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I'm struggling with a simple question of the homework.

Consider a set P s.t $$P = \{x \in R^{3} : x_1 + 2x_2 + 3x_3 = 1\}$$

Then it was asked to prove that this set is an affine set. I heard that one way to answer this question is to select two points $s_1$, $s_2$ s.t $s_1,s_2 \in P$ and show that $x(\theta) = \theta*x_1 + (1 - \theta) * x_2 \ \forall \ \theta$, but still I don't know how to start. For some way, I have a thought that induction is needed, or that I am missing some concepts.

Andrés E. Caicedo
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José Joaquim
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    What is your definition of "affine set"? – Alex Kruckman Apr 29 '19 at 16:14
  • Finish the phrase starting with with "to select two ..." with "is in $P$". Then prove that and you won't need the "but still ...". No induction needed. – Ethan Bolker Apr 29 '19 at 17:58
  • @AlexKruckman My main reference is Lay's et al. book, "Linear Algebra and Its Applications", 5th ed. It says: "A set is affine if p, q in S implies that (1 - t)p + tq is in S for each real number t". – José Joaquim Apr 29 '19 at 18:31
  • @EthanBolker But why is it sufficient to select two particular points that satisfy the condition to answer the question? Doesn't the "for all" operator imply that, in easy words, all pairs must be verified? – José Joaquim Apr 29 '19 at 18:36
  • Yes, that's what "for all" means. Since you do not identify the two points other than saying that they are in $P$ you are in fact dealing with all pairs of points. Perhaps you'd be more comfortable replacing "particular" with "arbitrary". – Ethan Bolker Apr 29 '19 at 20:08

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There's an issue here about the meaning of "select two points" - to verify the definition (given in the comments), it doesn't suffice to pick two particular points $p$ and $q$ in $P$ and verify that $(1-t)p + tq$. We have to verify this for all pairs of points. But often people will say "select two points $p,q\in P$" when they want to reason about arbitrary points.

Anyway, this comes down to a simple computation verifying the definition. Let $p,q\in P$. Let's write $p = (a_1,a_2,a_3)$ and $q = (b_1,b_2,b_3)$. We know $a_1+2a_2+3a_3 = 1$ and $b_1 + 2b_2 + 3b_3 = 1$.

Now we need to show $(1-t)p + tq \in P$ for any $t\in \mathbb{R}$. Let's call this point $r = (c_1,c_2,c_3)$. Let's write out the coordinates of this point: \begin{align*} r &= (1-t)p + tq\\ &= (1-t)(a_1,a_2,a_3) + t(b_1,b_2,b_3)\\ &= ((1-t)a_1, (1-t)a_2, (1-t)a_3)) + (tb_1,tb_2,tb_3)\\ &= ((1-t)a_1 + tb_1,(1-t)a_2 + tb_2,(1-t)a_3 + tb_3) \end{align*}

Now let's check if it's in $P$: \begin{align*} c_1 + 2c_2 + 3c_3 &= ((1-t)a_1 + tb_1) + 2((1-t)a_2 + tb_2) + 3((1-t)a_3 + tb_3)\\ &= (1-t)(a_1 + 2a_2 + 3a_3) + t(b_1 + 2b_2 + 3b_3)\\ &= (1-t)(1) + t(1) \\ &= 1 \end{align*} So it is!

You can easily abstract from this argument to show that for any $a,b,c,d\in \mathbb{R}$, the set $$P = \{(x,y,z)\in \mathbb{R}^3\mid ax + by + cz = d\}$$ is affine.

Alex Kruckman
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