That says pretty much what I said.
"That said, there still is a way to "represent" T by a matrix."
The "T" referred to is the operation T(p)= p(0)x+ 3xp'(x) applied to the space or all polynomials of degree 3 or less.
That itself is NOT a matrix product but can be represented by one in a given basis. For example we can take as basis \(\displaystyle v_1= 1\), \(\displaystyle v_2= x\), \(\displaystyle v_3= x^2\), and \(\displaystyle v_4= x^3\).
Applying T to the basis vectors \(\displaystyle T(v_1)= T(1)= 1x+ 3x(0)= x\), \(\displaystyle T(v_2)= T(x)= 0x+ 3x(1)= 3x= 3v_2\), \(\displaystyle T(v_3)= T(x^2)= 0x+ 3x(2x)= 6x^2= 6v_3\) and \(\displaystyle T(v_4)= T(x^3)= 0x+ 3x(3x^2)= 9x^3= 9v_4\).
We can
represent the basis "vector" 1 as \(\displaystyle \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}\), x as \(\displaystyle \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}\), \(\displaystyle x^2\) as \(\displaystyle \begin{bmatrix} 0\\ 0 \\ 1 \\ 0 \end{bmatrix}\), and \(\displaystyle x^3\) as \(\displaystyle \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}\).
Write \(\displaystyle \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}\) as the matrix "representing" T.
Representing \(\displaystyle T(v_1)= T(1)= x= v_2\) we have \(\displaystyle \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}[\begin{bmatrix}1 \\ 0 \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix}a_{11} \\ a_{21} \\ a_{31} \\ a_{41}\end{bmatrix}= \begin{bmatrix}0 \\ 1 \\ 0 \\ 0\end{bmatrix}\) so we must have \(\displaystyle a_{11}= 0\), \(\displaystyle a_{21}= 1\), \(\displaystyle a_{31}= 0\) and \(\displaystyle a_{41}= 0\).
Representing \(\displaystyle T(v_2)= T(x)= 3x= 3v_2\) we have \(\displaystyle \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 0 \\ 0 \end{bmatrix}= \begin{bmatrix} a_{21} \\ a_{22} \\ a_{23} \\ a_{24} \end{bmatrix}= \begin{bmatrix} 0 \\ 3 \\ 0 \\ 0 \end{bmatrix}\) so we must have \(\displaystyle a_{21}= 0\), \(\displaystyle a_{22}= 3\), \(\displaystyle a_{23}= 0\), and \(\displaystyle a_{24}= 0\).
Representing \(\displaystyle T(v_3)= T(x^2)= 6x^3= 6v_3\) we have \(\displaystyle \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}=\)\(\displaystyle \begin{bmatrix}a_{31}\\ a_{32} \\ a_{33}\\ a_{34}\end{bmatrix}=\)\(\displaystyle \begin{bmatrix}0 \\ 0 \\ 6 \\ 0 \end{bmatrix}\) so \(\displaystyle a_{31}= 0\), \(\displaystyle a_{32}= 0\), \(\displaystyle a_{33}= 6\), and \(\displaystyle a_{34}= 0\).
Representing \(\displaystyle T(v_4)= T(x^3)= 9x^3= 9v_4\) we have \(\displaystyle \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}\begin{bmatrix}0 \\ 0 \\ 0 \\ 1 \end{bmatrix}= \begin{bmatrix}a_{41} \\ a_{42} \\ a_{43}\\ a_{44}\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 9 \end{bmatrix}\) so that \(\displaystyle a_{41}= 0\), \(\displaystyle a_{42}= 0\), \(\displaystyle a_{43}= 0\), and \(\displaystyle a_{44}= 9\).
Putting that all together the matrix representing T,
in this basis, is
\(\displaystyle \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 6 & 0 \\ 0 & 0 & 0 & 9 \end{bmatrix}\(\displaystyle .\)\)
