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I need help starting with the following questions:

Let P3 be the vector space of real valued polynomials of degree ≤ 3. Find finite spanning sets for the following subspaces of P3.

(a) T = {p ∈ P3 | (0) = 0}

(b) U = {p ∈ P3 | p'(0) = 0}

(c) V = {p ∈ P3 | p(1) = p(2)}

I do understand how to tell if a vector spans but how do I find the spanning sets?

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    $\begingroup$ Basically to choose spanning sets, you just need to pick "enough" elements from given subspace so that they span the entire subspace. $\endgroup$ – lebesgue Apr 20 '18 at 12:44
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Well a general polynomial in $P_3$ takes the form $f(x) = ax^2 + bx + c$. Now let's solve (a), we want to have a more explicit description of $P$. If $f \in T$ then $$ f(x) = 0 \iff c = 0 $$ Therefore $$ P = \{ax^2 + bx : a, b \in \mathbb R \} $$ Then by observation a spanning set for this is $\{x, x^2\}$. (b) and (c) can be solved similarly, by finding what the spaces $U$ and $V$ actually look like.

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I'll tackle the third case:

A generic element of $P_3$ is $p(x)=ax^3+bx^2+cx+d$

If $p(1)=p(2)$ then $a+b+c+d=8a+4b+2c+d$

With this condition we can eliminate one of the variables by solving for it. I choose to solve for $c$.

$c=-7a-3b$

As long as this condition is met, we will have a third degree polynomial satisfying the property that $p(1)=p(2)$.

Substituting back into the original polynomial gives a function of the form: $p(x)=ax^3+bx^2+(-7a-3b)x+d$

We can then group the $a$'s $b$'s and $d$'s together to get our spanning set: $p(x) = a(x^3-7x) +b(x^2-3x)+d=span \left[ x^3-7x,x^2-3x,1\right] $

In general, if you want to find a particular basis for a subspace of any given vector space, you can follow this procedure:

  1. Write out a generic vector in the larger space.
  2. Substitute in the defining property or properties of the subspace you are dealing with to eliminate parameters.
  3. Group the terms of the result in terms of the remaining parameters, and the result will be a linear combination of your basis vectors.
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