A subset U of V is a subspace of V if and only if U satisfies the following conditions:
additive identity ${latex.inline0 \in U}
closed under addition ${latex.inlineu,w \in U \implies u + w \in U}
closed under scalar multiplication ${latex.inlinea \in F,\ u \in U \implies au \in U}
${latex.inline\Rightarrow} Suppose U is a subspace of V. Then U satisfies the three conditions above by the nature of it being a vector space. This is by definition ie: the subspace has the same additive identity, addition, and scalar multiplication and is still a vector space.
\({latex.inline[\Leftarrow](\Leftarrow)} Suppose U satisfies the three conditions. The first condition means that 0 is in U. The second condition says that addition makes sense on U, and the third that scalar multiplication makes sense on U. Since \){latex.inlineu \in U, -u \in U} because ${latex.inline-u = (-1)u} per 1753094630 - Axler 1.32 -1 * v = -v|Axler 1.32 and we have closure under scalar mult. The rest of the conditions hold because U is a subset of V with the same exact addition and scalar multiplication.
${latex.inline\Rightarrow} Suppose U is not a subspace of V. Then there is no guarantee that the additive identity in U. Also, it could be a totally random subset with different operations.
${latex.inline\Leftarrow} Suppose the additive identity is not in U. Then U can’t be a subset by definition.
Suppose addition is not closed in U. Then it’s not a vector space which means it cannot be a subspace.
Same with scalar multiplication.