# Show set of vectors is a subspace of R^3

Problem:

Determine which of the following are subspaces of ${\bf R}^3$.

All vectors of the form $(a, b, c)$, where $b = a + c + 1$.

My answers: Thought process: to show if a set of vectors called W is a subspace, it must follow Axiom 1 and 6, closed under addition and scalar multiplication respectively.

$(a, b, c)$, where $b = a + c + 1$.

$(a,a+ c + 1, c) + (a,a + c + 1, c) = (2a,2a + 2c + 2, 2c)$

No. It is not closed under addition because the form completely changed specifically the y component of the vector completely changed.

$2(a, a + c + 1,c) = (2a, 2a + 2c + 2, 2c)$.

No. It is not closed under scalar multiplication either because the form completely changed specifically the y component of the vector completely changed.

Since it is not closed under addition and scalar multiplication, I can say it is not a subspace of ${\bf R}^3$.

Is this the right way to think about figuring out if a set of vectors is a subspace by thinking of the final result has the same form as the original vector?

• Your idea is correct, but it is enough to show that one of your axioms is not satisfied. So you could stop after showing that the set is not closed under addition. Another idea: Every vector space contains the null vector, but your set doesnt. Why? – sranthrop Apr 10 '14 at 0:05
• I never heard of a null vector. Do you mean a "zero space that contains a zero vector"? This was an example from a textbook so I have no control over having a null vector in the vector space. – Nicholas Apr 10 '14 at 0:10
• null vector = zero vector = (0, 0, 0) – sranthrop Apr 10 '14 at 0:14

Generally, the best way to show that a set isn't a subspace is with a specific example of one of the necessary properties failing. In this case, it is easy to see that $(0,0,0)$ isn't in the set, and so it isn't a subspace. What you've done is okay though.
• Yes, your thought process is okay, but you might have been better served by using a numerical example. For instance, show that $(1,3,1)$ is in the set, but not $2 \cdot (1,3,1)$. Note that, if it is closed under scalar multiplication then $(0,0,0) = 0 \cdot (x,y,z)$ must be in the set for any $(x,y,z)$ in the set, and as we saw $(0,0,0)$ is not in the set. By the way, it is okay to check only that closure under addition and closure under scalar multiplication hold, so long as you also assume that the subset is nonempty. – user139388 Apr 10 '14 at 0:16