Question about "weird" transformation For polynomials $\mathcal P_2(\Bbb R)$ the transformation $\psi$ is defined as follows:
$$\psi:\mathcal P_2(\Bbb R)\rightarrow\mathcal P_2 (\Bbb R), \space \space (\psi p)(t)=-t^2\cdot \frac{d^2p(t)}{dt^2}+t \cdot \frac{d p(t)}{dt}$$


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*Prove that $\psi$ is a $\Bbb R$-linear transformation

*Calculate ker($\psi$), img($\psi$) and dim($\psi$)

*Find a transformation matrix $M_{\mathcal B}^{\mathcal B}(\psi)$ with respect to $\mathcal B=${$1,t,t^2$}
This is what I have tried so far:


*

*In order for a transformation to be linear it needs to satisfy $T(a+b)=T(a)+T(b)$ and $T(\lambda a)=\lambda T(a)$. I wasn't sure what a "$\Bbb R$-linear" transformation is. Is it just a normal linear transformation?


$$\begin{align} (\psi p)(a+b)&=-(a+b)^2\cdot \frac{d^2p(a+b)}{dt^2}+(a+b) \cdot \frac{d p(a+b)}{dt} \\
& = -(a^2+2ab+b^2) \cdot (\frac{d^2 p(a)}{dt^2}+\frac{d^2 p(b)}{dt^2})+(a+b) \cdot (\frac{d p(a)}{dt}+\frac{d p(b)}{dt} ) \\
& = -(a^2+2ab+b^2)\frac{d^2 p(a)}{dt^2}-(a^2+2ab+b^2)\frac{d^2 p(b)}{dt^2}+a\frac{d p(a)}{dt}+b\frac{d p(b)}{dt} \\
&\not = (\psi p)(a)+(\psi p)(b) \end{align}$$
Am I doing something wrong here? The transformation seems to violate the first criterion.


*Is it true that in order to determine ker($\psi$) I need to solve this equation: 
$$-t^2 \frac{d^2p(t)}{dt^2}+t\frac{d p(t)}{dt}=0$$


for $t$?
Is there some way to show that this mapping is surjective? (sort of the easy way to determine the image)


*I have no idea what to do here. Maybe someone can give me a hint.


Thanks in advance
 A: For linearity this is how you should check:
Let $a(t), b(t) \in \mathcal P_2(\Bbb R)$, then you need to verify if 
$$\psi(a(t)+b(t))=\psi(a(t))+\psi(b(t)).$$
For example, the left side should be interpreted as
$$\psi(a(t)+b(t))=-t^2\cdot \frac{d^2}{dt^2}[a(t)+b(t)]+t \cdot \frac{d }{dt}[a(t)+b(t)]$$
A: $P_2$ is the set of polynomials of degree at most $2$, so it is the set of polynomials $ax^2+bx+c$.  In the basis, that would be represented as $(c,b,a)^t$.
Calculate $\psi(ax^2+bx+c)$; represent that as a vector; and find what matrix sends $(c,b,a)^t$ to the new vector.
A: 1) Yes, you're wrong. I think you misinterpreted the transformation: $t$ is the variable, not the polynomial being transformed. Your computation should begin with:
$$\psi(a+b)=-t^2\frac{\mathrm d^2(a+b)}{\mathrm d t^2}+t\frac{\mathrm d(a+b)}{\mathrm d t}=\dotsm $$
Any way $\psi$ is a linear combination of first and second derivatives (which are linear transformations) with coefficients in the polynomial ring, so that it is linear.
2) Yes. Set $p(t)=at^2+bt+c$, whence $p'(t)=2at+b$, $p''(t)=2a$.
$$p(t)\in\ker\psi\iff -2at^2 +t(2at+b)=bt=0\iff b=0.$$
Thus $p(f))\in \ker\psi$ if and only if $p(t)$ has no linear term.
3) Compute $\psi(1), \psi(t), \psi(t^2)$ and use the results as the vector-columns of the matrix of $\psi$.
