Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
This is homework, but we don't actually submit anything. – Louis Dec 12 '12 at 2:37
up vote 1 down vote accepted

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!

share|cite|improve this answer
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.

share|cite|improve this answer
Thank you, I believe the book made a typo. – Louis Dec 12 '12 at 3:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.