Find $m \in \mathbb C$ such that $ f(p) = p'(t-m)-p(1)(t+m)^2 + p(0)t^2 $ is isomorphism I have solved this task, but I am not sure If I done this correctly - can you check my way and idea of solving this (calculations are not important)? 

Find $m \in \mathbb C$ such that linear transformation $f \in L(\mathbb C[x]_2,\mathbb C[x]_2)$: $$
 f(p) = p'(t-m)-p(1)(t+m)^2 + p(0)t^2$$
  is isomorphism
  

I know that: $ \dim Y = \dim im(f)$  and  $ \dim X = \dim im(f)$  but X=Y so $ \dim im(f) = 3 $ 
Firstly I calculate formula for $f(p)$ where $ p(t) = at^2+bt+c$
$$f(p) = ... = a(-2m-m^2-t^2) + \\b(-t^2-2tm-m^2+1) + \\c(-2tm-m^2)$$
so (after factor by $(-1)$)  $$f(p) = span([1,0,m^2+2m]^T,[1,2m,m^2-1]^T[0,2m,m^2]^T) $$
But I know that $ \dim im(f) = 3 $  so columns must be linearly independent so I do RREF and I get $$([1,0,0]^T,[0,1,0]^T,[0,0,1]^T) $$ for $ m \in \mathbb C \setminus \left\{ 0,-4\right\} $ otherwise  $$[1,0,m^2+2m]^T,[1,2m,m^2-1]^T[0,2m,m^2]^T $$ are linearly dependent.  ps: I am adding tag solution verification but system changes it to proof-verification :(
 A: We need only to find for wich $m$ given linear trasformation is injective i.e. $\ker(f) = \{0\}$, so for which $m$, the equation $f(p)=0$ forces $p=0$.
Write $p(x) = ax^2+bx+c$, then $p'(x)= 2ax+b$. Since we have $$p'(t-m)-p(1)(t+m)^2 + p(0)t^2 =0\;\;\;\;\forall t$$
we have $$-2am+b-m^2p(1)+(2a-2mp(1))t+(c-p(1))t^2=0$$

So $c=p(1)=a+b+c\implies a=-b$ and 
$a = m(a+b+c)\implies a=mc$ and
$m^2(a+b+c)+2am=b \implies cm^2+2am+a=0\implies 3am+a=0\implies \boxed{m=-{1\over 3}}$ or $a=0$. But later is true if $mc=0$ so $\boxed{m=0}$ (or $c=0$).
A: I'm getting the same result as @greedoid. Namely, $f$ is an isomorphism if and only if $m \ne 0, -\frac13$.
Just calculate the images of the basis $1, t, t^2$:
$$f(1) = -2mt-m^2$$
$$f(t) = 1-t^2-2mt-m^2$$
$$f(t^2) = 2t-2m-t^2-2mt-m^2$$
Now $f$ is an isomorphism if and only if $\{f(1), f(t), f(t)^2\}$ is linearly indepedendent, which is equivalent to 
$$\{-f(1), f(t)-f(1), f(t)^2-f(t)\} = \{m^2+2mt,1-t^2,2t-2m-1\}$$
being linearly independent.
Since $\deg(1-t^2) = 2$, the only way this set can be linearly dependent is if $m^2+2mt = \lambda(2t-2m-1)$ for some $\lambda$, which gives $\lambda = m$ and $m(3m+1) = 0$, or $m \in \left\{0, -\frac13\right\}$.
