# Axler's proof of dimension of sum, step unclear

In Axler's proof of the dimension of sum formula (page 47 of Linear Algebra Done Right), there is a step that requires showing that $$u_1,...,u_m,v_1,...,v_j,w_1,...w_k$$ is a basis of $$U_1+U_2$$.

Now, I understand that first I have to show that this set of vectors spans $$U_1+U_2$$. However, he says:

"Clearly span($$u_1,...,u_m,v_1,...,v_j,w_1,...w_k$$) contains $$U_1$$ and $$U_2$$, and hence equals $$U_1+U_2$$."

Why does that chain of logic lead to $$U_1+U_2 =$$ span($$u_1,...,u_m,v_1,...,v_j,w_1,...w_k$$)? Isn't it supposed to lead to $$U_1+U_2 \subseteq$$ span($$u_1,...,u_m,v_1,...,v_j,w_1,...w_k$$)? What about showing that span($$u_1,...,u_m,v_1,...,v_j,w_1,...w_k$$) $$\subseteq$$ $$U_1+U_2$$? What does it exactly mean to say that $$u_1,...,u_m,v_1,...,v_j,w_1,...w_k$$ spans $$U_1+U_2$$?

I'll add a little more. You are right that one technically has to show both inclusions.

The since every vector $$u_{i}$$, $$v_{j}$$, $$w_{k}$$ is in $$U_{1}+U_{2}$$, the span of them is in $$U_{1}+U_{2}$$ because it's a vector subspace (closed under vector addition and scalar multiplication). Now every vector $$v\in U_{1}+U_{2}$$ is the sum $$v_{1}+v_{2}$$ for $$v_{i}\in U_{i}$$. Now the $$u_{i}$$ with the $$v_{j}$$ give a basis for $$V_{1}$$ so $$v_{1}=\sum a_{i}u_{i} +\sum b_{j}v_{j}$$ and similarly, $$v_{2} = \sum\tilde{a}_{i}u_{i}+\sum\tilde{b}_{i}w_{j}$$. Puting these together yeilds \begin{align}v&=\sum a_{i}u_{i} +\sum b_{j}v_{j}+ \sum\tilde{a}_{i}u_{i}+\sum\tilde{b}_{i}w_{j}\\ &=\sum (a_{i}+\tilde{a}_{i})u_{i}+\sum b_{j}v_{j}+\sum\tilde{b}_{i}w_{j} \end{align}

So you have the reverse inclusion as well.

$$U_1+U_2$$ is the subspace of elements which can be written in the form $$a+b$$ with $$a\in U_1$$ and $$b\in U_2$$.

Though it's not written, I assume that the given vectors are chosen so that

• $$u_1,\dots,u_m$$ forms a basis for $$U_1\cap U_2$$,
• $$u_1,\dots,u_m,v_1,\dots,v_j$$ forms a basis for $$U_1$$ and
• $$u_1,\dots,u_m,w_1,\dots,w_k$$ forms a basis for $$U_2$$.

Consequently, each of the given vectors is in particular an element of either $$U_1$$ or $$U_2$$, in any case it's in $$U_1+U_2$$.
Since the latter is a subspace, the span of all these vectors (i.e. the set of all their linear combinations) is a subset of $$U_1+U_2$$.

On the other hand, because of the basis property, every $$a\in U_1$$ can be written as a linear combination of vectors $$u_i$$ and $$v_i$$. Similarly, any $$b\in U_2$$ can be written by vectors $$u_i$$ and $$w_i$$.
So, any element $$a+b\,\in U_1+U_2$$ can be written as a linear combination of all of $$u_1,\dots,u_m,v_1,\dots,v_j,w_1,\dots,w_k$$.

Equality is pretty clear, from the definitions of span and the sum of two vector spaces. That is, the latter is just the set of all sums.

There couldn't be anything in the span of the given vectors (though you didn't say what they are) that isn't a sum of elements of $$U_1$$ and $$U_2$$. This follows if all the basis vectors are elements of $$U_1$$ or $$U_2$$.

Axler evidently thought the other inclusion a little more difficult, and that's understandable.