Which of the following subsets of $R^3$ are actually subspaces?
(a) The plane of vectors $(b_1 , b_2 , b_3)$ with $b_1 = b_2$
(b) The plane of vectors with $b_1 = 1$.
(c) The vectors with $b_1b_2b_3 = 0$.

Is my answer for (a) correct and I don't understand what (b) and (c) mean?
(a) this is a subspace because if you have three vectors in $R^3$ and two of them are equal that means you have a plane in $R^3$ and that is closed under addition and subtraction.


b) is not a subspace, $(1,2,3) + (1,2,3) = (2,4,6)$

c) is not s subspace, it describes your coordinate axes. $(1,0,0) + (0,1,1) = (1,1,1)$, which is not on an axis.

  • $\begingroup$ Is the way I stated (a) correct? $\endgroup$ – idknuttin Feb 26 '16 at 23:17
  • $\begingroup$ Yes, you need to check for closure under scalar multiplication, addition and that it contains the zero vector, which it does. $\endgroup$ – Kevin Sheng Feb 26 '16 at 23:18
  • $\begingroup$ for (a) an example would be $(2, 2, 3) + (3, 3, 6) = (5, 5, 9)$ and $x(b_1, b_2, b_3)=xb_1, xb_2, xb_3$ I know it is closed under addition and scalar multiplication $\endgroup$ – idknuttin Feb 26 '16 at 23:20
  • $\begingroup$ You need to check for closure in a more general case (giving an example is not enough). But yes, the examples you give are in the right direction. However, you always need to check that $0$ is there. The empty set is closed under addition and scalar multiplication (vacuously) but is not a vector space. $\endgroup$ – Kevin Sheng Feb 26 '16 at 23:21

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