# Bound on the dimension of sum of subspaces (Axler)

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.

• Can you do the case $m=2$? – Lord Shark the Unknown Jul 11 '18 at 5:03
• From what you write, there seems to be no reason to suppose that $V$ is finite-dimensional. – Lord Shark the Unknown Jul 11 '18 at 5:04
• True, it doesn't say that V is finite-dimensional in the problem, just that $U_1,\ldots,U_m$ are finite-dimensional subspaces of V. So I'd probably need to use the smallest containing subspace in all of my assertions, not V. – ksankar Jul 11 '18 at 5:09
• The case $m=2$ makes sense since it translates to the formula from Thm 2.43. And $\dim U_1 + \dim U_2 \geq \dim U_1+\dim U_2-\dim(U_1 \cap U_2)$. – ksankar Jul 11 '18 at 5:11

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.

• Okay, that makes sense. So I'd have to use the fact that smallest containing subspace (as you defined $U$) has dimension equal to the sum of the individual dimensions from equation (3). I'd then use the dimension of a subspace bound to show that the LHS of equation (2) is less than or equal to equation (3) (since the sum of the $U_i$s is a subspace of $U$). Would this be the correct line of reasoning? – ksankar Jul 11 '18 at 4:51
• No, the smallest subspace containing all the $U-I$ has dimension less than or equal to the sum of the dimensions. That's what you're proving. I've having trouble fitting this in a comment; I'll update my answer. – saulspatz Jul 11 '18 at 5:20

The argument is made easier by working with spanning sets and bases.

By definition the dimension of a vector space is finite if it is spanned by a finite set of vectors. In this case any two bases of the vector space have the same number of vectors, and this number is called the dimension of the vector space.

Note that this implies that if $k$ vectors span $W$ then $\dim W \leq k$, because a basis of $W$ has no more vectors than a spanning set.

If the dimensions of $U_{i}$ are finite, then each is spanned by a finite collection of vectors. By definition of the sum $U_{1} + \dots + U_{m}$ the union of these finite spanning sets spans the sum $U_{1} + \dots + U_{m}$, so the sum is finite-dimensional. If the set spanning $U_{i}$ is a basis of $U_{i}$, then its cardinality is $\dim U_{i}$; the union of these bases spans the sum $U_{1} + \dots + U_{m}$ but it is not necessarily linearly independent, so it contains at least as many vectors as any basis of $U_{1} + \dots + U_{m}$. It contains $\sum_{i = 1}^{m}\dim U_{i}$ vectors, so $\sum_{i=1}^{m}\dim U_{i} \geq \dim(U_{1} + \dots + U_{m})$.