- #1
Adyssa
- 203
- 3
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!