Clarification needed for the proof of $\dim(U_1+U_2)=\dim(U_1)+\dim(U_2)-\dim(U_1 \cap U_2)$

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:

$\dim(U_1+U_2)=\dim(U_1)+\dim(U_2)-\dim(U_1 \cap U_2)$

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.

To prove this, suppose:

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

$\implies c_1w_1+...+c_kw_k = -(b_1v_1+...+b_jv_j+a_1u_1+...+a_mu_m)$

This shows that $c_1w_1+...+c_kw_k \in U_1$ which is evident looking at the right hand side of the previous equality.

However, after this the book says: "All the $w$'s are in $U_2$, so this implies that $c_1w_1+...+c_kw_k \in U_1\cap U_2$"

I am having trouble understanding this statement. We had to extend $(u_1,u_2,...)$ to $(u_1,u_2,...,u_m,w_1,w_2,...,w_k)$ in order to cover the region of whole region of $U_2$ (i.e. include the regions outside of $U_1\cap U_2$ in $U_2$). So it should mean that $(w_1,w_2,...,w_k)$ is basis of $U_2-U_1\cap U_2$. But they claim that $c_1w_1+...+c_kw_k \in U_1\cap U_2$ which seems contradictory!

Am I making any conceptual error?

• Do you agree that the $w$'s are in $U_2$? Since $U_2$ is a vector space, then any linear combination of the $w$'s must also be in $U_2$. [Note that the statement "$c_1w_1 + ... + c_kw_k \in U_1 \cap U_2$ just means that $c_1w_1 + ... + c_kw_k \in U_1$ and $c_1w_1 + ... + c_kw_k \in U_2$]. – Michael Dec 23 '17 at 22:01
• @Michael My problem is that $(w_1,w_2,...,w_k)$ was initially introduced to span the region outside of $U_1\cap U_2$ but in $U_2$ (as $(u_1,u_2,..,u_m)$ could only span $U_1\cap U_2$ and not the whole $U_2$). So how can a linear combination of $(w_1,w_2,...,w_k)$ be an element of $U_1\cap U_2$ ? – user400242 Dec 23 '17 at 22:06
• I think Bernard gave an answer to that. This is kind of like trying to "work towards a contradiction" in proof by contradiction, then getting confused at the end because we ended up with something contradictory and forgetting that is what we wanted to do. Here, we are not working towards a contradiction, just working towrads showing the big sum of vectors is really just 0. – Michael Dec 23 '17 at 22:25

...it should mean that $(w_1,w_2,...,w_k)$ is basis of $U_2-U_1\cap U_2$.

This is meaningless, as $U_2-U_1\cap U_2$ is not a subspace. $(w_1,w_2,...,w_k)$ is a basis of a complement of $U_1\cap U_2$ w.r.t. $U_2$, that's all. A complement of a subspace is not the difference set.

• If $(w_1,w_2,...,w_k)$ is a basis of the complement of the subspace $U_1\cap U_2$ w.r.t $U_2$, then how can it's linear combination $c_1w_1+...+c_kw_k$ be an element of $U_1\cap U_2$ ? It would be helpful if you address this concern a bit... – user400242 Dec 23 '17 at 22:11
• Precisely, it can't, unless it's $0$. – Bernard Dec 23 '17 at 22:14
• The complement subspace is a subspace $V\subset U_2$ such that $U_2=U_1\cap U_2\oplus V$. For instance , in $\mathbf R^2$, a complement of the $x$-axis is any vector line with basis a vector not collinear with the vector $(1,0)$, whereas the difference set is the set of all vectors $(x,y)$ such that $y\ne 0$ – this is not even a vector (sub)space. – Bernard Dec 23 '17 at 22:38

Since $w_1,\ldots,w_k\in U_2$, $c_1w_1+\cdots c_kw_k\in U_2$. But you also know that $c_1w_1+\cdots c_kw_k\in U_1$. Therefore, $c_1w_1+\cdots c_kw_k\in U_1\cap U_2$.

There is a conceptual problem in what you wrote. You wrote that $\{w_1,\ldots,w_k\}$ is a basis of $U_2\setminus U_1\cap U_2$. How can that be? $U_2\setminus U_1\cap U_2$ is not a vector space.

• I'm not sure why you got a downvote. I gave you +1. – Michael Dec 23 '17 at 22:13
• @Michael I also don't understand it. – José Carlos Santos Dec 23 '17 at 22:21
• It seems to have corrected itself. Likely a mis-hit. – Michael Dec 23 '17 at 22:23
• @Michael Right; it was corrected. – José Carlos Santos Dec 23 '17 at 22:27