# Question in proof that $\dim(U_1+U_2)=\dim U_1+\dim U_2-\dim(U_1 \cap U_2)$

In the proof that Dim$$(U_1+U_2)=DimU_1+DimU_2-Dim(U_1 \cap U_2)$$

In "Linear Algebra Done Right" by Sheldon Axler:

Theorem 2.18:

If $$U_1$$ and $$U_2$$ are subspaces of a finite dimensional vector space, then:

$$\dim(U_1+U_2)=\text{dim}(U_1)+\text{dim}(U_2)-\text{dim}(U_1 \cap U_2)$$

PROOF (in short):

Let $$(u_1,u_2,...)$$ be a basis of $$U_1\cap U_2$$. This can be extended to a basis $$(u_1,u_2,...,u_m,v_1,v_2,...,v_j)$$ of $$U_1$$. Also, it can be extended to a basis $$(u_1,u_2,...,u_m,w_1,w_2,...,w_k)$$ of $$U_2$$.

Clearly $$\text{span}(u_1,...,u_m,v_1,...,v_j,w_1,...,w_k)$$ is $$U_1 + U_2$$. To show that this list is a basis of $$U_1+U_2$$ we just need to show that it is linearly independent.

I cannot understand why $$\text{span}(u_1,...,u_m,v_1,...,v_j,w_1,...,w_k)$$ is $$U_1 + U_2$$. Can someone help explain this to me?

In very simple terms:

• by construction every vector in $$U_1$$ is a linear combination of $$(u_1,...,u_m,v_1,...,v_j)$$ and every vector in $$U_2$$ is a linear combination of $$(u_1,...,u_m,w_1,...,w_k)$$.

• By definition, every vector in $$U_1+U_2$$ is of the form $$u+u^\prime$$ with $$u\in U_1$$ and $$u^\prime\in U_2$$.

Putting things together every vector in $$U_1+U_2$$ can be written as a linear combination of $$(u_1,...,u_m,v_1,...,v_j,w_1,...,w_k)$$.

This is an instance of a more general fact:

$$\operatorname{span}(A\cup B)=\operatorname{span}(A) + \operatorname{span}(B).$$

which may be proven by noting that if $$v_a\in \operatorname{span}(A)$$ and $$v_b\in \operatorname{span}(B)$$ then clearly $$v_a+v_b$$ can be written as a linear combination of elements in $$A\cup B$$ by simply expanding $$v_a$$ as a combination in $$A$$ and $$v_b$$ as a combination in $$B$$. Conversely, if $$v\in \operatorname{span}(A \cup B)$$, then $$v$$ can be written as a linear combination of elements in $$A\cup B$$, which can be split as a sum of a linear combination of elements of $$A$$ plus a combination of elements of $$B$$.

Here, we could take $$A$$ as the basis of $$U_1$$ and $$B$$ as the basis of $$U_2$$ to see that $$\operatorname{span}(A\cup B)= U_1+U_2$$.

• Should I think of taking $x=c_1u_1+\dots +c_mu_m+d_1v_1+\dots+d_jv_j \in U_1$and $y=a_1u_1+\dots+a_mu_m+g_1w_1+\dots + g_kw_k \in U_2$ and adding $x$ and $y$? and then because $x+y$ is a linear combination of all these basis vectors it is in the sum since the sums of subspaces are closed under taking linear combinations? – user736276 Jan 11 '20 at 2:15
• @68e1515 Well, you might think of expanding as you've done, then combining the terms to see that $x+y$ is in the larger span, which shows that the larger span contains $U_1+U_2$. You would also do the reverse of writing $z=\sum c_iu_i + \sum d_iv_i + \sum g_iw_i$ from the larger span and breaking it as the sum of $(\sum c_iu_i + \sum d_iv_i)\in U_1$ and $\sum g_iw_i \in U_2$ - there's sort of two inclusions to show to establish the equality. – Milo Brandt Jan 11 '20 at 2:20
• I have one more question about this.If you were trying to prove the two set inclusions both ways is the way I proved it above exactly right? Then to finish off the other direction I will prove it your way? – user736276 Jan 11 '20 at 3:47
• @68e1515 Yes, I think that's the case. – Milo Brandt Jan 11 '20 at 5:17