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I am having some issues with this problem in my Linear Algebra textbook. The goal is to either show that the given set, W, is a vector space, or to find a specific example to the contrary:

\begin{Bmatrix} \begin{bmatrix} a\\ b\\ c\\ d \end{bmatrix} : \begin{matrix} 3a + b = c\\ a + b + 2c = 2d \end{matrix} \end{Bmatrix}

I understand the basic properties of Vector Spaces - such as having to contain the zero vector, being closed under addition, and being closed under scalar multiplication. I have no problem proving when these sets are not vector spaces, for example if they do not contain the zero vector.

This set appears to contain the zero vector (if you plug in 0 for a, b, c, and d, the equations are consistent). But I'm not quite sure how to prove that this set is a vector space, or how to prove that it is closed under addition and scalar multiplication.

Thanks for your help.

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    $\begingroup$ Have you learnt about vector subspaces? You can prove that this space is a vector space by exhibiting it as a vector subspace of a known vector space, for example, $\mathbb{R}^4$. $\endgroup$ – Moss Oct 27 '13 at 20:24
  • $\begingroup$ Yes, I have learned about subspaces. Could you elaborate on this? I'm guessing this is an easier alternative to showing closure under addition and multiplication. $\endgroup$ – Matt Vukas Oct 27 '13 at 20:46
  • $\begingroup$ To show a subset is a vector subspace you need to check that the set is closed under addition and scalar multiplication and it contains 0. The answers below show this. $\endgroup$ – Moss Oct 27 '13 at 20:57
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If you have a vector satisfying the two constraints then multiplying by $k$ you get $$3a+b=c \implies 3(ka)+(kb)=(kc)$$ $$ a + b + 2c = 2d \implies (ka) + (kb) + 2(kc) = 2(kd)$$ so you have closure under scalar multiplication. Similarly for addition $$3a_1+b_1=c_1 \text{ and }3a_2+b_2=c_2 \\ \implies 3(a_1+a_2)+(b_1+b_2)=(c_1+c_2)$$ $$a_1 + b_1 + 2c_1 = 2d_1 \text{ and }a_2 + b_2 + 2c_2 = 2d_2 \\ \implies (a_1+a_2) + (b_1+b_2) + 2(c_1+c_2) = 2(d_1+d_2)$$

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  • $\begingroup$ This one seems to be the most tangible and straightforward answer for me, so I'll pick it. Thanks $\endgroup$ – Matt Vukas Oct 27 '13 at 20:50
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Verify all conditions that define a vector space one by one. For example, you have to verify that if $u$ and $v$ are two vectors that satisfy the given equations and if $\alpha$ is a constant (an element of the underlying field) then $\alpha \times u$ is a solution and $u+v$ is a solution.

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Algorithmic approach: If the given set was to contain vectors u, v and constant c Then go down the list below and make sure all constraints in your problem are met through trial.

Can u and v exist in this space?

  1. Addition:

(a) u + v is a vector in V (closure under addition).

(b) u + v = v + u (Commutative property of addition ).

(c) (u+v)+w = u+ (v+w) (Associative property of addition).

(d) There is a zero vector 0 in V such that for every u in V we have (u + 0) = u (Additive identity).

(e) For every u in V , there is a vector in V denoted by −u such that u + (−u) = 0 (Additive inverse).

  1. Scalar multiplication:

(a) cu is in V (closure under scalar multiplication0.

(b) c(u + v) = cu + cv (Distributive propertyof scalar mult.).

(c) (c + d)u = cu + du (Distributive property of scalar mult.).

(d) c(du) = (cd)u (Associate property of scalar mult.).

(e) 1(u) = u (Scalar identity property).

Remark. It is important to realize that a vector space consisits of four entities:

  1. A set V of vectors.

  2. A set of scalars. In this class, it will alawys be the set of real numbers R. (Later on, this could be the set of complex numbers C.)

  3. A vector addition denoted by +.

  4. A scalar multiplication.

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