Show that V is a subspace of P2

[rayne1]

$$V = \{({x}^{2}-1)p(x) | p(x) \in {P}_{2}\}$$ show that V is a subspace of ${P}_{2}$I tried:
$({x}^{2}-1)(0) = 0$ so 0 is in ${P}_{2}$ (axiom 1 is satisfied). If p(x) and q(x) are in ${P}_{2}$, then $({x}^{2}-1)p(x) + ({x}^{2}-1)q(x) = ({x}^{2}-1)(p(x)+q(x))$ and since $p(x)+q(x) \in {P}_{2}$, axiom 2 is satisfied. Finally, if $p(x) \in {P}_{2}$ then $ap(x)$ (a is a scalar) is also in ${P}_{2}$. Since all three axioms are met, V is a subspace of ${P}_{2}.$

Is what I did correct?

 

[Deveno]

Well you showed it is a subspace of *something* but it's *not* a subspace of $P_2$, since we have:

$x^2 \in P_2$, but certainly $x^4 - x^2 = (x^2 - 1)x^2 \not\in P_2$, since it has degree $4$.

 

[rayne1]

Well you showed it is a subspace of *something* but it's *not* a subspace of $P_2$, since we have:

$x^2 \in P_2$, but certainly $x^4 - x^2 = (x^2 - 1)x^2 \not\in P_2$, since it has degree $4$.

Ok, then is it still possible to find a basis and dimension of V if it is not a subspace of ${P}_{2}$?

 

[Deveno]

Sure. If your underlying field is $\Bbb R$, then $V$ would be a subspace of $\Bbb R[x]$. Otherwise, if its some *other* field $F$, then $V$ is a subspace of $F[x]$.

But a set need not be a subspace of some *other* vector space in order to be a vector space-it's just that we have fewer axioms to check if we're checking for a subspace (only 3 conditions), instead of the 8,9 or 10 axioms you often see listed in textbooks.

EVERY VECTOR SPACE HAS A BASIS.

I cannot stress enough the importance of this. The proof involves the axiom of choice for "arbitrary" vector spaces, but for finite-dimensional spaces, it's practically "true by definition" (since finite-dimensional *means* we have a finite basis).

Of course, $\Bbb R[x]$ (or $F[x]$ for that matter) is not finite-dimensional. But the subspace of polynomials of degree at most $n$ (for any positive integer $n$) *is* finite-dimensional, with dimension $n+1$ (so $P_2$ has dimension 3, one possible basis is $\{1,x,x^2\}$).

In your case, "your" $V$ has dimension at most $5$ (it is a subspace of $P_4$). If $\{x^2 - 1, x^3 - x,x^4 - x^2\}$ are linearly independent over your field, then $V$ has dimension at least $3$ (can you see this?).

So you have two cases (perhaps) to rule out (if you prove my statement above)-can you do this?