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**obd2** Please excuse my previous post as I accidentally pressed the submit button instead of review button. English is not my first language but I'm doing my best. I hope that the following helps, I like to avoid using any vectors whatsoever in proofs like this.

Ok. first (i) $\displaystyle \Rightarrow$(ii). Given (i) then you know that $\displaystyle V \subseteq U \oplus W$. We also know that $\displaystyle dim(U\oplus W) = dim(U) + dim(W)$. It is known that the dimesion of a subspace is lesser than or equal to the dimesion of the larger space. Hence we have shown that

$\displaystyle $$dim(V) \le dim(U\oplus W) = dim(U ) + dim(W)$$$.

Now $\displaystyle (ii) \Rightarrow (i)$. Given (ii) we automatically know that

$\displaystyle $$ dim(V) \le dim(U) + dim(W)$$$, and we know that $\displaystyle U\cap W = \{0\}$. Next, we know that since $\displaystyle U$ and $\displaystyle W$ are both subspaces of $\displaystyle V$ then their sum (that is $\displaystyle U+W$) is also a subspace of $\displaystyle V$, hence the following holds:

$\displaystyle $$ dim(U+W) \le dim(V)$.

Finally, since we have the equation $\displaystyle dim(U+W) = dim(U) + dim(W) - dim(U\cap W)$, which gives us that $\displaystyle dim(U+W) = dim(U) + dim(W)$ since $\displaystyle dim(U\cap W) = dim(\{0\}) = 0$. Effecetively we have shown that

$\displaystyle $$dim(V) \ge dim(U) + dim(W)$$$

also holds. Then we must have $\displaystyle dim(V) = dim(U) + dim(W) = dim(U + W) $. This gives us that $\displaystyle V = U+W$ and, since $\displaystyle U\cap W = \{0\}$, therefore $\displaystyle V = U \oplus W$ which is equivalent to (i).