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## Homework Statement

Determine if the set is a subspace:

S = {(a, b, c) 2 R3 | a − 2b = 0 and 2a + b + c = 0};

## Homework Equations

as above

## The Attempt at a Solution

It is a subspace, I'm just not 100% sure how to write up the proof. So far I have this:

The set is non-empty, as it contains the zero vector (0, 0, 0).

0 - (2)0 = 0 and (2)0 + 0 + 0 = 0

Let X = (2, 1, -5) and Y = (4, 2, -10) (element of) S and z be any real number.

X + Y = (2 + 4, 1 + 2, -5 + -10) = (6, 3, -15) and 6 - (2)3 = 0 and (2)6 + 3 + -15 = 0

(z)X = (z)(2, 1, -5) = ((z)2, (z)1, (z)(-5)) and (z)2 - (2)(z)1 = 0 and (2)(z)2 + (z)1 + (z)(-5) = 0

Is it alright to use actual numbers in my proof? All the examples we did in class used letters only, and I wonder if I'm only proving it for two vectors, and not

*all*the possible vectors in the subspace? Is it enough to show it like this? Also I'm a bit sketchy with my example of closure under scalar multiplication, I know it works for any z, but do I need a concrete example?

Proofs are a bit confusing!