Show that, if $X+Y = X⊕ Y$ then $A \cup B$ is a basis for $X+Y$ 
Let $X$ and $Y$ be subspaces of $V$, and let $A$ and $B$ be bases of $X$ and $Y$ respectively. Show that, if $X+Y = X⊕ Y$ then $A \cup B$ is a basis for $X+Y$

This part is about obtaining new spaces from existing ones. Firstly, we give a couple of definitions.
Let $X$ and $Y$ be two subspaces of of a vector space $V$ . 
Their sum $X + Y$ is defined by
$$
X + Y = \{x + y : x ∈ X \text{ and } y ∈ Y \}
$$
If, in addition, each vector in $X + Y$ is written uniquely as $x + y$ where x ∈ X and y ∈ Y
then we say that $X + Y$ is the direct sum of $X$ and $Y$ , written $X ⊕ Y$.
 A: Hint: In order for $A \cup B$ to be a basis of $X \oplus Y$, there are two requirements.
Using the fact that $A$ spans $X$ and $B$ spans $Y$, show that $A \cup B$ spans $X \oplus Y$.  Using the fact that $A$ and $B$ are linearly independent sets and the definition of $X \oplus Y$, show that $A \cup B$ is a linearly independent set.
A: Let $X$and $Y$ be subspaces of vector space $V$ and $ {{x_1,...x_n}} =A $, a basis for $X$ and $y_1,...y_n = B$, a basis for Y.
Because $X\oplus Y$ for some $x \in X$ and $y \in Y$, $x+y$ can be uniquely written as 
$$a_1x_1+b_1y_2+a_1x_2+b_2y_2+ \cdots+ a_nx_n+b_ny_n .$$
because each $x= a_1x_1+\cdots+a_nx_n$ and each $y = b_1y_1+\cdots b_ny_n$ can be written so.
Now if our list is ${x_1,...x_n, y_1,...,y_n}$ linearly dependent then we have shown that it also spans  ($X+Y$) and therefore a basis of the sum of the two subspaces. 
If not then since $x_j$ is linearly independent in $ {{x_1,...x_n}}$ and $y_j$ is linearly independent in $y_1,...,y_n$, the list can be only linearly dependent if for some $x_j=c_jy_j$. Replace (using induction) all $k $ such linearly dependent vectors until we have a linearly independent list
$${x_1,...x_n, y_1,...,y_{n-k}}.$$
Therefore we have the linearly independent basis set $A  \ \cup B$ where there where $k$ elements in common.
This also spans $X+Y$ and therefore is a basis of the sum.
Let me know if you have any questions!
