Linear mapping matrix with paramters. I solved a linear mapping problem recently and it turns out no to be correct, although i thought it was a simple problem. The problem asks me to find real parameters $a,b,c$ such that linear mapping defined as $A(p) = p' (x+1)$ ( mapping from vector space of $2^{nd}$ order polynomials to that same space), where $p'$ is first derivative with respect to $x$ has a linear mapping matrix as shown (with respect to some basis $B$ of vector space of 2nd order polynomials):
$$A=\begin{bmatrix}
a && 1 && 0 \\
b && 0 && 1 \\
c && 0 &&  0
\end{bmatrix}$$
My soluton:
Let's assume basis $B=\{ e_1, e_2,e_3 \}$, such that
$$e_1 = \alpha x^2 + \beta x + \gamma $$
$$e_2 = \delta x^2 + \epsilon x + \eta $$
$$e_3 = \theta x^2 + \kappa x + \lambda $$
,assuming all parameters are such that these vectors are linearly independant ( for example, for $\beta =0, \gamma =0, \delta =0, \eta =0, \theta =0, \kappa =0$ we get standard base)
Then, if we apply linear mapping $A$ to basis vectors:
$$A(e_1) = (2 \alpha x + \beta ) (x+1) = (2\alpha x^2 + (2\alpha+\beta)x + \beta )  $$
$$A(e_2) = (2 \delta x + \epsilon ) (x+1) = (2\delta x^2 + (2\delta+\epsilon)x  +  \epsilon )  $$
$$A(e_3) = (2 \theta x + \kappa ) (x+1) = (2\theta x^2 + (2\theta+\kappa)x  + \kappa )  $$
From here, linear mapping matrix is defined as:
$$M=
\begin{bmatrix}
2 \alpha  &&  2 \delta && 2 \theta \\
2\alpha + \beta && 2 \delta + \epsilon && 2 \theta + \kappa \\
\beta && \epsilon && \kappa
\end {bmatrix}
$$
Putting $A=M$ gives us:
$$\kappa = 0 $$
$$2 \theta =0$$
$$2 \theta + \kappa = 1 $$
which is clearly impossible, therefore there are no such $a,b,c$ that satisfy this condition. 
I would be very grateful if you helped me find what my mistake is. Thank you.
 A: Using the work done by @Joao you can find $b$ as follows. First, find the matrix of the linear transformation in the canonical basis. It is
$$
M = 
\left[
\begin{array}{ccc}
0 & 1 & 0\\
0 & 1 & 2\\
0 & 0 & 2
\end{array}
\right]
$$
Let $B$ be the matrix that transforms the unknown basis to the canonical basis. Then we have, $BAB^{-1} = M \iff BA = MB$. Let
$$
B = 
\left[
\begin{array}{ccc}
\alpha_1 & \beta_1 & \gamma_1\\
\alpha_2 & \beta_2 & \gamma_2\\
\alpha_3 & \beta_3 & \gamma_3\\
\end{array}\right]
$$
Then it follows that
$$
\left[
\begin{array}{ccc}
3\alpha_1 + b\beta_1 & \alpha_1 & \beta_1\\
3\alpha_2 + b\beta_2 & \alpha_2 & \beta_2\\
3\alpha_3 + b\beta_3 & \alpha_3 & \beta_3\\
\end{array}\right]
=
\left[
\begin{array}{ccc}
\alpha_2 & \beta_2 & \gamma_2\\
\alpha_2 + 2\alpha_3 & \beta_2 + 2\beta_3 & \gamma_2 + 2\gamma_3\\
2\alpha_3 & 2\beta_3 & 2\gamma_3\\
\end{array}\right]
$$
From this equality we find that $\alpha_3(1 + b/2) = 0$. Now suppose that $\alpha_3 = 0$. Then it follows that $\alpha_3 = \beta_3 = \gamma_3 = 0$. However, this would imply that $B$ has a row of zeros which is not possible since $B$ is invertible. Therefore, $\alpha_3 \neq 0$ and so we find that $b = -2$.
A: You have used the whole greek alphabet on this problem haha.
Realize that you written the coordinates on M on the canonical basis, instead on the B basis. This correct way then will lead you to a lot of boring work.
$A$ isn't an isomorphism (the kernel are the constant polinomials), and because $det A = 0$ follows that $c=0$ (calculating the determinant).
Moreover, the trace of this transformation is 3 (calculated on the canonical basis), and so $a=3$. Just $b$ is missing now.
