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Let S be the collection of vectors $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ in $\mathbb{R}^3$ such that $|x-y|=|y-z|$. Either prove that S forms a subspace of $\mathbb{R}^3$ or give a counterexample to show that it does not.

(from Poole, Linear Algebra - A Modern Introduction 2nd ed., exercise 3.5.8)

I'm working on this book by myself, so if you can just give me a hint to show me the right direction, that would be great. I have been trying to solve this by going through the three requirements of a subspace.

  1. The zero vector is in S.
  2. If u and v are in S, then u + v is in S.
  3. If u is in S and c is a scalar, then cu is in S.

It's easy to see the first one holds. For the third one I noticed that by adding c as multiplier on both sides of $|x-y|=|y-z|$ and fiddling a bit it becomes $|$c$u_x$-c$u_y|$ = $|$c$u_y$-c$u_z|$, which is the form I need to show it is closed under scalar multiplication.

But how to show the second one I have no idea. Stuff I tried tended to split into multiple cases because absolute values are taken of the left and right sides, it didn't feel like the right direction. Is there something really easy I'm missing here?

I also tried imagining what the subspace would look like, but it's too difficult to visualize.

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up vote 5 down vote accepted

What about vectors $\begin{bmatrix} 3\\ 7\\ 11 \end{bmatrix}$ and $\begin{bmatrix} 2\\ 8\\ 2 \end{bmatrix}$? These are both in $S$, but $\begin{bmatrix} 5\\ 15\\ 13 \end{bmatrix}$ is not in $S$.

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How did you create this counterexample? – Bemmu Feb 16 '11 at 7:00
@Bemmu If you strictly increase the entries in both vectors by some fixed value, you find that their sum is again in $S$. So to switch it up, I chose a second vector where I increase by $6$, and then decrease by $6$, on a hunch that it would give a counter example. If I had chosen instead $[2,8,14]$, notice that $[5,15,25]$ is in $S$. I think the key is to 'build' the vectors differently and a counterexample will pop out. – yunone Feb 16 '11 at 7:07

An easy way to find a counter-example is to take a vector of form $\begin{bmatrix} a\\ b\\ c \end{bmatrix}$ where $|a-b|=|b-c|$ holds, but, for example, $a=b+c$ (or $c=a+b$). Then add some vector of form, say, $\begin{bmatrix} x\\ 0\\ x \end{bmatrix}$ to your first vector to derive a counterexample.

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This may be the geometric angle for why you don't get a subspace:

Break down your problem into the 4 possible cases for absolute values:

One case would be:

x-y>=0 y-z>=0

etc. And check what happens where the 4 regions overlap

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The steps is the following 1. find a [x y z] where /x-y/ = /y-z/ For example, [y-4 y y+4 ] , that is if y is any integer, say 7? so, the first vector is [ 3 7 11]. 2. you need find another vector, let y=8, hence [y-6 y y+6 ]=[2 8 14]. 3. test, [3 7 11] + [2 8 14] = [5 15 25], ops... this satisfies /x-y/=/y-z/ 4. so let's try if [y-6 y y-6] works, let y=8, then we have [ 2 8 2 ] 5. test, [3 7 11] + [ 2 8 2] = [ 5 15 13], this does not satisfy /x-y/ =/y-z// 6. yes. we find a counter example

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I suggest reformatting your formulas to make them easier to read: – Joonas Ilmavirta Nov 29 '14 at 21:42

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