# Math Help - Vector Spaces

1. ## Vector Spaces

Hi everyone, I have a question,
I hope some two can help me.
Here tjey are

Let V be the set of all polynomials of degree 2 with the definitions of addidtion and scalar multiplication.
1) Show that V is not a closed under addition.
2) Show that V is closed under scalar multiplication.

Thanks

2. If you mean the set of all polynomials strictly of degree two then consider:
$\left( {x^2 - 3x} \right) + \left( { - x^2 - 3x} \right)$.
But if the problem is about polynomials of degree at most two, then part a is false.

Part b should be very easy. What have you done?

3. Originally Posted by Plato
If you mean the set of all polynomials strictly of degree two then consider:
$\left( {x^2 - 3x} \right) + \left( { - x^2 - 3x} \right)$.
But if the problem is about polynomials of degree at most two, then part a is false.

Part b should be very easy. What have you done?
No I need to use the Real Vector Space definition, which is a set of V of elements and have two operations (+) and (*) and have 8 properties.

4. Originally Posted by MarianaA
No I need to use the Real Vector Space definition, which is a set of V of elements and have two operations (+) and (*) and have 8 properties.
Did you understand any part of my first response?
Part a asks you to show closure. Period.

5. Originally Posted by Plato
Did you understand any part of my first response?
Part a asks you to show closure. Period.
I know is asking to show clusure but I dont know how to do it,
thats why I am asking for help.

6. 1 Says to show that V is NOT closed.
If that is the set of all polynomials strictly of degree two then
$\left( {x^2 - 3x} \right) + \left( { - x^2 - 3x} \right)= -6x$
shows that it is not closed because $-6x$ is not of degree 2.

However, if the problem is about polynomials of degree at most two, then it is false.

7. Originally Posted by Plato
1 Says to show that V is NOT closed.
If that is the set of all polynomials strictly of degree two then
$\left( {x^2 - 3x} \right) + \left( { - x^2 - 3x} \right)= -6x$
shows that it is not closed because $-6x$ is not of degree 2.

However, if the problem is about polynomials of degree at most two, then it is false.
Ok, I got it