# Can Somebody Check My Proofs Please

1. Feb 6, 2010

### mmmboh

Hi, would someone be able to check my proofs for me and tell me if they are right and if not what is wrong please?

So for the first one I said let u=p(x) and v=b(x)
T(u+v)=p(x)+b(x)=p(5)x2+b(5)x2=Tu+Tv
and T(ku)=(kp)(x)=kp(5)x2=kTu
So it is a linear transformation.

For the second I said T(u+v)=p(x)+b(x)=x2p(1/x)+x2b(1/x)=Tp(x)+Tb(x)
and T(ku)=x2kp(1/x)=k(x2p(1/x))=kTp(x)

So it is also a linear transformation.
For the third I said T(kp)(x)=xkp'(x)kp''(x)=k2xp'(x)p''(x) which does not equal KTp(x)

So it is not a linear transformation

Did I do these right?

Thanks for any help :)

2. Feb 6, 2010

### talolard

for the first one, what would happen if you took A= x^2 and B=-x^2?

3. Feb 6, 2010

### Staff: Mentor

You haven't used the given information that P2(x) is the space of polynomials of degree 2 or less. Every function in this space is of the form p(x) = ax2 + bx + c, for some constants a, b, and c.

4. Feb 6, 2010

### mmmboh

So am I suppose to write it out as T(ax2 + bx + c+dx2+ex+f)=[25(a+d)+5(b+e)+(c+f)]x2?

I am a bit confused

5. Feb 6, 2010

### mmmboh

Hm I did it that way and I still get that it is a linear transformation.

Well either way I get zero.

6. Feb 6, 2010

### vela

Staff Emeritus
You have some problems with your notation. For instance, you have u=p(x) and v=b(x), so $T(u+v) = T[(p+b)(x)] \ne p(x)+b(x)$.

7. Feb 6, 2010

### mmmboh

Right I made that mistake, but I don`t think it affected the wrongness of my answer.

8. Feb 6, 2010

### mmmboh

T(u+v)=T[p+b](x)=(p+b)(5)x2. Does this not equal p(5)x2+b(5)x2? I don't understand why not.

9. Feb 6, 2010

### vela

Staff Emeritus
I think your conclusions are right.