Is my linear algebra proof valid? (Using Axler’s book)

[grim knoll]

Theorem. Suppose $U$ and $W$ are subspaces of $V$. Then $U + V$ is a direct sum iff $U \cap V = {0}$.
\begin{proof}
Suppose that $U + V$ is a direct sum, and that $u \in U$ and $m \in U \cap V$. Let $s = u + m$ be their sum; that is, $u - s + m = 0$. Since $u \in U$ and $m \in U$, it follows that $s = u + m \in U $, thus $u - s \in U$ as well. Since $U + V$ is a direct sum and $u - s \in U$ and $m \in V$, by Theorem 1.45 (see image), $u - s =0$ and $m = 0$. Therefore, $U \cap V = {0 }$. \
Conversely, suppose $U \cap V = {0}$ and $u, u' \in U$ and $v, v' \in V$. Let $u + v$ be an arbitrary sum of $U+V$ s.t. it is equal to another sum $u' + v'$. Thus, $u - u' = v - v'$. Since $U$ and $V$ are closed under addition and $U \cap V = {0}$, $u - u' = v - v'$; that is, $u = u'$ and $v = v'$. Therefore $U + V$ is a direct sum.
\end{proof}

[summer hamlet]

herrperſon

[sage flame]

yes; as a note the first direction can be done in an easier/clearer way, and you have a few typos in the second, and u-u' = v-v' isn't quite correct

[grim knoll]

Lemme fix the second part

[sage flame]

oh wait

I misread your definition

why does u=u' and v=v' mean U+V is a direct sum? as in how do you get to 0 is uniquely 0 + 0

[grim knoll]

Yeah, it should be u - u' = v' - v

[grim knoll]

I meant: “Since U and V are closed under addition and their intersection, is equal to {0}, u - u' = v - v' = 0. Thus u = u' and v' = v. It follows that u + v can only be expressed as a sum in only one way, therefore it is a direct sum.”

[sage flame]

but the definition is that 0 can only be expressed as 0+0

at least I'm assuming that "condition 1.45" is your definition of direct sum in the absence of any other information

oh

oh that's just the reference in your proof

nvm

yeah that's fine then

[grim knoll]

That's not actually how my book defines a direct sum