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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?

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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 $\operatorname{span}(A\cup B)= U_1+U_2$.

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  • $\begingroup$ 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? $\endgroup$ – user736276 Jan 11 at 2:15
  • $\begingroup$ @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. $\endgroup$ – Milo Brandt Jan 11 at 2:20
  • $\begingroup$ 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? $\endgroup$ – user736276 Jan 11 at 3:47
  • $\begingroup$ @68e1515 Yes, I think that's the case. $\endgroup$ – Milo Brandt Jan 11 at 5:17
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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 $(u_1,...,u_m,v_1,...,v_j,w_1,...,w_k)$.

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