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

These are my two questions! I don't even know how to set up these two questions!

For each of the following problems, is $\mathcal{W}$ a subspace of the given vector space $V$ over the given scalar field $\mathbb{F}$? Explain your reasoning!

$$(a)\quad V = \mathbb{P, \ F = R},\ \mathcal{W}=\{ax^3 + bx^2 + cx + d:\quad c + d = a + b\}$$

$$\text{($\mathbb{P}$ is the vector space for all polynomials over $\mathbb{R}$)}.$$

$$(b)\quad V = \mathbb{R^3, F =R}, \mathcal{W} = \left\{\begin{bmatrix} x\\ y\\ z\\\end{bmatrix} : 3x + 5y + 4z - 3 = 0\right\} $$

share|cite|improve this question
Is something missing? – Amzoti Feb 27 '13 at 1:13
There are two operations defined on a vector space and one distinguished element. So, given a subset of a vector space, you have to show that it is closed under both operations (and you will automatically get alignment of distinguished elements). – peoplepower Feb 27 '13 at 1:30
@peoplepower I am still not following :/ – Emily Taylor Feb 27 '13 at 1:39

In general, given a vector space $V$ and a subset $W$, $W$ is a subspace of $V$ provided 1) $W$ is not empty, and 2) for all $p$ and $q$ in $W$, and all real numbers $\alpha$ and $\beta$, $\alpha p+\beta q$ is in $W$.

So, in (a), first show $W$ is not empty, then let $p$ and $q$ be typical elements of $W$, let $\alpha$ and $\beta$ be real numbers, and see if you can prove that $\alpha p+\beta q$ is in $W$.

(b) is a bit easier --- there is one element that is guaranteed to be in every vector space, and you should know what that element is, and you should be able to tell whether it's in $W$.

share|cite|improve this answer
Thank you so much! – Emily Taylor Feb 27 '13 at 2:21

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.