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I'm studying about vectors and there's a question that equates 2 expressions of the same vector and requires me to find k and r:

(1-k)b + 3ka = (1-r)a + 3rb

k and r are scalar values.

How do I definitively know that the only way for the two sides to be equal is for the coefficients of a and b to be equal on both sides? Is there a way to prove it? How do you know that there is no other combination of values that will allow both sides to equate?

I'm sorry if my phrasing of the question sounds weird, I can't seem to find the right way to articulate my issue.

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    $\begingroup$ It's definitely not true in general; for example we might have ${\bf a}={\bf b}={\bf 0}$, so that all values of $k$ and $r$ work! The fact we would need, to confirm that equating coefficients is the only way for the sides to be equal, is for $\{{\bf a},{\bf b}\}$ to be a linearly independent set. $\endgroup$ Oct 30, 2021 at 17:34
  • $\begingroup$ What else do you know about the vectors $\mathbf a$ and $\mathbf b$? When they were presented to you, were you told they were linearly independent? Were you told something else? What is the entire context of the question, the parts that you're not telling us about? That's probably where the answer is. $\endgroup$
    – David K
    Oct 31, 2021 at 12:08
  • $\begingroup$ a and b are on the same plane, but are not collinear $\endgroup$
    – prata
    Oct 31, 2021 at 21:15

2 Answers 2

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  1. $$(1-k)\mathbf b + 3k\mathbf a = (1-r)\mathbf a + 3r\mathbf b\\\iff\\ 1-k=3r\;\:\text{ and }\;\:3k=1-r$$ iff $\mathbf a$ and $\mathbf b$ are not collinear.
  2. $$p\mathbf a+q\mathbf b+r\mathbf c=\mathbf 0\\\iff\\p,q,r=0$$ iff $\mathbf a, \mathbf b,$ and $\mathbf c$ are not coplanar.
  3. In general: $$p_1\mathbf a_1+p_2\mathbf a_2+\ldots+p_n\mathbf a_n=\mathbf 0\\\iff\\p_1,p_2,\ldots,p_n=0$$ iff $\mathbf a_1, \mathbf a_2, \ldots, \mathbf a_n$ are linearly independent.

Is there a way to prove it? How do you know that there is no other combination of values that will allow both sides to equate?

The third point above is actually the definition of linear independence.

A more intuitive characterisation is this: $\{\mathbf a_1, \mathbf a_2, \ldots, \mathbf a_n\}$ is linearly dependent iff some vector in the set is a linear combination of the others.

Notice that in Euclidean space $\mathbb R^m,$

  • when $n=2,$ the condition “some vector in the set is a linear combination of the others” is equivalent to “the two vectors are collinear” (i.e., one of them is a scalar multiple of the other); and
  • when $n=3,$ the same condition is equivalent to “the three vectors are coplanar”.

When the above boldfaced conditions are not met, “equating coefficients” is invalid. For example, $$p\begin{pmatrix} 1 \cr 0\end{pmatrix}+q\begin{pmatrix} 2 \cr 0\end{pmatrix}=-3\begin{pmatrix} 1 \cr 0\end{pmatrix}+4\begin{pmatrix} 2 \cr 0\end{pmatrix}\\\kern.6em\not\kern-.6em\implies p=-3,\;q=4;$$ the unknown tuple $(p,q)$ could well have been $(1,2),(3,1),$ etc.

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  • $\begingroup$ I'm sorry, I think you may have misunderstood my question, either that or I don't understand your answer. Let's simplify this. In the question I'm doing, the a and b are not collinear, but they are on the same plane, which is point 1. My question is how do you know that you can just equate the coefficients like that? I'm not asking for the answer to the question. $\endgroup$
    – prata
    Oct 31, 2021 at 7:29
  • $\begingroup$ @prata My points 1 and 3 are directly relevant to your exercise. Point 1 says that given two non-collinear vectors in $\mathbb R^n,$ equating coefficients works. Point 3 explains that this is because two non-collinear vectors automatically form a linearly-independent set of vectors. $\endgroup$
    – ryang
    Oct 31, 2021 at 7:44
  • $\begingroup$ @prata Note that if we are instead dealing with three vectors, then we require a stronger condition in order to justify that equating coefficients is valid: we now require that that the three vectors be non-coplanar (instead of merely non-collinear). Point 3 explains that this is due to three non-coplanar vectors automatically forming a linearly-independent set of vectors. $\endgroup$
    – ryang
    Oct 31, 2021 at 7:44
  • $\begingroup$ I think maybe what I'm asking is why does a linearly independent set vectors allow us to equate coefficients. Cus I still don't really understand. $\endgroup$
    – prata
    Oct 31, 2021 at 9:20
  • $\begingroup$ @prata maybe what I'm asking is why does a linearly independent set vectors allow us to equate coefficients This is by definition, as I said in the last line of the original Answer. I think your real question is this: Why is it that, for a set of two vectors, linear dependence is equivalent to collinearity; why is it that, for a set of three vectors, linear dependence is equivalent to coplanarity? I've just expanded my Answer to address this; the more we dig, the more we delve into linear algebra. $\endgroup$
    – ryang
    Oct 31, 2021 at 11:10
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Geometrically: thinking of $(1-k) \mathbf b + (3k) \mathbf a$ as $(1-k) \mathbf b + k (3\mathbf a)$, we see that as $k$ varies over $\mathbb R$, this gives us the equation of a line through the points $\mathbf b$ and $ 3\mathbf a$. As $k$ increases, the point moves in the direction of the vector $3\mathbf a - \mathbf b$. So two different values give two different points along the line.

Algebraically: we can think of $(1-k)\mathbf b + (3k)\mathbf a = (1-r)\mathbf b + 3r\mathbf a$ as a sequence of $n$ equations (if $\mathbf a, \mathbf b \in \mathbb R^n$) of the form $(1-k)b_i + 3k a_i = (1-r) b_i + 3r a_i$. We can always simplify this to $k(3a_i - b_i) = r(3a_i - b_i)$, and if $3a_i - b_i \ne 0$ (for at least one $i$) then we conclude $k=r$.

In either case, we see that if $3\mathbf a = \mathbf b$, then we can't conclude that $k=r$. In the first case, because the line is degenerate and becomes a single point; in the second case, because we cannot divide by $3a_i - b_i$ for any $i$.

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