Let U and V be two subspaces of a vector space E such that $U ∩ V = \{0\}$. Prove that if $X ⊆ U$ and $Y ⊆ V$ are two linearly independent sets then so is $X \cup Y$.
So I believe I have a good start, but it may not hold (I'm still a beginner at conducting proofs..). Feel free to critique wherever necessary!
Origonal Proof:
$X$ and $Y$ are linearly independent. Thus, they do not contain the $0$ element.
$X$ is a subset of $U$ and $Y$ is a subset of $V$, and $V \cap U = \{0\}$, so $X \cap Y = \{\emptyset\}$. (So, for any $ x \in X \cup Y$, $x \neq 0 $)
Suppose $X\cup Y$ is linearly dependent.
So, there exists $x \in X\cup Y$ and $y \in X\cup Y$ such that $x = y$ and they exist in the linear combination of other elements $\in X \cup Y$.
i.e. $ x = (\lambda a + \mu b) =y $ for $a, b \in X \cup Y$ and scalers $\lambda$ and $\mu$ in the Field.
If $ (\lambda a + \mu b) = 0$, then we are done, for $x \neq 0 \neq y$.
Let $ (\lambda a + \mu b) \neq 0$. (Note: $a \neq 0 \neq b$, and $\lambda \neq 0 \neq \mu$)
Then $a$ and $b$ are in a linearly independent set, so $X \cup Y$ is linearly independent.
Thank you for your help!
Best Answer
Suppose $X \cup Y$ is not linearly independent.
So, there exists $x_1, ... ,x_n \in X $ and $y_1, ..., y_n \in Y$ s.t. $\lambda_1 x_1 + ... + \lambda_n x_n + \alpha_1 y_1 +...+ \alpha_n y_n = 0$.
$\lambda_1 x_1 + ... + \lambda_n x_n = -(\alpha_1 y_1 +...+ \alpha_n y_)$
$(\lambda_1 x_1 + ... + \lambda_n x_n) \in$ $span(X)$, which is contained in $U$, since U is a subspace.
$-(\alpha_1 y_1 +...+ \alpha_n y_) \in$ $span(Y)$ which is in $V$, similarly.
But $span(U \cap V)=\{0\}$, so the only solution to $\lambda_1 x_1 + ... + \lambda_n x_n = -(\alpha_1 y_1 +...+ \alpha_n y_)$ is if all of the scalers, $\lambda$ and $\alpha$ are $=0$
thus shows linear independence in $X \cup Y$