0
$\begingroup$

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 :)

  • 0
    @hqt To display {} in math-mode you have to add backslash `$\{...\}$`.2012-11-04

1 Answers 1

2

$U$ 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 $0\notin U$.