In Axler's Linear Algebra Done Right the theorem given is
Suppose U1,…,Um are subspaces of V. Then U1+⋯+Um is the smallest subspace of V containing U1,…,Um. I can see that the sum will be a subspace of V.
However, I did not find any reasonable intuitive answer for the following: Why do we have to choose 2 subspaces (say V and W) contained in a third one (say L) and show that their sum (V+W) is also contained in L.
Why does this proposition assert that V+W is what gives us the SMALLEST SUBSPACE containing both V and W? Why does the SMALLEST is defined this way?
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$\begingroup$In this context, a set $A$ is “smaller” than (or equal to) $B$ if $A\subseteq B$.
$U_1+U_2$ satisfies the following three conditions:
- $U_1+U_2$ contains both $U_1$ and $U_2$ (as sets);
- $U_1+U_2$ is a subspace of $V$;
- If $W$ is any subspace of $V$ that contains both $U_1$ and $U_2$, then $W$ also contains $U_1+U_2$ (as sets).
So: $U_1+U_2$ contains $U_1$ and $U_2$ (by 1), and is a subspace (by 2), and it is contained in any subspace that contains both $U_1$ and $U_2$. So it is the smallest subspace that contains $U_1$ and $U_2$, where we say that one subspace is “smaller” than another if and only if the one is contained in the other.
(Compare to the definition of “greatest common divisor”: we say $d$ is a greatest common divisor of $a$ and $b$ if and only if (i) it is a common divisor of $a$ and $b$; and (ii) if $e$ is any common divisor of $a$ and $b$, then $e$ divides $d$. So it is “greatest” under the order given by divisibility; for subspaces, ‘smallest’ is defined in terms of containment of sets)
(See also this for a general discussion of this kind of “closure”)
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