1.5 Exercise 21

Let $S_1$ and $S_2$ be disjoint linearly independent subsets of $V$. Prove that $S_1 \cup S_2$ is linearly dependent if and only if $span(S_1) \cap span(S_2) \neq \{0\}$.

Proof:
$\Rightarrow$: Since $S_1 \cup S_2$ is linearly dependent, then there exists vectors $u_1,u_2,...,u_n$ in $S_1$, the vectors $w_1,w_2,...,w_m$ in $S_2$ and not all zero scalars $a_1,...a_n,b_1,...,b_m$ such that

$$ \begin{align*} a_1u_1 + a_2u_2 + ... + a_nu_n + b_1w_1 + b_2w_2 + ... + b_nw_m &= \bar{0}. \\ \end{align*} $$

Re-arranging the terms

$$ \begin{align*} a_1u_1 + a_2u_2 + ... + a_nu_n &= - (b_1w_1 + b_2w_2 + ... + b_nw_m). \end{align*} $$

Let $w = a_1u_1 + a_2u_2 + ... + a_nu_n = - (b_1w_1 + b_2w_2 + ... + b_nw_m)$. We claim that $w$ is a non-zero vector. Why? suppose for the sake of contradiction that it was the zero vector. We said earlier that not all the scalars are zero. This is a contradiction because $S_1$ and $S_2$ are linearly independent. This means that the solution to $a_1u_1 + a_2u_2 + ... + a_nu_n = - (b_1w_1 + b_2w_2 + ... + b_nw_m) = \bar{0}$ must only be the trivial solution where all the scalars are zero. Therefore we must have $w$ be a non-zero vector.

Futhermore, $w = (a_1u_1 + a_2u_2 + ... + a_nu_n)$ is a linear combination of the elements in $S_1$ and so it’s a vector in $Span(S_1)$ by the defintion of the span of a set. Similarly, $w = - (b_1w_1 + b_2w_2 + ... + b_nw_m)$ is a vector in $Span(S_2)$. From this, we see that $w$ is in both the span of $S_1$ and $S_2$ and so $w \in Span(S_1) \cap Span(S_2)$.

$\Leftarrow$: Suppose now that $span(S_1) \cap span(S_2) \neq \{\bar{0}\}$. Let $v$ be a non-zero vector in both $Span(S_1)$ and $Span(S_2)$. This means we can write $v$ a linear combination of the elements of $S_1$,

$$ \begin{align*} v = a_1u_1 + a_2u_2 + ... + a_nu_n. \end{align*} $$

And we can also write $v$ as a linear combination of the elements in $S_2$ as follows,

$$ \begin{align*} v = b_1w_1 + b_2w_2 + ... + b_nw_m. \end{align*} $$

From this we see that,

$$ \begin{align*} &a_1u_1 + a_2u_2 + ... + a_nu_n = b_1w_1 + b_2w_2 + ... + b_nw_m. \\ &a_1u_1 + a_2u_2 + ... + a_nu_n - (b_1w_1 + b_2w_2 + ... + b_nw_m) = \bar{0}. \end{align*} $$

Since $v$ is non-zero and both $S_1$ and $S_2$ are linearly independent, then all of these scalars are non-zero and so this means that $S_1 \cup S_2$ is linearly dependent by the definition of linear dependence. $\blacksquare$

References:

Linear Algebra 5th Edition