# Linear Algebra - Linearity of a transformation

1. Jun 10, 2017

### SetepenSeth

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

A transformation is linear if

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

And

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

3. 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.

Last edited: Jun 10, 2017
2. Jun 10, 2017

### 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?

3. Jun 10, 2017

### SetepenSeth

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

4. Jun 10, 2017

### LCKurtz

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.

5. Jun 10, 2017

### SetepenSeth

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