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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 :)

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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
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 $0\notin U$.

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