# The sum of subspaces is the smallest subspace containing all the summands

In Axler's Linear Algebra Done Right the theorem given is

Suppose $U_1,\ldots, U_m$ are subspaces of $V$. Then $U_1+\cdots+ U_m$ is the smallest subspace of $V$ containing $U_1,\ldots, U_m.$

I can see that the sum will be a subspace of $V$. What I don't understand is the following paragraph:

"Clearly $U_1,\ldots, U_m$ are all contained in $U_1+\cdots+ U_m$ (to see this consider sums $u_1 + \cdots+u_m$ where all except one of the $u$s is $0$). Conversely every subspace of V containing $U_1,\ldots, U_m$ must contain $U_1+\cdots+ U_m$ (because subspaces must contain all finite sums of their elements). Thus, $U_1+\cdots+ U_m$ is the smallest subspace of $V$ containing $U_1,\cdots, U_m."$

This question was addressed here: Misunderstanding in the proof that the sum of subspaces is the smallest containing subspace. but the answer wasn't clear enough for me.

I can see how the first part is true, that $U_1,\ldots, U_m$ are all contained in $U_1+\cdots+ U_m$ and the second part, $U_1,\ldots, U_m$ must contain $U_1+\cdots+ U_m$ but I don't understand how together they prove the theorem. Can someone please take the trouble of giving an intuitive explanation, or a geometric one if it is possible? Thanks a lot.

• Let $W = U_1 + \dots + U_m$. There are three basic parts to the statement. (1) $W$ is a subspace. (2) $W$ contains each subspace $U_1, \dots, U_m$. (3) If $V$ is any subspace that contains $U_1, \dots, U_m$, then we must have $W \subseteq V$. Do you agree that these three parts together show that $W$ is the smallest subspace containing $U_1, \dots, U_m$? – David Jan 5 '16 at 17:06
• Deep down this proof is of the kind "Show that two sets $A$ and $B$ are equal by showing $A\subseteq B$ and $B\subseteq A$". The two sets in question are the vector space $U_1+\cdots+U_m$ and "the smallest subspace of $V$ containing all the $U_i$" – Arthur Jan 5 '16 at 17:07

[…] and second part, $U_1, \dotsc, U_m$ must contain $U_1 + \dotsb + U_m$.

This is wrong. Consider for example $V = \mathbb{R}^2$ and the subspaces $$U_1 = \{(x,0) \mid x \in \mathbb{R}\} \quad\text{and}\quad U_2 = \{(0,y) \mid y \in \mathbb{R}\}.$$ Then $U_1 + U_2 = \mathbb{R}^2$, but neither $U_1$ nor $U_2$ contain $\mathbb{R}^2$.

The statement that we really want is the following:

$U_1 + \dotsb + U_n$ is a subspace containing $U_1, \dotsc, U_n$, and if $W \subseteq V$ is any subspace with $U_1, \dotsc, U_n \subseteq W$, then we already have $U_1 + \dotsb + U_n \subseteq W$.

You already understand the first part of this statement, that $U_1 + \dotsb + U_n$ is a subspace containing $U_1, \dotsc, U_n$.

For the other part suppose $W \subseteq V$ is some subspace containing $U_1, \dotsc, U_n$. For all $u_1 \in U_1, \dotsc, u_n \in U_n$, we then have $u_1, \dotsc, u_n \in W$. Because $W$ is a subspace it follows that also $u_1 + \dotsb + u_n \in W$ (because $W$ is closed under finite sums). Therefore $$U_1 + \dotsb + U_n = \{u_1 + \dotsb + u_n \mid u_1 \in U_1, \dotsc, u_n \in U_n\} \subseteq W.$$

I'm not sure whether it is clear to me what you are asking, but this is the classical case where you show that two sets $X,Y$ are equal by showing that both $X\subset Y$ and $Y\subset X$ hold true.