Linear Transformation Proofs: Check My Work and Correct Errors | Math Help

[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 KT_p(x)

So it is not a linear transformation

Did I do these right?

Thanks for any help :)

 

[talolard]

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

 

[Mark44]

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.

 

[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

 

[mmmboh]

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

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

Well either way I get zero.

 

[vela]

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

 

[mmmboh]

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

 

[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.

 

[vela]

I think your conclusions are right.