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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 $U_1$ and $U_2$, and we write $V=U_1\oplus U_2$. 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 $v_2 \in U_2$

My Proof: Let $V$ be a vector space. Let every vector in vector space $V$ be unique s.t. for any vector $u \in V$, if $w_1 +w_2 = u = v_1+v_2$ form $w_1,v_1 \in U_1$ and $w_2,v_2 \in U_2$ then $v_1 = w_1$ and $v_2 = w_2$. Thus since we know that the vectors $v_1 = w_1$ and $v_2 = w_2$ the subspace does not overlap besides at $0_V$ because all subspaces contain the zero vector.

Any tips on how I can fix this proof so it is correct? Really new to proofs so there are probably dumb mistakes.

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    Sorry, I am new around here. I am really having trouble with proofs so I added "fake proofs", but my question was where did I go wrong in this proof and how can I fix it. – 2017-02-07
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    No problem, it's better now. – 2017-02-07

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I don't know the purpose of the proof-verification, but in my opinion the last sentence is not rigorous enough to show that $U_1\cap U_2=\{0\}$.

To show this more rigorously, you should assume that $u\in U_1\cap U_2$, then show that $u=0$ using the unique representability.

Additionally, you also have to show the reverse direction, i.e. $V=U_1\oplus U_2$ implies each vector in $V$ has a unique representation as a sum of vectors from $U_1$ and $U_2$.