$A(e)\in M_3(\mathbb R)$ represents $A\in L(\mathcal P_2)$ in canonic basis. $$A(e)=\begin{bmatrix}-1&2&a\\1&2&c\\b&0&1\end{bmatrix}$$
$(a)$ Find $a,b,c\in\mathbb R$ s.t. $A(f),$ basis $f=\{1+t,t-t^2,1+t+t^2\}$ is symmetric.
$(b)$ Is there any basis $f$ s.t. $A(f)$ is skew-symmetric? Give an argument.
My attempt:
$(a)$ $$A(1)=\begin{bmatrix}-1\\1\\b\end{bmatrix},At=\begin{bmatrix}2\\2\\0\end{bmatrix},A\left( t^2\right)=\begin{bmatrix}a\\c\\1\end{bmatrix}$$
Since $A$ is linear, by definition: $$A(1+t)=A(1)+At=\begin{bmatrix}1\\3\\b\end{bmatrix}$$ $$A\left(t-t^2\right)=At-A\left(t^2\right)=\begin{bmatrix}2-a\\2-c\\-1\end{bmatrix}$$ $$A(1+t+t^2)=A(1)+At+A\left(t^2\right)=\begin{bmatrix}1+a\\3+c\\b+1\end{bmatrix}$$
$$A(e')=\begin{bmatrix}1&2-a&1+a\\3&2-c&3+c\\b&-1&b+1\end{bmatrix}$$ $A(e')=A^{\tau}$ if: $$2-a=3\implies a=-1\\1+a=b=0\\3+c=-1\implies c=-4$$
$(b)$ $\underline{\text{There is no basis}}$ $f$ s.t. $A(f)$ is skew-symmetric because $A(e)$ and $A(f)$ have to be similar in order to have the same characteristic polynomial. Similar matrices have equal traces, as I proved here. Trace of a skew-symmetric matrix equals $0$ and $\operatorname{trace}(A(e))=2$.
Is that correct? If so, how can I improve my answer and reasoning? Thank you in advance!
Your solution for part (b) is fine. Let’s check part (a): Setting $$P=\begin{bmatrix}1&0&1\\1&1&1\\0&-1&1\end{bmatrix}$$ with your solution we have $$P^{-1}AP = \begin{bmatrix}-1&1&0\\2&3&-1\\2&2&0\end{bmatrix},$$ which isn’t symmetric. What went wrong? Well, when you computed $A(1+t)=(1,3,b)^T$, it seems that you overlooked the fact that the right-hand side consists of the coordinates of $A(1+t)$ relative to the standard basis $e$, but you need its coordinates relative to the basis $f$ instead. The same mistake was made in computing the other two basis vector images. In effect, you solved the equation $AP=(AP)^T$ instead of the correct equation $P^{-1}AP=(P^{-1}AP)^T$.