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Definition. A subspace of a vector space V is a subset H of V that has three properties

(a) The zero vector of V is in H.

(b) $\forall$ u, v $\in H$, we have u + v $\in H$.

(c) $\forall$ u $\in H$, c scalar, we have c u $\in H$.

For $n \geq$ 0, denote $\mathbb{P}_n$ to be the set of all polynomials of degree at most $n$. Which of the following are subspaces of $\mathbb{P}_n$ for an appropriate value of $n$? Support your answer with reason.

(a) All polynomials of the form $p(t) = a + t^2$, where $a \in \mathbb{R}$.

(b) All polynomials of degree at most 3, with integers as coefficients.

Solution:

Both the statements (a) and (b) above are not subspaces of $\mathbb{P}_n$.

Reason:

For part (a), All polynomials of the form $p(t) = a + t^2$, where $a \in \mathbb{R}$, is not a subspace because it failed to meet the standards of the first property stated above, that is, the coefficient of the $t^2$ term is always a 1, therefore the zero vector is not in this subspace of polynomials.

For part (b), All polynomials of degree at most 3, with integers as coefficents is also not a subspace becuase it as well failed to meet the standards of the third property, that is, integer coefficents, when multiplied by a real scalar, will not necessarily be of integer type leading to integer coefficents. Thusly, the resulting polynomial will not be in the subset of polynomials $\mathbb{P}_n$. $\Box$

Can someone please look at this and see if I have the right idea and approach to this question. Thank You

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Why did you remove the second question and your reasoning, which was correct? –  Ross Millikan Mar 19 '11 at 18:41
    
This looks like homework. Why else would someone be wondering this (and phrase the question in such a standard textbook style). To have other people solve this question for you is really not very educational. –  KCd Mar 19 '11 at 18:48
    
I've rolled back the question. –  Arturo Magidin Mar 19 '11 at 21:11
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1 Answer

Yes, you are right in both cases.

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