There is this Corollary that says:
Let $U_1$ and $U_2$ be subspaces of vector space $V$, then $U_1+U_2$ is also a subspace of $V$.
I looked for proof and i found this:
$$
U_1=(x_1, y_1)
$$
$$
U_2=(x_2, y_2)
$$
$$
x=(x_1+x_2) x_1\in U_1, x_2\in U_2.
$$
$$
y=(y_1+y_2) y_1\in U_1, y_2\in U_2
$$
$$
x+y=x_1+x_2+y_1+y_2=(x_1+y_1)+(x_2+y_2) \in U_1+U_2
$$
I know that subspaces are closed under addition but I can't relate it to what i found. Any explanation?
Best Answer
The proof you found is not complete. You have to check all axioms for a linear space.
The lines just show that $U_1 + U_2$ is closed under addition. Set $U=U_1 + U_2$ and let $a,b\in U$. We have to show $a+b\in U$. Since $a\in U$, there exists $a_1\in U_1$ and $a_2\in U_2$ such that $a=a_1 + a_2$. Similar for $b$. Hence we can write $$a+ b = a_1 + a_2 + b_1 + b_2.$$ Since $a_i, b_i \in U_i$ and the operation $+$ is commutative we further have $$a_1 + a_2 + b_1 + b_2 = a_1 + b_1 + a_2 + b_2$$ and since the $U_i$ are linear spaces, $a_1 + b_1 \in U_1$ and $a_2 + b_2 \in U_2$. But this shows $a+b \in U$.
All you have to do now is to carefully check the remaining axioms. You will see that the properties of $U_1$ and $U_2$ transfer to $U$.
Hint: Instead of checking all axioms you should use a so called subspace criterium/definition. See here.