# Sum of two subspaces is also a subspace

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?

• Another approach would be to show that $U_1 + U_2$ is the image of the linear map $A : V \to V$ defined by $A(v) = P_{U_1}(v) + P_{U_2}(v)$, where $P_{U_1}$ and $P_{U_2}$ are projections onto $U_1$ and $U_2$ respectively. The image of a linear map is always a subspace. – Bungo Sep 26 '17 at 19:26

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 $U$.