I'm having trouble answering this question that is based on linear independence: Let $V$ be a vector space over a field $F$ and let $W$ and $Y$ be subspaces of $V$ such that $W \cap Y = \{ \vec{0} \}$. Suppose $w⃗_1,w⃗_2,…,w⃗_n$ are linearly independent vectors in $W$ and $\vec{y}_1,\vec{y}_2,…,\vec{y}_m$ are linearly independent vectors in $Y$. Show that $\{ w⃗_1, w⃗_2, . . . , w⃗_n,\vec{y}_1,\vec{y}_2, . . . ,\vec{y}_m \}$ is linearly independent in $V$.
I know that since both $w⃗_1,w⃗_2,…,w⃗_n$ and $\vec{y}_1,\vec{y}_2,…,\vec{y}_m$ are linearly independent, I can say that they both equal $0$ if there exists some $p$ and $q$ such that $pw⃗_1,pw⃗_2,…,pw⃗_n$ and $q\vec{y}_1,q\vec{y}_2,…,q\vec{y}_m$. Does this mean that $\{ w⃗_1, w⃗_2, . . . , w⃗_n,\vec{y}_1,\vec{y}_2, . . . ,\vec{y}_m \}$ is then linearly independent or do I need to prove something else?
Best Answer
We must have $$\sum_i \alpha_i w_i + \sum_j \beta_j y_j = 0$$ with at least one $\alpha_i \ne 0$ and at least one $\beta_j \ne 0.$
Write this as $w + y = 0$ so that $w=-y \in Y$ (since $y\in Y$ and $Y$ is a subspace).
Contradiction, because $W \cap Y = \{ 0 \} \Rightarrow w \notin Y $.