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I have a set of $3$ vectors in $\mathbb{R}^{2}$. The vectors are $v_{1} = (1, 3)$, $v_{2} = (5, 15)$ and $v_{3} = (69, 207)$.

The vector $v_{1}$ can be multiplied by $5$ to get $v_{2}$, and $v_{1}$ can be multiplied by $69$ to get $v_{3}$. I know this means that $v_{3}$ and $v_{1}$ are linear combinations of each other and that $v_{2}$ and $v_{1}$ are linear combinations of each other, but what does this mean for $v_{2}$ and $v_{3}$?

There is no scalar value that can be multiplied to $v_{2}$ to get $v_{3}$ and vice versa, so does this mean $v_{3}$ and $v_{2}$ are NOT linear combinations of each other?

Átila Correia
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Alex S.
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    Scalars don't have to be integers. You can multiply by, say, $69/5$; that's still a scalar. – Brian Tung May 01 '22 at 00:50
  • Additionally: if a set of vectors contains a linearly dependent pair, then the set contains a vector that is a linear combination of the others; so, by definition, the entire set must be linearly dependent. – ryang May 01 '22 at 05:34

1 Answers1

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Brian Tung essentially answered in his comment, but I'll provide a more detailed response here.

As you correctly stated:

$v_2 = 5v_1$ and $v_3 = 69v_1$

We can explicitly solve for $v_1$ in terms of $v_2$ using our first equation, and then substitute in our second equation to get $v_3$ in terms of $v_2$.

Specifically, we get $v_3 = \frac{69}{5}v_2$, so $v_3$ and $v_2$ are indeed linear combinations of each other.

Note: Formally, we would need to give a field, such as $\mathbb{R}$ or $\mathbb{C}$, that our scalar coefficients are members of, to avoid such ambiguities. Unless explicitly stated, there is no reason to restrict your scalars to integers. Any real number, such as $\frac{69}{5}$, is perfectly acceptable.