Patrick Michalik

Finding the Matrix of a Linear Transformation

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


and so we have

T(abc)=(2ab).T\begin{pmatrix} a\newline b\newline c \end{pmatrix}=\begin{pmatrix} 2a\newline b \end{pmatrix}.



the matrix of TT 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(abc)=(t1t2t3t4t5t6)(abc)=(2ab)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 t1t_1 through t6t_6. We can see that

t1a+t2b+t3c=2at4a+t5b+t6c=b,\begin{aligned} t_1a+t_2b+t_3c&=2a\newline t_4a+t_5b+t_6c&=b, \end{aligned}


t1a+t2b+t3c=2a+0b+0ct4a+t5b+t6c=0a+1b+0c.\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 t1=2t_1 = 2, t5=1t_5 = 1, and the remaining tt’s are 0. This gives us

T(abc)=(200010)(abc),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


As a closing note, we could just as well have TT be a mapping from R3\mathbb{R}^3 to R3\mathbb{R}^3, in which case the transformation matrix would be 3-by-3, and one of its rows would contain only zeros.

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