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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?

Caio Velasco
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1 Answers1

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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:

  1. $U_1+U_2$ contains both $U_1$ and $U_2$ (as sets);
  2. $U_1+U_2$ is a subspace of $V$;
  3. 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”)

Arturo Magidin
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  • Thank you very much @ArturoMagidin. I am new to Mathematical Analysis, and now I see that the textbook did not make it clear that $W$ was any subspace of $V$. Therefore, it is very clear to me now that if you take an arbitrary $W$ from $V$, then yes, if both $U_1$ and $U_2$ are contained in $W$, because $W$ is arbitrary, their sum is contained in W. Thus, it follows tha their sum is the smallest subspace that contains both $U_1$ and $U_2$ of $V$. – Caio Velasco Apr 23 '20 at 13:02