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**Drexel28** Ok. That's fine. So let's go through an example, ok?

A subspace must be closed under scalar multiplication and vector addition. So, for example F. is not a subspace since if we let $\displaystyle V=\left\{(x,y,z)\in\mathbb{R}^3:x+y+z=1\right\}$ then surely $\displaystyle (1,0,0)\in V,(0,1,0)\in V$ but since $\displaystyle (1,0,0)+(0,1,0)=(1,1,0)$ has the quality that $\displaystyle 1+1+0\ne 1$ we see that $\displaystyle (1,0,0)+(0,1,0)\notin V$. But, this HAS to happen in a subspace. So, $\displaystyle V$ is not a subspace.