Internal Direct Sum Linear Algebra Proof Question

Question

Let V be a vector space. If $U_1$ and $U_2$ are subspaces of $V$ such that $U_1 + U_2 = V$ and $U_1\cap U_2 = {0_v}$, then we say that $V$ is the internal direct sum of $U_1\oplus U_2$. Show that V is the internal direct sum of $U_1$ and $U_2$ if and only if every vector in $V$ may be written uniquely in the form $v_1 + v_2$ with $v_1 \in U_1$ and $v2 \in U_2$

So my understanding is that subspace $U_1$ and $U_2$ contain all of the vector space $V$ and their only intersection is the $0_v$. Thus since the subspaces don't have any elements in common besides the zero vector, $v_1 \in U_2$ and $v_2 \in U_1$ s.t. $v_2+v_1$ exists. Is this a correct way to think or am I missing something

Answer 1

subspace $U_1$ and $U_2$ contain all of the vector space $V$

No, for example, you have $\mathbb{R} = V_1 \oplus V_2$ where $V_1$ is x-axis and $V_2$ is y-axis, but you can see that $V_i$ only contains vectors $\alpha e_i$, $e_1 = (1,0)$ and $e_2=(0,1)$.

Answer 2

No U1 and U2 wont contain all the elements of V, it just only says that every element of V is a sum of two elements each from U1 and U2.

Suppose V is the internal direct product of U1 and U2, then from the definition you can write every vector in V as sum of some u1 and u2 from U1 and U2 respectively, the uniqueness is as follows suppose we have u1+u2=v and v1+v2=v then u1-v1+u2-v2=0, which imply that u1-v1=0=-u2+v2 because the intersection is 0 and linear independence of u1-v1 with u2-v2.