Show direct sum the two linear spans makes the basis for $V$ Suppose that $V = M ⊕ N$ and that {$x_1, · · · , x_r$} is a basis for $M$ and {$y_1, · · · , y_s$} is a basis for $N$.
Show that {$x_1, · · · , x_r, y_1, · · · , y_s$} is a basis for $V$.
I'm not really sure how to prove this all out. I think I have an idea of what the prove is about, but I'm having a hard time putting it together. So, I think I need to show that there is some element $v$ of $V$, where $v = x+y$, for all $v$ elements of $V$ and $x$ of $M$, $y$ of $N$. I think that's that general idea but I need help showing it. I'm looking for a pretty simple proof, nothing too fancy, since I'm a beginnner.
All help is much appreciated. Thanks!
 A: To show that {$x_1, · · · , x_r, y_1, · · · , y_s$} it is base, you need to show that this set is linearly independent and that it span $V$. See that the set is linearly independent: if
$$  \alpha_1 x_1+ \cdots \alpha_r x_r+ \beta_1 y_1+ \cdots+ \beta_s y_s =0 $$
then
$$  \alpha_1 x_1+ \cdots \alpha_r x_r= -\beta_1 y_1- \cdots- \beta_s y_s $$
but on the right side, we have an element of $N$, and on the left an element of $M$. By definition of $M ⊕ N$ ($M \cap N = \{ 0 \}$), it follows that
$$  \alpha_1 x_1+ \cdots \alpha_r x_r= 0 \mbox{ and } \beta_1 y_1+ \cdots+ \beta_s y_s=0. $$
How {$x_1, · · · , x_r$} is a basis for $M$ and {$y_1, · · · , y_s$} is a basis for $N$, then $\alpha_1 = \alpha_2 = \cdots  = \alpha_r = \beta_1=\beta_2= \cdots = \beta_s=0.$
Now  {$x_1, · · · , x_r, y_1, · · · , y_s$} span $V$. In fact, how $V = M ⊕ N$ , if $v \in V$, then $v=m+n$, with $m \in M$ and $n \in N$. How {$x_1, · · · , x_r$} is a basis for $M$ and {$y_1, · · · , y_s$} is a basis for $N$, $m = \alpha_1 x_1+ \cdots \alpha_r x_r$ and $n = \beta_1 y_1+ \cdots+ \beta_s y_s$.Therefore
$$   v=m+n = \alpha_1 x_1+ \cdots \alpha_r x_r+ \beta_1 y_1+ \cdots+ \beta_s y_s. $$
so {$x_1, · · · , x_r, y_1, · · · , y_s$} is basis for $V$.
