This question is from the Third Edition of Axler's Linear Algebra Done Right (2.C, #14)
Suppose $U_1,\ldots,U_m$ are finite-dimensional subspaces of vector space V. Prove that $U_1 + \cdots + U_m$ is finite-dimensional and
$$\text{dim($U_1 + \cdots + U_m$)} \leq \text{dim $U_1 + \cdots +$ dim $U_m$.} \tag{1}\label{eq1}$$
I understand how to show that $ U_1 + \cdots + U_m $ is finite-dimensional since the sum of subspaces of V is a subspace of V (Thm. 1.39) and any subspace of a finite-dimensional vector space is finite-dimensional (Thm. 2.26). However, I'm stuck on proving the bound.
Here's what I've got so far:
From the previous statements, it follows that
$$\text{dim($U_1 + \cdots + U_m$)} \leq \text{dim }V \tag{2}\label{eq2} $$
from Thm 2.38. Now, assume $ v_1, \ldots, v_n $ is a basis for V. Then there exists a subset of these basis vectors that is a basis for each of $U_1,\ldots,U_m$. Summing these basis vectors together is the same as adding the dimensions of the subspaces together, resulting in
$$\text{dim $U_1 + \cdots +$ dim $U_m$.} \tag{3}\label{eq3}$$
I'm not sure how to connect Equations \eqref{eq2} and \eqref{eq3} to produce \eqref{eq1}. I'm also not sure if my method of summing basis vectors is valid or the best way to go about this problem.
Best Answer
You're on the right track in considering a basis of each $U_i,$ but equation $(2)$ is not quite what you need. Instead of $V,$ consider the subspace $U$ spanned by all the bases of the all $U_i.$ (This is the smallest subspace containing all the $U_i.$) Then I think you'll have no trouble.
EDIT
Since all the $U_i$ are subspaces of $U,$ you can substitute $U$ for $V$ in your equation $(2)$.$$\dim(U_1 + \cdots + U_m) \leq \dim U $$ But we have a set of vectors of cardinality $\text{dim} U_1 + \cdots + \text{dim} U_m$ that spans $U,$ namely all the basis vectors of all the $U_i,$ so $\dim U$ can't be more than that.