Proving the direct sum of two particular subspaces of $\mathbb{R}^2$ holds iff a related system of equations has an unique solution. 
Let $a,b,c,d \in \mathbb{R}$ (such that $a$ and $b$ cannot be $0$ at the same time neither $c$ and $d$ cannot be zero at the same time), $U= \lbrace (x,y) \in \mathbb{R}^2 \mid ax+by=0 \rbrace$ and $V= \lbrace (x,y) \in \mathbb{R}^2 \mid cx+dy=0 \rbrace$. Show that $U$ and $V$ are subspaces of $\mathbb{R}^2$ and that $U \oplus V = \mathbb{R}^2$ if and only if the system
$$
\left\lbrace
  \begin{aligned}
    ax + by &=  0 \,, \\ 
    cx + dy &=  0 \,,
  \end{aligned}
\right.
$$
has an unique solution.

Proving that $U$ and $V$ are subspaces of $\mathbb{R}^2$ is straightforward. Proving the required equivalence is not trivial. I have this hunch that assuming the system as a unique solution implies $U \oplus V = \mathbb{R}^2$ is the easy implication. That means we need to prove $U \cap V = \lbrace 0 \rbrace$ and $U + V = \mathbb{R}^2$ my idea was to show that  every $(u,v) \in \mathbb{R}^2$ can be represented uniquely as a sum $(u,v)= (x,y) +(x',y')$ where $(x,y) \in U$ and $(x',y') \in V$. But I see this. For the other implication I've run out of ideas.
 A: First suppose that the system of equations has a unique solution. In particular we know that this solution is $(0, 0)$ as this is a valid solution and our hypothesis is that there is only one solution. Thus, we have that $U\cap V=\{0\}$. As if this were not the case, then the system would not have a unique solution. Recall that $U+V$ is a direct sum if and only if $U\cap V=\{0\}$. Thus, we have that $U+V$ is a direct sum. To show that $U\oplus V=R^2$, note that $U$ is non-empty as $(1, -\frac a b)$ is a solution for $$ax+by=0$$ Assuming without loss of generality that $b\ne 0$. Similarly we have that $V$ is non-empty. This implies that each subspace must have dimension at least $1$. We know that neither subspace can have dimension $2$ as that would imply that that subspace is $R^2$, which would contradict our previous result that $U\cap V=\{0\}$. Thus, each subspace has dimension $1$(they can't have zero dimension as they are non-empty). Now use the result that $$\dim(U+V)=\dim(U)+\dim(V)-\dim(U\cap V)$$ to conclude that $\dim(U\oplus V)=2$. Which implies that $U\oplus V=R^2$.
For the other direction, suppose that $U+V$ is a direct sum and that $U\oplus V=R^2$. Because $U+V$ is a direct sum, we have that $U\cap V=\{0\}$. This then implies that the system of equations $$ax+by=0\\ cx+dy=0$$ has the only solution $(0, 0)$.
