Originally Posted by

**Bruno J.** Can you show it's a subspace?

Assuming that you can, if you want to show that the dimension is at least 2, consider the following two cases :

**(1)** $\displaystyle \vec a = (a_1, a_2, a_3)$ and $\displaystyle \vec b =(b_1, b_2, b_3)$ are linearly independent : then the subspace contains the span of $\displaystyle \vec a, \vec b$, which is two dimensional (in fact, try to show that it's *equal* to the span of $\displaystyle \vec a, \vec b$);

**(2)** $\displaystyle \vec a, \vec b$ are linearly dependent : then the space is all of $\displaystyle k^3$! (Can you see why?)

Can you generalize to higher dimensions?