$\begin{array}{l}{\text { A basis of } V :=\Pi_{2}(\mathbb{R}) \text { is } \mathcal{M} :=\left(m_{1}, m_{2}, m_{3}\right)} \\ {\text { here is } m_{1}(t) :=1, m_{2}(t) :=t \text { und } m_{3}(t) :=t^{2}} \\ {\text { a) show that } \mathcal{B} :=\left(b_{1}, b_{2}, b_{3}\right) \text { is also a basis of } V \text {. }} \\ {\text { Here is } b_{1}(t) :=3+t-2 t^{2}, b_{2}(t) :=-5-5 t+3 t^{2} \text { und } b_{3}(t) :=3-2 t^{2} \text { . }} \\ {\text { }(V, \mathcal{M}) \text { with }\left(\mathbb{R}^{3}, E_{3}\right)} \\ {\qquad }\end{array}$
$$ \begin{array}{l}{\text { b) The linear mapping } F : V \rightarrow V \text { is defined by: }} \\ {\qquad \begin{aligned} F & : V \rightarrow V \\ p(t) & \mapsto t\left(p^{\prime}+p^{\prime \prime}\right)(t) \\ \text { } p^{\prime} \text { is the first derivative and } p^{\prime \prime} \text { the second derivative of } p \text { } \\ \text { Find the representation Matrix } F_{\mathcal{M}}^{\mathcal{M}} \text { of } F \text { with basis } \mathcal{M} \text { and } \mathcal{M} \end{aligned}}\end{array} $$
No clue how to solve this
To check whether a set is a basis for a vector space, you have to check two things:
If you can verify $(b_1, b_2, b_3)$ satisfies these two things, you will have answered a).
Find $F(m_1)$, $F(m_2)$, and $F(m_3)$. Each of these polynomials can be represented by a vector containing the coefficients when expressing the polynomial as a linear combination of $(m_1, m_2, m_3)$. Combining these three column vectors into a matrix gives you your representation matrix.
For example, $F(m_2)(t) = t(1+0) = t = 0 \cdot m_1(t) + 1 \cdot m_2(t) + 0 \cdot m_3(t) $ is represented by the vector $\begin{bmatrix}0\\1\\0\end{bmatrix}$, and this will be the second column of your matrix.