Show a subspace and find a spanning set

[slugbunny]

Homework Statement

Let W be the subset of all P₃ defined by

W={p(x) in P₃: p(1)=p(-1) and p(2)=p(-2)}

Show that W is a subspace of P₃, and find a spanning set for W.

Homework Equations

The Attempt at a Solution

This is homework we can correct. I've attached my work. The first one with red and purple writing is what I submitted. The black and yellow one is my redo attempt.

On my original one, the TA wrote in red and I wrote random stuff in purple.

In my redo, I first attempted to prove W is a subspace using the closure, multiplication, and zero vector properties.
I don't know how to find the spanning set. Is that a way to get it without guessing? I understand the answer but I don't know how to get it.

 

[lanedance]

well first do you know the dimension of your subspace? that tells you how many vectors you're looking for. The dimension of P^3 is 4, so it will be less than or equal to that.

Note there is no unique set of spanning vectors as when you have a basis any vector within it can be be replaced by a linear combination of the orginal vectors and any other basis vectors.

so first write an arbitary element in your set P^3 as
$$ax^3 + bx2 + cx +d$$
now it must satisfy p(1) = p(-1) so
$$a + b + c +d = (-a) + b + (-c) +d$$
and p(2) = p(-2) so
$$8a + 4b + 2c +d = (-8a) + 4b + (-2c) +d$$

use these constraints to find the basis vectors

 

[HallsofIvy]

Any "vector" in P3 can be written in the form $ax^3+ bx^2+ cx+ d$. The requirement that p(1)= p(-1) means that $a+ b+ c+ d= -a+ b- c+ d$ or $a+ c= 0$. The requirement that p(2)= p(-2) means that $8a+ 4b+ 2c+ d= -8a+ 4b- 2c+ d$ or $a+ c= 0$. You really have just one condition.

Knowing that c= -a means you can write $ax^3+ bx^2+ cx+ d$ as what?

 

[lanedance]

wait Halls so
$a+ b+ c+ d= -a+ b- c+ d$ gives $2a+ 2c= 0$
$$\implies a+ c= 0$$

but doesn't
$8a+ 4b+ 2c+ d= -8a+ 4b- 2c+ d$ gives $16a+ 4c= 0$
$$\implies \textbf{4}a+ c= 0$$

which leads to a=c=0?

 

[slugbunny]

Thanks for both of your help.

wait Halls so
$a+ b+ c+ d= -a+ b- c+ d$ gives $2a+ 2c= 0$
$$\implies a+ c= 0$$

but doesn't
$8a+ 4b+ 2c+ d= -8a+ 4b- 2c+ d$ gives $16a+ 4c= 0$
$$\implies \textbf{4}a+ c= 0$$

which leads to a=c=0?

Thanks so much! Since a=c=0, I'm left with

p(x) = 1 + x²

So the basis is 1, x² which is the spanning set.
I really appreciate it, so much.

 

[HallsofIvy]

wait Halls so
$a+ b+ c+ d= -a+ b- c+ d$ gives $2a+ 2c= 0$
$$\implies a+ c= 0$$

but doesn't
$8a+ 4b+ 2c+ d= -8a+ 4b- 2c+ d$ gives $16a+ 4c= 0$
$$\implies \textbf{4}a+ c= 0$$

which leads to a=c=0?

Yes, you are right. Guess my eyes went fuzzy there!