subsets and subspaces: Show U is subspace of P3

[Clifford]

Let P3 denote the vector space containing polynomials with a degree of 3 or less as well as the 0 polynomial. Let U be a subset of P3 consisting of those polynomials p in P3 for which 1 is a root. Prove that U is in fact a subspace.

I really have no idea how to go about this type of question.

For the subset of U, I know that p(1)=0 since in the question it tells that 1 is a root for those polynomials p in P3. But I have no idea whatsoever about how to prove that it is a subspace

 

[pka]

Re: subsets and subspaces

There are several equivalent ways to prove that a subset of a vector space is a subspace. Basically you need to show that the set is closed with respect to vector addition and scalar multiplication.
You must answer these questions.
Given that polynomials P & Q with 1 as a root:
Does P+Q have 1 as a root?
If r is a scalar, does rP have 1 as a root?

 

[Clifford]

Re: subsets and subspaces

How would you go about it for answering the first question in general?

I know for instance you could do like x^2 -1 and x^2 - 3x + 2.

The addition of these have root 1.

edit: Could you just say that P + Q is some linear combination of P, thus it will always have a root of 1?

If you multiplied x^2-1 by a scalar r, you would get rx^2 - r then you could factor out the r, so it would be r ( x^2 - 1 ) and then you could factor that so it's r (x-1)(x+1).

 

[pka]

Re: subsets and subspaces

How would you go about it for answering the first question in general?

Do not take any offence at this because none is intended. But yours is an example where the lack of basic ideas and abstractions hurts you.

If each of P & Q is a polynomial in \(\displaystyle P_3\) the you know that \(\displaystyle P(1) = 0\,\& \,Q(1) = 0\).
So \(\displaystyle \left[ {P + Q} \right]\left( 1 \right) = P(1) + Q(1) = 0 + 0 = 0\).
And \(\displaystyle r \in \Re \quad \Rightarrow \quad \left[ {rP} \right]\left( 1 \right) = r\left[ {P(1)} \right] = r\left( 0 \right) = 0\).

That is the whole proof.