How can I determine if W is a subspace of V
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How can I determine if W is a subspace of V

[From: ] [author: ] [Date: 11-12-20] [Hit: ]
-deg(p) = 2means that p(t) is a quadratic: p(t) = at^2 + bt + c. Also a must not = 0.Otherwise the degree of p(t) is 1 or 0.W is the set of all quadratic polynomials.The polynomial z(t) = 0 is not quadratic.So W does not contain 0.......
V = P_2
W = {p(t) | deg(p) = 2}

I understand I have to show that W is closed by linear combinations and contains the zero vector. However, I am not certain how this is shown for polynomials? When it says such that deg(p) = 2 does it mean simply x²?

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deg(p) = 2 means that p(t) is a quadratic: p(t) = at^2 + bt + c. Also a must not = 0.
Otherwise the degree of p(t) is 1 or 0.

W is the set of all quadratic polynomials. The polynomial z(t) = 0 is not quadratic.
So W does not contain 0. Therefore W is not a linear subspace of P_2 .

How about S = { p(t) : p(0) = 0} . Is S a linear subspace of P_2 ?

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S = { p(t) : p(0) = 0} is not the trivial subspace. If p(t) = at^2 + bt + c, then p (0) = c, so S is the polynomials of P_2 that have 0 for a constant term. It's the set generated by . Also, be careful about closure, W wasn't. There was no reason why a_2 + b_2 could not be 0.

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It means any polynomial where the highest degree term has degree 2.
p(t) could be something like t^2, or 3t^2 - 1, or 5t^2 + 8t + 2. However, p(t) does not include polynomials of degree 1, i.e., 2x, x + 1, or any other polynomials with even lower degree.

So for showing that W is a subspace: Is W closed under linear combinations? Will the linear combination of two polynomials of degree 2 always be of degree 2 (and not less)? If still not done, then is the zero polynomial a member of W?
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