Writing out the span of this polynomial vector space?

[halo31]

Homework Statement

The problems states "All polynomials of the form p(t)= at^2, where a is in R."
I'm supposed to see if it is a subspace of P_n. I've already done that but the book's answer is that it spans P_n by Theorem 1, because the set is span{t^2}

Homework Equations

Theorem states "1 If v₁...v_p are in a vector space V , then Span {v₁... v_p.} is a subspace of V ."

The Attempt at a Solution

I've already checked the three properties showing that it is a subspace of P₂.
I checked that it was closed under addition and scalar multiplication as well as containing the zero vector. But I am not clear how the book comes up with the span{t^2} part? If I am correct would the reason be that since a is already a scalar then t^2 is the part that generates the vectors in P₂.

 

[tiny-tim]

hi halo31!

… the book's answer is that it spans P_n by Theorem 1, because the set is span{t^2}
…
Theorem states "1 If v₁...v_p are in a vector space V , then Span {v₁... v_p.} is a subspace of V ."

yes … the definition of eg span{P,Q,R} is all elements of the form aP + bQ + cR

when there's only one element in the original set eg span{P}, that's just all elements of the form aP

here, P = t² ​

 

[HallsofIvy]

I recommend that you go back and look at the basic definitions. The set of all polynomials of the form $ax^2$ is a vector space because
1) the 0 polynomial is of this form with a= 0.
2) the product of a real number, p, with such a polynomial is of this form: $p(ax^2)= (ap)x^2$ which is of that form with a replaced by ap.
3) the sum of two such polynomials, $ax^2$, $bx^2$ is of the same form- $ax^2+ bx^2= (a+ b)x^2$.
4) the "additive inverse" of $ax^2$ is $-ax^2= (-a)x^2$ which is again of that form.

A basis for a vector space is a set of vectors that both span the space and are independent. Every vector in this space can be written as $ax^2= (a)x^2$ so that each vector is a linear combination (actually a multiple) of $x^2$ so $\{x^2\}$ alone spans the space. And, of course, a set consisting of a single non-zero vector is always independent.