# Internal Direct Sum Linear Algebra Proof 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

• What is your definition of internal direct sum? – angryavian Feb 6 '17 at 7:22
• Sorry I left that sentence out. – Xerxes Feb 6 '17 at 16:53

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 $e_2=(0,1)$.
• I wish you can help, How can I go, in proving directs sums, from having that ($v_1+v_2=0$ with $v_1 \in V_1$ and $v_2 \in V_2$ $\implies v_1=v_2=0$) to the fact that $V_1 \cap V_2=\{0\}$ ? – Fareed Abi Farraj Mar 26 '19 at 10:35