some $\vec v\in B_1\cap B_2,$ then $\vec v\in U_1\cap U_2.$ What then can we conclude by definition of $B_0$? – Cameron Buie Sep 14 '16 at 0:46
• Ok, then $v$ must be a linear combination of $B_{0}$. But, this forces both $B_{1}$ and $B_{2}$ to be linearly dependent, right? – Jonathan Hebert Sep 14 '16 at 0:53

Cameron Buie's answer does answer the original question sufficiently. However, I'm adding the formal proof of the theorem in context (taken from "Linear Algebra Done Right" by Sheldon Axler).

If $$U_1$$ and $$U_2$$ are subspaces of a finite dimensional vector space then: $$\dim(U_1+U_2)=\dim U_1 + \dim U_2 - \dim(U_1 \cap U_2)$$

Let $$u_1,...,u_m$$ be a basis of $$U_1\cap U_2$$; thus $$\dim (U_1\cap U_2)=m$$. Because $$u_1,...,u_m$$ is a basis in $$U_1\cap U_2$$, it is linearly independent in $$U_1$$. Hence this list can be extended to a basis $$u_1,...,u_m,v_1,...,v_j$$ of $$U_1$$. Thus, $$\dim U_1 = m+j$$. Also extend $$u_1,..,u_m$$ to a basis $$u_1,...,u_m, w_1,...,w_k$$ of $$U_2$$. $$\dim U_2 = m +k$$.

We will show that $$u_1,...,u_m,v_1,...,v_j,w_1,...,w_k$$ is a basis of $$U_1+U_2$$. This will complete the proof because then we will have $$\dim(U_1+U_2)=m+j+k=\dim U_1 + \dim U_2 - \dim (U_1\cap U_2)$$

We just need to show that the list $$u_1,...,u_m,v_1,...,v_j,w_1,...,w_k$$ is linearly independent. To prove this, suppose:

$$a_1u_1+...+au_m+b_1v_1+...+b_jv_j+c_1w_1+...+c_kw_k=0$$

where all $$a,b,c$$'s are scalars. We need to show that all the $$a,b$$ and $$c$$'s are $$0$$.

The equation can be rewritten as

$$c_1w_1+...+c_kw_k=-a_1u_1 - ... -a_mu_m -b_1v_1 - ... - b_j v_j$$

Which shows that $$c_1w_1+...+c_kw_k\in U_1$$. But actually all $$w$$'s are in $$U_2$$. So the LHS must be an element of $$U_1\cap U_2$$.

$$c_1w_1+...+c_kw_k=d_1u_1+...+d_mu_m$$ for some choice of scalars $$d_1,d_2,...,d_m$$. But $$u_1,...,u_m,w_1,...,w_k$$ is linearly independent. So our last equation implies that all the $$c$$'s equal $$0$$.

Thus our original equation involving $$a,b,c$$ becomes

$$a_1u_1+...+a_mu_m+b_1v_1+...+b_jv_j=0$$

But we already knew that the list $$u_1,...,u_m,v_1,...,v_j$$ is linearly independent. This equation implies that all the $$a$$'s and $$b$$'s are $$0$$. We now know that all $$a,b$$ and $$c$$'s are $$0$$, hence proving our original claim.