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Let $H$ be:

$$H=\{(x, y, z, w)^T : x^2-y^2+3w^3=0\}$$

I know it has to be closed under addition and scalar multiplication, and it does contain the 0 vector. But how do I prove it's closed under the first two? Do I just keep finding examples for what $x, y, z$, and $w$ could be and just keep scaling them and adding them? There's gotta be a more efficient way than that. Help please?

HipsterMathematician
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user103068
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  • This is not a vector space. It's not closed under scalar multiplication. – lhf Nov 02 '13 at 19:00
  • The most efficient way here is to show that $H$ is not closed under either addition or scalar multiplication (assuming the base field is $\mathbb R$ or similar) – Hagen von Eitzen Nov 02 '13 at 19:00

2 Answers2

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Well, it is not a subspace. The way you can know beforehand (and if you prove the following then you will always know) is: when given as a function of the coordinates, a subset of $\;\Bbb R^n\;$ is a subspace iff it is a homogeneous equation in the $\;n\;$ coordinates of degree $\;1\;$, i.e. iff it is of the form

$$a_1x_1+...+a_nx_n=0\;,\;\;a_i\in\Bbb R$$

DonAntonio
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Hint: Is $(1,1,0)\in H$? is $(1,-1,0)\in H$? Is their sum $\in H$ (this assumes that $2\ne 0$; as a matter of fact, $H$ would be a subspace of $\mathbb F_2^4$)?

Hagen von Eitzen
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