Question regarding dimensions of vector spaces How can one prove the following ? 
Let $U,V$ be two vector spaces.
$$\dim (U + V) + \dim (U \cap V) = \dim U + \dim V
$$
 A: Hint
Let $\mathcal B$ be a basis for the linear subspace $U\cap V$ of both $U$ and $V$. We can extend this basis such that $\mathcal B_U = \mathcal B \cup \mathcal U$ is a basis of $U$ where $\mathcal U$ is a basis for $(U\cap V)^\perp$ in $U$ and to a basis $\mathcal B_V = \mathcal B \cup \mathcal V$ where $\mathcal V$ is a basis for $(U\cap V)^\perp$ in $V$.
$\mathcal B \cup \mathcal V\cup \mathcal U$ must now be a basis for $U+V$ (why?) and thus by inclusion-exclusion principle
$$\dim (U+V) + \dim (U\cap V) = |\mathcal B \cup \mathcal U \cup \mathcal V| + |\mathcal B| = |\mathcal B| + |\mathcal U| + |\mathcal B| + |\mathcal V| = \dim U + \dim V$$
A: A more abstract approach:
Consider the map
$$
\Phi : U \times V \to U+V, (u,v) \mapsto u+v.
$$
Show that this is a (well-defined) linear map with ${\rm ker}(\Phi) = U\cap V$.
EDIT: Ok, the kernel of $\Phi$ is not really $U\cap V$, but it is
$$
{\rm ker}(\Phi)=\{(u,-u)\mid u \in U\cap V\},
$$
which is isomorphic to $U\cap V$ via $U \cap V \to {\rm ker}(\Phi), u \mapsto (u,-u)$, so both spaces have the same dimension. 
By the first isomorphism  theorem, this yields
$$
U+V \cong (U \times V) / {\rm Ker}(\Phi). 
$$
Now it is easy to see $\dim(U\times V) = \dim(U) + \dim(V)$.
The only thing you have to know now (which is useful in general) is that
$$
\dim(X/Y) = \dim(X) - \dim(Y),
$$
if $Y$ is a subspace of $X$.
But this follows by choosing a basis $(y_1, \dots, y_k)$ for $Y$, extending it to a basis $(y_1, \dots, y_k, x_1, \dots, x_\ell)$ of $X$ and noting that $(x_1 + Y, \dots, x_\ell + Y)$ is a basis of $X/Y$.
