Are all vectors of the form $(a, b, c)$, where $b = a + c$, subspaces of $R^3$

I have to use the theorem which states:

If W is a set of one or more vectors in a vector space V, then W is a subspace of V if and only if the following conditions hold.
a) If u and v are vectors in W, then u + v is in W.
b) If k is any scalar and u is any vector in W then k*u* is in W.

to find if all vectors of the form $(a, b, c)$, where $b = a + c$, subspaces of $R^3$.

I understand how to do it for the first two problems which were of the form $(a, 0, 0)$ and $(a, 1, 0)$, but don't understand for this form.

For example for $(a, 1, 0) + (d, 1, 0) = (a + d, 2, 0)$, which is not in the correct form.

But for $(a, b, c)$ I am not sure what to make of it.

I get $(a, b, c) + (d, e, f) = (a + d, b + e, c + f)$
$(b+e) = (a+c) + (d+f)$, or $(a+d) + (c+f)$

I eventually get $(a, b, c) + (d, e, f) = (a+d, [(a+d) + (c+f)], c+f)$

I see a pattern there, but I no longer recognize the form and don't understand what it's telling me.

The answer key in the book says that it is not a subspace of $R^3$.

How should I be using Part A of the theorem?

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This is homework, but we don't actually submit anything. – Louis Dec 12 '12 at 2:37

First, let's give a clear statement of the problem: we want to prove that if $$W=\{{(a,b,c):b=a+c\}}$$ then $W$ is a subspace of ${\bf R}^3$.

So: let $u$ and $v$ be in $W$. Then $u=(d,e,f)$ with $e=d+f$, and $v=(g,h,i)$ with $h=g+i$, right? If you don't understand that, stop and think about it until you get it.

Now let $w=u+v$. We want to prove $w$ is in $W$. That means, if $w=(j,k,l)$, we want to prove $k=j+l$, right? Again, don't go on until you get this.

But we know $u=(d,e,f)$, and we know $v=(g,h,i)$, so we can write down the components of $w$, that is, we can write down $j,k,l$ in terms of $d,e,f,g,h,i$, right? So, do it!

Now, see whether you can prove $k=j+l$, using what you already know about $d,e,f,g,h,i$.

That will settle part a) of the problem. You can work out part b) in a similar fashion. When you have it all worked out, write it up, and post it as your answer to your question!

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Thank you for helping set it up in my head properly. – Louis Dec 12 '12 at 3:20

According to the theorem you have to show two things:

$1$. If $(a,b,c)$ is such that $b=a+c$ and $(d,e,f)$ is such that $e=d+f$, then does $(a,b,c)+(d,e,f)=(a+d,b+e,c+f)$ satisfy $b+e=a+d+c+f$?

$2$. If $(a,b,c)$ satisfies $b=a+c$, does $k(a,b,c)=(ka,kb,kc)$ sastisfy $kb=ka+kc$?

If so, the theorem says it is a subspace.

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Thank you, I believe the book made a typo. – Louis Dec 12 '12 at 3:17