Subspace of Polynomials: \vec{}p(t)=a+t^2, where a is real

[renob]

Homework Statement

Determine if all polynomials of the form \vec{}p(t)=a+t^2, where a is in real, are a subspace of \Re_{}n.

The Attempt at a Solution

the correct answer says that p(t) is not a subspace since the zero vector is not in the set.

im trying to work this out and got P(0)=a+0=a
is this correct? if so, why couldn't I just assume that a=0 and call that the zero vector? I'm a bit confused here.

 

[spamiam]

Remember, the polynomials themselves are the vectors: you shouldn't have to evaluate them at points. The zero vector in this case would be 0 or 0 + 0t + 0t². So yes, you can set a=0, but what is the coefficient of t²?

 

[renob]

Oh so since the coefficient of t² is 1, t² will never be multiplied by the scalar 0, and thus the entire vector never equals 0?
What confuses me is that why can't we assume both a=0 and t=0? or alternatively, why can't we multiply both sides of P(t)=a + t² by the zero vector?

 

[poochie_d]

Homework Statement

Determine if all polynomials of the form \vec{}p(t)=a+t^2, where a is in real, are a subspace of \Re_{}n.

Are you sure the problem is asking for a subspace of \mathbb{R}^n? Because the polynomials in question are not vectors in \mathbb{R}^n, but lie in \mathcal{F} = {the set of all polynomials \vec{p} : \mathbb{R} \rightarrow \mathbb{R}}.

im trying to work this out and got P(0)=a+0=a
is this correct? if so, why couldn't I just assume that a=0 and call that the zero vector? I'm a bit confused here.

It may help to remind yourself of the definition of the zero vector.

 

[bll5z6]

I know it's an old thread but can someone clear this up? How is the zero vector not in the set? If t=0 and a=0, you get the zero vector. I can see that polynomials of the form a+b^2 is not closed under scalar multiplication or vector addition but I don't see the zero vector argument.