Using the first direct sum test to determine if $V=W_{1}\oplus W_{2}$. The definition says:

Let $B_1$ and $B_2$ be disjoint bases for subspaces $W_1$ and $W_2$, respectively, of a finite-dimensional vector space $V$. If $B_1 \cup B_2$ is a basis for $V$, then $V=W_1 \oplus W_2$.

So I have basis for ($B_1$ and $B_2$), but also for vector space $V$ ... how would one determine if $V=W_1\oplus W_2$?
 A: Let $B_1=(b_1,\ldots,b_n)$ and $B_2=(c_1,\ldots,c_m)$ the basis of $W_1$ and $W_2$ respectively. Let $v\in V$ and since $B_1\cup B_2$ is a basis for $V$ then $v$ can be written:
$$v=\underbrace{\lambda_1 b_1+\cdots +\lambda_n b_n }_{\in W_1}+\underbrace{\lambda'_1 c_1+\cdots +\lambda'_n c_m }_{\in W_2}$$
hence
$V\subset W_1+W_2$
and the other inclusion is clear so
$$V= W_1+W_2$$
Now if $v\in W_1\cap W_2$ then $v$ is linear combination of the vectors $b_i$ and $c_i$:
$$v={\lambda_1 b_1+\cdots +\lambda_n b_n }={\lambda'_1 c_1+\cdots +\lambda'_n c_m }$$
so
$$0={\lambda_1 b_1+\cdots +\lambda_n b_n }-{\lambda'_1 c_1-\cdots -\lambda'_n c_m }$$
and since $B_1\cup B_2$ is a basis then $\lambda_i=\lambda'_i=0$ and then $v=0$ hence
$$V=W_1\oplus W_2$$
A: Check that the union is linearly independent by setting a linear combination of the vectors in the union to the zero vector and solving.
Also, if V is given to you ahead of time (and not just as the span of the union) then you have to check that the union spans V.
