Let $P_3(\mathbb{R})$ denote the vector space of real polynomials:
$f=aX^2+bX+c$ of degree $\leq 2$ (less than, including, 2).
Show that the function
$L:P_3(\mathbb{R})\rightarrow P_3(\mathbb{R}),$
$f \mapsto X\cdot f'-f$
is a linear transformation
I'm not sure I've done the calculations correct, but I do think I've understood the the concept of linear transformation. If I've done some minor calculation error (or typos) or even completely missed the task please let me know.Here is what I've tried:
By definition L is a linear transformation if it satisfy the following two requirements:
Let $L:V\rightarrow W$
If for all $u,v\in V$ and $\alpha\in W$ we have:
(a) $L(u+v)=L(u)+L(v)$
(b) $L(\alpha\cdot v)=\alpha\cdot L(v)$.
So L is in our case a linear transformation if: $u,v\in f$ and (a), (b) is satisfied.
We choose $u(x)=ax^2+bx+c$ and $v(x)=px^2+qx+r$
(a): $L(u+v)=L((a+b)x^2+(b+q)x+c+r)=(a+p)\cdot 2x+(b+q)-((a+p)x^2+(b+q)x+(c+r))=2xa-ax^2+2px-px^2+b-bx-q+qx-c-r=L(\frac{d}{dx}(u(x))-u(x)+\frac{d}{dx}(v(x)))-v(x))=L(u)+L(v)$
Therefore L satisfies (a).
Now for (b)
$L(\alpha u)=L(\alpha (ax^2+bx+c))=L(\alpha ax^2+\alpha bx + \alpha c)=\alpha (L(\alpha 2x+b-(x^2+bx+c))=\alpha L(u))$
As it satisfies both (a) and (b) L is a linear transformation.
You did it the rightest and the fastest way possible for this kind of problem (returning to the definition and try if the definition suits to the function).
The only weird thing is that in the question you use $X$ in the polynoms but in your solution you use $x$. This is not a big matter, but just decide yourself which notation you use and do not change, or you can make some silly mistakes.