Given a linear map $T: P_3(\mathbb{R}) \longrightarrow P_2(\mathbb{R})$ find a basis for $P_3$ and other for $P_2$ Let $T:P_3(\mathbb{R})→P_2(\mathbb{R})$ be given by $T(f(x))= f′(x)$ I need to find a basis of $P_3(\mathbb{R})$ and other for $P_2(\mathbb{R})$ such that the transformation matrix of T with respect these basis is the following:
$$    \begin{pmatrix}
        1 & 0 & 0 & 0 \\
        0 & 1 & 0 & 0 \\
        0 & 0 & 1 & 0
    \end{pmatrix}    $$
I don't even know where to start.
 A: Let's assume that the basis for $P_3(\mathbb{R})$ and $P_2(\mathbb{R})$ are the monomial basis: $(x^3, x^2, x, 1)$ and $(x^2, x, 1)$. So the polynomial $a_3x^3a_2x^2+a_1x+a_0$ can be written as the vector
$
\begin{bmatrix}
a_3 \\
a_2 \\
a_1 \\
a_0 \\
\end{bmatrix}
$
Let's look at that what the transformation matrix does to this input vector:
\begin{align*}
    \begin{bmatrix}
    1 & 0 & 0 & 0 \\
    0 & 1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    \end{bmatrix}
    \cdot 
    \begin{bmatrix}
    a_3 \\
    a_2 \\
    a_1 \\
    a_0 \\
    \end{bmatrix}
    = 
    \begin{bmatrix}
    a_3 \\
    a_2 \\
    a_1 \\
    \end{bmatrix}
\end{align*}
Now we write the output vector as a polynomial in respect of our basis: $a_3x^2+a_2x+a_1$. Even though this is not the derivative, it is pretty close!
The last coeficciant got dropped and all the powers got one smaller. The correct solution would be $f'(x)=a_33x^2+a_22x+a_1$ but the transformation matrix did not change the coefficients. So we need to find a different basis for $P_3(\mathbb{R})$, where writing the polynomial as a vector in respect of that basis already gives us the coefficients of the derivative. Since those need to be $3x^3, 2x^2, 1x$ our basis has to be $(\frac{x^3}{3}, \frac{x^2}{2}, x, 1)$.
