Take the 2-minute tour ×
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.

Given : $ U = \{f(x)| f(x) \in P_{3}, \operatorname{deg} f(x) = 3\}$

Does: U is a subspaces of $P_{3}$

I think the answer is yes. But in my textbook, they say no. And explain that zero is not in set, and scalar multiplication and addition are not closed under U.

Please explain for me.

Thanks :)

share|improve this question
2  
The zero polynomial is not in $U$, because the condition for entry into $U$ is being of degree 3, and the zero polynomial is not of degree 3. A subspace must contain the zero element of the space of which it is a subspace. –  Gerry Myerson Nov 4 '12 at 5:42
    
@hqt To display {} in math-mode you have to add backslash $\{...\}$. –  Martin Sleziak Nov 4 '12 at 7:19
add comment

1 Answer

up vote 2 down vote accepted

$U$ is not closed under addition. For example, take $x^3$ and $-x^3$. Adding these gives the polynomial $0$, which does not have degree $3$, whence $0\notin U$.

share|improve this answer
add comment

Your Answer

 
discard

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.