Union of a Basis is a basis Suppose that $V$ is the direct sum of $S$ and $T$, that is, $V=S\oplus T$. If $A$ and $B$ are bases of $S$ and $T$, respectively, then $A \cup B$ is a basis of $V$.
I already verified that $A \cup B$ spans $V$. 
Any hint how to show its linear independence?
 A: Let $A = \{\mathbf{u_1},\ \cdots,\ \mathbf{u_n}\}$ and $B = \{\mathbf{v_1},\ \cdots,\ \mathbf{v_m}\}$. Consider the linear independence equation of the basis
$$a_1\mathbf{u_1} + \cdots + a_n\mathbf{u_n} + b_1\mathbf{v_1} + \cdots + b_m\mathbf{v_m} = \mathbf{0}$$
The vectors of $A$ sum to a vector $\mathbf{u}\in S$
$$\mathbf{u} = a_1\mathbf{u_1} + \cdots + a_n\mathbf{u_n}$$
and similarly the vectors in $B$ sum to a vector $\mathbf{v} \in T$
$$\mathbf{v}=b_1\mathbf{v_1} + \cdots + b_m\mathbf{v_m}$$
Therefore you are solving
$$\mathbf{u} + \mathbf{v} = \mathbf{0}$$
where $\mathbf{u}\in S$ and $\mathbf{v}\in T$. Show that this implies $\mathbf{u}$ and $\mathbf{v}$ are both zero (hint: what's the intersection of $S$ and $T$?) so that you are solving the linear independence equations for both basis simultaneously.
A: Suppose that $A=\{a_1,\ldots,a_n\}$ and $B=\{b_1,\ldots,b_m\}$.
If $\lambda_1a_1+\ldots+\lambda_na_n+\mu_1b_1+\ldots+\mu_mb_m=0$ then 
$\lambda_1a_1+\ldots+\lambda_na_n=-\mu_1b_1-\ldots-\mu_mb_m\Rightarrow \\ \lambda_1a_1+\ldots+\lambda_na_n=-\mu_1b_1-\ldots-\mu_mb_m \in S\cap T \Rightarrow \\ \lambda_1a_1+\ldots+\lambda_na_n=-\mu_1b_1-\ldots-\mu_mb_m=0 \ (\text{why ?})\Rightarrow \\ \lambda_1=0, \ \ldots\lambda_n=0, \ \mu_1=0, \ \ldots\mu_m=0 (\text{why ?}).$
