Explain this relationship to me and prove it? Let $T: \mathbb{R}_{\leq 3}[X] \rightarrow \mathbb{R}_{\leq 2}[X]$ be a linear map defined as $T(f(x)) = f'(x)$, and let $\beta$ and $\gamma$ be the standard ordered bases for resp. $\mathbb{R}_{\leq 3}[X]$ and $\mathbb{R}_{\leq 2}[X]$. Then the matrixrepresentation of the linear map is given as \begin{align*} A = [T]_{\beta}^{\gamma} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}. \end{align*} Let $\phi_{\beta}: \mathbb{R}_{\leq 3}[X] \rightarrow \mathbb{R}^4$ and $\phi_{\gamma}: \mathbb{R}_{\leq 2}[X] \rightarrow \mathbb{R}^3$ be linear maps that represent de coordinates of the respective polynomials as columnvectors. We search a relationship between $L_A, \phi_{\beta}, \phi_{\gamma}$ and $T$, where $L_A$ is the left-multiplication transformation. Choose the polynomial $g(x)$, defined as $g(x) = 2 + x - 3x^2 + 5x^3 $. Then $\phi_{\beta} = \begin{pmatrix} 2 \\ 1 \\ -3 \\ 5 \end{pmatrix}$. Hence we have \begin{align*} L_A \circ \phi_{\beta} (g(x)) = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \\ -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 1 \\ -6 \\ 15 \end{pmatrix}. \end{align*} Because $T(g(x)) = g'(x) = 1 -6x + 15x^2$, we have also \begin{align*} \phi_{\gamma}\circ (T(g(x))) = \phi_{\gamma} (g'(x)) = \begin{pmatrix} 1 \\ -6 \\ 15 \end{pmatrix}. \end{align*} So we can conclude that \begin{align*} L_A \circ \phi_{\beta} = \phi_{\gamma} \circ T. \end{align*} How should I understand this now? And how can I prove it in a rigorous manner? Some explanation would be helpful.
 A: First let's write the domain and codomain of the relevant maps. I'll write $\mathbb{P}_n$ for the space of polynomials of degree at most $n$.
$$T : \mathbb{P}_3 \to \mathbb{P}_2 \\
\phi_\beta : \mathbb{P}_3 \to \mathbb{R}^4 \\
\phi_\gamma : \mathbb{P}_2 \to \mathbb{R}^3 \\
A : \mathbb{R}_4 \to \mathbb{R}^3.$$
The equality 
$$A=\phi_\gamma \circ T \circ \phi_\beta^{-1}$$
can be thought of as the definition of the equality
$$A=[T]_\beta^\gamma.$$ 
That is, this equality can be read as follows. The map $A$ proceeds in these steps:


*

*Given a vector $x$ in $\mathbb{R}^4$, map to the polynomial $p \in \mathbb{P}_3$ defined by $p(y)=x_0 + x_1 y + x_2 y^2 + x_3 y^3$, as described by the basis $\beta$. In other words, apply $\phi_\beta^{-1}$ to $x$.

*Map $p$ to its derivative $p'$. In other words, apply $T$ to $\phi_\beta^{-1} x$.

*Map $p'$ to its representation in the basis $\gamma$. In other words, apply $\phi_\gamma$ to $(T \circ \phi_\beta^{-1}) x$.

A: In my opinion, it is better to see how this relatin holds by looking at general case. I'll try to make similar notation.


*

*When you have linear transformation $F$ from $R^n$ to $R^m$ then there exists exactly one matrix $A\in M(n\times m)$ such that $Fx$=$Ax.$ And $A_{i}^j=F^j(e_i),$ where $F^j(x)$ is a scalar from formula $F(x)=\Sigma_{j=1}^m F^j(x)e_j$

*Similarly. When $A\in M(n\times m),$ then there exist exactly one linear transformation $L_A$ from $R^n$ to $R^m$ such that $L_{A}v$=$Av.$ And $L_A$ is just a left multiplication transformation.

*If $V$ is a vector space of dimention $n$ and you specify a ordered base $\beta=(v_{1},\dots,v_{n})$ then you have a linear isomorphism $\phi_\beta:V\rightarrow R^n$ that sends $v_i$ to $e_{i}.$


Combining previous you get the following.
If $T:V\rightarrow W$ is linear and you choose basis $\beta=(v_{1},\dots,v_{n})$ and $\gamma=(w_{1},\dots,w_{m})$ in $V$ and $W$ respectively. Then by 3 you have pair of linear isomorphisms $\phi_\beta:V\rightarrow R^n$ and $\phi_\gamma:W\rightarrow R^m.$ Consider linear transformation $F:R^n\rightarrow R^m$ defined by the formula $F=\phi_\gamma\circ T\circ\phi_\beta^{-1}.$ From 1 there exist exactly one matrix $A$ such that $Ax=\phi_\gamma\circ T\circ\phi_\beta^{-1}x.$ (You just has to compute the A) Now by 2 you get that $L_A=\phi_\gamma\circ T\circ\phi_\beta^{-1}.$ I left arguments due to the fact that now we are only working with linear transformations. Sinsce $\phi_\beta$ is an isomorphism we get that
$$L_A\circ\phi_\beta=\phi_\gamma\circ T.$$
