- #1

- 39

- 2

## Homework Statement

Is {a0 + a1x + a2x^2 + a3x^3 | a0a3 - a1a2 = 0} a subspace of P3? Why or why not?

*The digits should be in subscript.

How would I go about answering this?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter dmitriylm
- Start date

- #1

- 39

- 2

Is {a0 + a1x + a2x^2 + a3x^3 | a0a3 - a1a2 = 0} a subspace of P3? Why or why not?

*The digits should be in subscript.

How would I go about answering this?

- #2

Mark44

Mentor

- 34,976

- 6,729

## Homework Statement

Is {a0 + a1x + a2x^2 + a3x^3 | a0a3 - a1a2 = 0} a subspace of P3? Why or why not?

*The digits should be in subscript.

How would I go about answering this?

Let's call your set as described above S. There are three things you need to verify to say that S is a subspace of P

- The zero polynomial is in S.
- If p
_{1}and p_{2}are any two polynomials in S, then p_{1}+ p_{2}is in S. - If c is any real constant and p
_{1}is in S, the cp_{1}is also in S.

- #3

- 39

- 2

Do I need to do anything with the conditions a0a3 – a1a2 = 0?

- #4

Mark44

Mentor

- 34,976

- 6,729

Absolutely! That equation describes the elements of your set.

- #5

- 39

- 2

How do I derive the elements of my set from the equation?

If I select a0=2, a1=3, a2=4, and a3=6, then these numbers meet the requirements of the equation.

Similarly, a0=-2, a1=-3, a2=-4 and a3=-6 would as well. Where do I go from there?

If I select a0=2, a1=3, a2=4, and a3=6, then these numbers meet the requirements of the equation.

Similarly, a0=-2, a1=-3, a2=-4 and a3=-6 would as well. Where do I go from there?

Last edited:

- #6

Mark44

Mentor

- 34,976

- 6,729

Let p

Is p

Is k*p

- #7

- 39

- 2

Thanks a lot for the help!

- #8

- 828

- 2

Let's call your set as described above S. There are three things you need to verify to say that S is a subspace of P_{3}:

- The zero polynomial is in S.
- If p
_{1}and p_{2}are any two polynomials in S, then p_{1}+ p_{2}is in S.- If c is any real constant and p
_{1}is in S, the cp_{1}is also in S.

He doesn't need to show (1); this follows directly from (3).

- #9

Mark44

Mentor

- 34,976

- 6,729

- #10

- 828

- 2

## Homework Statement

Is {a0 + a1x + a2x^2 + a3x^3 | a0a3 - a1a2 = 0} a subspace of P3? Why or why not?

*The digits should be in subscript.

How would I go about answering this?

I think that it might just be the polynomial thing that is causing you troubles; is that accurate? If it is, could you prove that a set of vectors in R^n is a subspace of R^n. That is, what if I asked this:

Is {(a_0),(a_1),(a_2),(a_3) | a0a3 - a1a2 = 0} a subspace of R^4, could you solve it? If so, you might want to prove (or disprove) that this is a subspace of R^4, then you can literally just take your proof and translate it to P^3. Do you understand what I am saying? I'm saying that R^4 and P^3 are essentialy the same (that is, they are isomorphic.) You can't prove it in R^4 and turn this in, but you can use it to get some insights.

Share: