Show that set of all solutions $(a, b, c)$ of the equation $a+b+2c=0$ is a subspace of a vector space $V^3(R)$.

I wonder if I should show that

(1) the solution set satisfies all the axioms of a vector space?

(2) for the solution set $W=\lbrace (a,b,c): a, b, c \in R \rbrace$, $\alpha(a_1, b_1, c_1)+\beta(a_2,b_2, c_2)\in W$ for all $\alpha,\beta\in R$, $(a_1, b_1, c_1),(a_2,b_2, c_2)\in W$?

(3) the zero vector exists in $W$ and it is closed under vector addition and scalar multiplication?

I am unable to proceed as I don’t know which conditions would I should prove.

  • $\begingroup$ You need to show that if $(a,b,c) in W$ then $\lambda (a,b,c) \in W$ for all scalars $\lambda$ and if $(a_k,b_k,c_k) \in W$ then $(a_1+a_2,b_1+b_2,c_1+c_2) \in W$. Being in $W$ in this case is tantamount to showing that the condition $(a+b+2c) = 0$ is satisfied. $\endgroup$ – copper.hat Aug 7 '18 at 4:01

You don't need to show that it satisfies all the axioms of a vectors space explicitly, because some are inherited from the surrounding space. For example, addition is commutative and associative, scalar multiplication distributes over addition, etc.

Proving $2)$ and $3)$ are almost the same, but not quite. As pointed out by @zipirovich, the problem with $2$) is that if there are no solutions, then $2)$ holds vacuously, but the empty set is not a vector space. So, you can prove that the solution set is not empty and $2)$ holds, or you can prove $3).$ Of course, $\mathbf 0$ is clearly a solution, so in practice, proving $2)$ and $3)$ are pretty much the same.

  • $\begingroup$ Your last claim isn't exactly true. The empty subset is closed under addition and multiplication by scalars, but it's not a vector subspace because it does not contain $\mathbf{0}$. Showing that a subset contains $\mathbf{0}$ may be replaced with showing that it's non-empty. But just the two conditions -- of being closed under addition and multiplication by scalars -- are not sufficient. $\endgroup$ – zipirovich Aug 7 '18 at 3:58
  • $\begingroup$ You are correct. I overlooked that. Thanks. $\endgroup$ – saulspatz Aug 7 '18 at 4:00

To show that a subset of a vector space is a subspace, it suffices to show that the zero vector is in the subset and it closed under vector addition and scalar multiplication.

That is your option (3) in your list.


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