February 3, 2023

Here I’ll show how to find the matrix of a linear transformation by doing so for the transformation corresponding to the differentiation of a second-degree polynomial. This transformation takes the coefficients of the polynomial to be differentiated and returns the coefficients of the derivative, which is a first-degree polynomial. We know that

$\frac{\textrm{d}}{\textrm{d}x}\left(ax^2+bx+c\right)=2ax+b,$and so we have

$T\begin{pmatrix} a\newline b\newline c \end{pmatrix}=\begin{pmatrix} 2a\newline b \end{pmatrix}.$Because

$T:\mathbb{R}^3\rightarrow\mathbb{R}^2,$the matrix of $T$ is 2-by-3. (The product of two matrices has as many rows as the first matrix, and as many columns as the second matrix. Also, two matrices can be multiplied only if the first matrix has as many columns as the second one has rows.) Thus,

$T\begin{pmatrix} a\newline b\newline c \end{pmatrix}=\begin{pmatrix} t_1&t_2&t_3\newline t_4&t_5&t_6 \end{pmatrix}\begin{pmatrix} a\newline b\newline c \end{pmatrix}=\begin{pmatrix} 2a\newline b \end{pmatrix}$for some $t_1$ through $t_6$. We can see that

$\begin{aligned} t_1a+t_2b+t_3c&=2a\newline t_4a+t_5b+t_6c&=b, \end{aligned}$or

$\begin{aligned} t_1a+t_2b+t_3c&=2a+0b+0c\newline t_4a+t_5b+t_6c&=0a+1b+0c. \end{aligned}$It is clear that $t_1 = 2$, $t_5 = 1$, and the remaining $t$’s are 0. This gives us

$T\begin{pmatrix} a\newline b\newline c \end{pmatrix}=\begin{pmatrix} 2&0&0\newline 0&1&0 \end{pmatrix} \begin{pmatrix} a\newline b\newline c \end{pmatrix},$with the transformation matrix being

$\begin{pmatrix}2&0&0\newline0&1&0\end{pmatrix}.$As a closing note, we could just as well have $T$ be a mapping from $\mathbb{R}^3$ to $\mathbb{R}^3$, in which case the transformation matrix would be 3-by-3, and one of its rows would contain only zeros.