# Vector Subspaces

1. Sep 4, 2012

### markovchain

Show that T = (x, y, z) : -1 ≤ x + y + z ≤ 1
is not a vector subspace of R3

Thanks!

2. Sep 4, 2012

### DonAntonio

$$(1,0,0)\in T\,\,\,but\,\,\,2(1,0,0)=(2,0,0)\notin T$$

Thus T cannot be v. subspace as it isn't closed under scalar multiplication

DonAntonio