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I need to confirm if the set of all triples of real numbers of the form $(0, y, z)$ where $y=z$ with standard operations of addition and scalar multiplication on $\mathbb{R}^3$ is a vector space. Any clues will be greatly appreciates.

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What have you tried? There is not actually any trick here, it is just a matter of applying the definitions. – Tobias Kildetoft Aug 8 '13 at 8:41
Well the "trick" is not computing all axioms but just checking if it is a subspace of $\mathbb{R}^3$ – Dominic Michaelis Aug 8 '13 at 8:42
You'll save yourself a lot of hesitating and problems if you can prove (and/or convince yourself) that a parametric set of vectors in $\,\Bbb R^n\,$ is a vector subspace iff each and every coordinate is given by an homogeneous polynomial on the $\,n\;$ variables of (homogeneous) degree $\;1\;$, i.e. of the form $\;a_1x_1+\ldots+a_nx_n\;,\;\; a_i\in\Bbb R\;,\;\;\forall\,i$ – DonAntonio Aug 8 '13 at 9:58
Thanks very much for all the responses. Please forgive my noobness for apparently posting an off-topic question. Will try to be more relevant in the future. – John G Aug 8 '13 at 11:32
Your question is not actually off-topic in the usual sense of the term: there's nothing wrong with the topic at all. Some people here are venomously opposed to questions that do not include some of your own work and seem to go out of their way to put such questions on hold. You can protect yourself from them by including some indication of what you've tried. Fortunately, you got some useful answers first. – Brian M. Scott Aug 8 '13 at 17:03

First check to see if the zero vector is in your subset. I am calling it a subset because we don't know if it is a subspace yet or not. Then check to see if you pick any two arbitrary vectors in your subset then their sum is also in the subset. Finally, the third thing to check is to see if you pick any arbitrary vector and then multiply it by any arbitrary real scalar then the result is also in your subset. If all three conditions hold, then you have a subspace. If any one of them fails then you don't have a subspace.

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Hint: For all $y \in \mathbb{R}$, $(0,y,y)=y \cdot (0,1,1)$.

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Three possible approaches, choose one: (Denote your set by $S$)

  1. Check, that $S$ fulfills the vector space axioms. (tedious)
  2. Show, that $S\subset\mathbb R^3$ is a subspace (provided you already know, that $\mathbb R^3$ is a vector space). This is simpler, you just need to check closure.
  3. Note, that $S\to\mathbb R, (x,y,z)\mapsto y$ is a linear bijection, hence an isomorphism (between the loose structure of a set with two binary operations).
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