Finite dimensional vectors spaces external direct products and linear maps. Suppose that V is a finite dimensional vector space over a field $\Bbb F$, and that $T \in L(V )$. Suppose also that $V = U_1 \oplus U_2 $, where $U_1 $ and $U_2$ are T-invariant. 
Let $\beta_1 $ and $\beta_2$ be bases for $U_1 $ and $U_2$,
respectively, and let $\beta = \beta_1 \cup \beta_2$. 
I have two questions, i can split them up into different questions if needed but they are fairly related.
Edit:
How to show that $\beta = \beta_1 \cup \beta_2$ is actually a basis for $V$?
 A: Part 2 is a bit easier, so let's answer that first:
By the definition of a direct sum, $U_1,U_2$ are such that every $v \in V$ can be uniquely written as a sum $v = u_1 + u_2$ with $u_1 \in U_1$ and $u_2 \in U_2$.  Suppose that $\beta_1 = \{v_1,\dots,v_m\}$, and $\beta_2 = \{w_1,\dots,w_m\}$.  It's easy to show that $\beta_1 \cup \beta_2$ spans $U$, but independence can be a bit tricky. 
We can show that the union is linearly independent as follows: suppose coefficients $c_i,d_i$ are such that
$$
\sum_{i=1}^m c_iv_i + \sum_{i=1}^n d_iw_i = 0
$$
We wish to conclude that all coefficients are equal to zero.  Noting that $\sum_{i=1}^m c_iv_i \in U_1$ and $\sum_{i=1}^n d_iw_i \in U_2$ are two vectors that add up to $0$, the definition of a direct sum allows us to conclude that $\sum_{i=1}^m c_iv_i = 0$ and $\sum_{i=1}^n d_iw_i = 0$.  From there, the linear independence of the individual bases allows us to conclude that all $c_i$ are $0$ and all $d_i$ are $0$, as desired.

Now, as for part 1: the key is to have a sufficiently clear picture of what $[T]_\beta$ (which I will write instead of $[T]_\beta^\beta$) means.  Here's the definition I'd prefer to work with: let $a_{ij}$ denote the entries of $A = [T]_\beta$.  Then these entries are chosen such that if $\beta = \{v_1,\dots,v_d\}$, we have
$$
T(v_j) = \sum_{i=1}^d a_{ij} v_i
$$
Now, a hint to get you started: using the $T$-invariance of the subspaces, which entries of $A$ can you guarantee will be zero?  
