Finding a basis

• February 28th 2009, 02:58 PM
smellatron
Finding a basis
Hey, this is for a intro to linear algebra class.

Find a basis for those that are subspaces. From P2 to P2.

{p(t): p'(1) = p(2)} (p' is the derivative)

The answer is that it is a subspace with a basis 1-t, 2-t^2 (t squared)

I just can't seem figure out how to get to this conclusion. Any help would be greatly appreciated.
• February 28th 2009, 05:31 PM
HallsofIvy
Quote:

Originally Posted by smellatron
Hey, this is for a intro to linear algebra class.

Find a basis for those that are subspaces. From P2 to P2.
{p(t): p'(1) = p(2)} (p' is the derivative)

The answer is that it is a subspace with a basis 1-t, 2-t^2 (t squared)

I just can't seem figure out how to get to this conclusion. Any help would be greatly appreciated.

[/quote]
I find this very hard to understand. I think you mean that you want to determine whether or not the subset of the vector space of all polynomials of degree two or less, with the property that p'(1)= p(2), is a subspace.

To show that it is a subspace, you must show that, if p1 and p2 both satisfy that equation, then so does ap1+ bp2 for a, b any numbers.
(ap1'+ bp2')(1)= ap1'(1)+ bp2'(1)= a(p1(2))+ b(p2(2)).

To find a basis, remember that any polynomial of degree 2 or less can be written as $p(x)= ax^2+ bx+ c$. Then $p'(x)= 2ax+ b$ so "p'(1)= p(2)" becomes 2a+ b= 4a+ 2b+ c or 2a+ b+ c= 0.We can choose two of those to be whatever we choose and solve for the third. For example, if we take a= 1, b= 0 we get 2+ c= 0 or c= -2. $x^2- 2$ is a basis vector. If we take a= 0, b= 1 we get 1+ c= 0 or c= -1. x- 1 is a We have no restriction on a and, whatever b is, c must be -b. x- 1 is a basis vector.

Those are the choices I would make and the basis I would determine. If we choose a= -1, b= 0, then c= 2 so $2- x^2$ is a basis vector. If we choose a= 0, b= -1, then c= 1 so $1- x$ is a basis vector. That is the basis your textbook gives.