Linear Algebra - Linearity of a transformation

[SetepenSeth]

Homework Statement

Let be T : ℙ2 → ℙ2 a polynomial transformation (degree 2)
Defined as

T(a+bx+cx²) = (a+1) + (b+1)x + (b+1)x²

It is a linear transformation?

Homework Equations

A transformation is linear if

T(p1 + p2) = T(p1) + T(p2)

And

T(cp1)= cT(p1) for any scalar c

The Attempt at a Solution

Let p1=(a+bx+cx²) and p2=(d+ex+fx²) degree 2 polynomials

T(p1+p2)= (a+d+1) + (b+e+1)x + (b+e+1)x²

However

T(p1) + T(p2)=[(a+1)+(d+1)] + [(b+1)+(e+1)]x + [(b+1)+(e+1)]x²
T(p1) + T(p2)=(a+d+2) + (b+e+2)x + (b+e+1)x²

So

T(p1+p2) ≠ T(p1) + T(p2)

Making it non linear transformation.

Yet, my answer key says it is linear, either the key is wrong or there is something here I am not understanding.

Any advise would be appreciated.

 

[LCKurtz]

Certainly, if $P^2$ is the second degree polynomials with the usual arithmetic, it is even easier to see your transformation is not linear because$$
T(0 + 0x + 0x^2) \ne 0 + 0x + 0x^2$$Are you sure the destination $P^2$ has the usual operations?

 

[SetepenSeth]

Indeed T(0) will not map it to P² zero, it will suffice to prove the answer key is wrong. Thank you.

 

[LCKurtz]

Indeed T(0) will not map it to P² zero, it will suffice to prove the answer key is wrong. Thank you.

I'm guessing that the answer key is not wrong and what is wrong is that you don't have the standard addition, additive identity, etc. in your destination. That is why I asked you about the operations in the destination of your mapping. Are you sure that your destination has the same operations as your domain? If I am right in my guess, you shouldn't call the destination simply $P^2$. You might be mapping $(P^2,+,\cdot)$ to $(P^2,\oplus,*)$ where the operations are different.

 

[SetepenSeth]

It is correct, both destination and domain have the usual operations.