Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 need to solve this question:

Decide which of these subsets are vector subspace of $\mathbb{R}^n$ ($\mathbb{R}$ is Real field):
$\{x \mid Ax = b\}$, where $A_{(m,n)} = 0$ and $b_{(m,1)} = 0$.

Could you please explain me and help me understand this question?

share|cite|improve this question
What is Rn? What is 'a "syntax" of vector subspace'? – Chris Eagle Dec 2 '12 at 20:12
Did you mean $\mathbb{R}^n$ ? $x$ an $n\times 1$ vector? – amWhy Dec 2 '12 at 20:15
up vote 1 down vote accepted

It is difficult to figure out what you are asking. So I assume that you have the following: $\mathbb{R}$ is the real numbers. You have the vector space $\mathbb{R}^n$ of dimension $n$. You then have a subset of $\mathbb{R}^n$ $$ Y = \{x\in \mathbb{R}^n : Ax = b\} $$ where $A$ is a fixed $n\times n$ matrix with entries from $\mathbb{R}$ and $b\in \mathbb{R}^n$. You assume that the matrix $A$ has the $(m,n)$th entry zero and that the $m$th coordinate of $b$ is also zero.

Now if you have $Ax = b$ and $Ay = b$, then $A(x+y) = Ax + Ay = b + b = 2b$. This means that $b$ has to be ... for $Y$ to be a vector space.

share|cite|improve this answer
Good mind reading – amWhy Dec 2 '12 at 20:18
@amWhy: Yeah, it took me a while :) – Thomas Dec 2 '12 at 20:19
@Thomas I would love to send you the HW work document but unfortunately, the PDF didn't wrote in English so that you won't be able to understand them .. and as you see, I don't know to explain the question well.. – Billie Dec 2 '12 at 20:50
@user1798362: I hope that my answer did help some. – Thomas Dec 2 '12 at 21:17
I believe that it will, after i'll understand this question (ill ask my teacher). but currently- it isn't. thank you anyway, for spend your time. – Billie Dec 2 '12 at 22:32

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.