Question: show that the set L={(x,y,z,w) is in $\displaystyle R^4$ | x+y-z+w=0} with operations from $\displaystyle R^4$ is a vector space.

Answer: show that it's closed under vector addition (between itself and an arbatrary vecotr) and closed under scalar multiplication (between itself and an arbitrary constant)

Confusion: I thought that's what's done to show something is a subspace? To show something is a vector space, the set must satisfy 10 properties including things like

$\displaystyle \mathbf{u} \oplus \mathbf{v} = \mathbf{v} \oplus \mathbf{u}$ for **u** and **v** in the vector space

and

$\displaystyle (c+d) \odot \mathbf{u} = c \odot \mathbf{u} \oplus d \odot \mathbf{u}$

so why is it in the answer they only showed that it was closed under scalar multiplication and vector addition and didn't need to show the other properties, for example like the two I list above?