I tried searching the web about this and came up with one question on StackExchange that no one answered.
So my question is, If I have a linear transformation $T:R_3[x]\rightarrow R_3[x]$ where T is defined like so: $$T(p(x)) = p(x+1) - xp'(x)$$ Where p'(x) is the polynomial derivative, what do I assign if I put any polynomial in $T$ in the right side?
I mean, if I need to solve for $T(1)$ or $T(x)$, where do I assign 1 or x in the transformation?
I thought that I just replace the $x$ in the $p$, (for example: $T(1) = p(1+1)-xp'(1)=2$, which is wrong... (By the way, I dont know why I need to leave the $x$ outside the $p$ alone)
What really confused me is that (and this is the answer from the exam I took): $$T(1) = 1$$ $$T(x) = (x+1) - x(1) = 1$$ $$T(x^2) = (x+1)^2 - x(2x) = -x^2+2x+1$$
How did they get to $T(1) = 1$? What did they assign to the transformation?
How did they get to $T(x) = 1$? How come here $(x+1)$ is left alone?
How did they get to $T(x^2) = -x^2+2x+1$? How come $(x+1)$ is squared?
I searched all over the internet with different text but didn't come to anything close to answering my question..
Thanks in advance!
$T(1)$ means that you take $p(x)=1$ So $$T(1)=1+0=1$$ and so on...