# In $\mathscr{V}$, let $X \subset \mathscr{V}$ be a set of $n$ vectors. $Y \subset X$ contains vectors all scalar multiples, $X$ linearly dependent.

I would just like to verify that my proofs are sound and receive any suggestions on rewording. (If relevant, I am self-studying and haven't done a serious proof in about a year.) $\mathscr{V}$ is a vector space, and $\mathscr{F}$ is the field containing the scalars.

Lemma. Let $X = \{x, kx\} \subset \mathscr{V}$, $k \in \mathscr{F}$. Then $X$ is linearly dependent.

Proof. Suppose there exist $a_1, a_2 \in \mathscr{F}$ such that $a_1x + a_2 kx = (a_1 + a_2k)x = 0$.

If $k = 0$, choose $a_2 \in \mathscr{F} - \{0\}$ and $a_1 = 0$. Hence $X$ is linearly dependent.

If $k \neq 0$, let $a_1 = -a_2k$. As long as $a_2 \neq 0$, we have satisfied the equality, hence $X$ is linearly dependent.

Theorem. Suppose $X \subset \mathscr{V}$ is a set of $n < \infty$ vectors. If there exists a linearly dependent set of $m \leq n$ vectors in $X$, then $X$ is linearly dependent.

Proof. Suppose $X = \{x_1, x_2, \dots, x_n\}$ is a set of linearly independent vectors. Then without loss of generality, consider the set $X_m = \{x_1, x_2, \dots, x_m\}$, $m \leq n$. Since $X$ is linearly independent, $$\sum\limits_{i=1}^{n}a_ix_i = 0 \implies a_i \equiv 0\text{.}$$ Thus, $$\sum\limits_{i=1}^{m}a_ix_i = 0 \implies \sum\limits_{i=1}^{m}a_ix_i + \sum\limits_{i=m+1}^{n}0x_i = 0\text{,}$$ and hence $a_i \equiv 0$ due to linear independence of $X$. Hence $X_m$ is linearly independent.

Corollary. Suppose $X \subset \mathscr{V}$ is a set of $n < \infty$ vectors. If there exists a set of $m$ vectors ($2 \leq m \leq n$) vectors in $X$, say $Y_m$, such that the vectors in $Y_m$ are scalar multiples of each other (that is, for each $y \neq z$, both in $Y_m$, we can write $y = kz$ for some $k \in \mathscr{F}$), then $X$ is linearly dependent.

Proof. Perform induction on $m$.

If $m = 2$, the Lemma states that a set with two vectors satisfying the given property is linearly dependent, and hence by the Theorem, $X$ is linearly dependent.

Now let $m = k$ be such that $2 \leq k \leq n$. Assume that a subset of $X$ with $k$ vectors satisfying the property given in the statement of the Corollary is linearly dependent, and hence $X$ is linearly dependent by Theorem.

We show that a subset of $X$ with $k+1$ vectors satisfying the property given in the statement of the Corollary is linearly dependent. Remove one of the vectors from this set of $k+1$ vectors. Then this set of $k$ vectors is linearly dependent by the induction hypothesis, and by Theorem, the subset of $X$ with $k + 1$ vectors is linearly dependent. Hence $X$ is linearly dependent by Theorem. $\square$

If $k = 0$, choose $a_2 \in \mathscr{F} - \{0\}$ and $a_1 = 0$. Then $a_1x + a_2kx = 0$. Hence $X$ is linearly dependent. If $k \neq 0$, let $a_1 = -a_2k$, $a_2 \in \mathscr{F} - \{0\}$. Thus $a_1x + a_2kx = 0$, hence $X$ is linearly dependent.
In $Y_m$, pair two of the vectors together. By Lemma, this pair of vectors forms a linearly dependent set, and by Theorem, $X$ is linearly dependent.