Let $U,U'$ be linear subspaces of the $\mathbb{K}$-vectorspace $\,V$, with $\,\,U \cap U'=0$.
Show that if $x_1,...,x_r \in U$ and $y_1,...,y_s \in U'$ are both linear independent systems, so
the system $x_1,...,x_r,y_1,...,y_s$ is also linear independent in $V$
My attempt:
Suppose the system $x_1,...,x_r,y_1,...,y_s$ is linear dependent.
This means (w.l.o.g we choose an element of the system in regards of $U$)
$$\exists 1\le i \le r:x_i= \sum\limits_{l \in L}\alpha_lx_l+\sum\limits_{k \in K}\beta_ky_k$$
with $L\subseteq \lbrace 1,...,r\rbrace \setminus \lbrace i\rbrace, K\subseteq\lbrace 1,...,s \rbrace$ and $\alpha_l,\beta_k \in \mathbb{K}$
Obviously $\sum\limits_{l \in L}\alpha_lx_l \ne 0$ or else $x_i \in U'$ and obviously $\sum\limits_{k \in K}\beta_ky_k \ne 0$ or else the system
in regards to $U$ would be linear dependent.
$$x_i= \sum\limits_{l \in L}\alpha_lx_l+\sum\limits_{k \in K}\beta_ky_k \Longleftrightarrow x_i-\sum\limits_{l \in L}\alpha_lx_l=\sum\limits_{k \in K}\beta_ky_k$$
But since $x_i-\sum\limits_{l \in L}\alpha_lx_l \ne 0 \Longrightarrow x_i-\sum\limits_{l \in L}\alpha_lx_l \in U'\Longrightarrow U \cap U'=\lbrace0,x_i-\sum\limits_{l \in L}\alpha_lx_l\rbrace$
which is a contradiction!
$\Box$
Would be great if someone could look over it and verify my solution, or give me feedback to solve the issues :)
You could also give a direct proof. Suppose
$$(\star) \hspace{0.5cm} \sum_{i = 1}^{r} \alpha_{i}x_{i} + \sum_{i = 1}^{s} \beta_{i}y_{i} = 0. $$ Then
$$ \sum_{i = 1}^{r} \alpha_{i}x_{i} = - \sum_{i = 1}^{s} \beta_{i}y_{i} . $$
From this, we can deduce that $\sum_{i = 1}^{r} \alpha_{i}x_{i} \in U \cap U^{'}$, so
$$ \sum_{i = 1}^{r} \alpha_{i}x_{i} = 0 .$$
Since $x_{1}, \ldots, x_{r}$ are linearly independent, it follows that $\alpha_{1} = \ldots = \alpha_{r} = 0$, so $(\star)$ becomes
$$ \sum_{i = 1}^{s} \beta_{i}y_{i} = 0. $$
Using the fact that $y_{1}, \ldots, y_{s}$ are linearly independent, we deduce that $\beta_{1} = \ldots = \beta_{s} = 0$ and we are done.