Let \begin{equation*} A= \begin{bmatrix} 3&-1&8&1\\ 2&0&5&-2\\ 1&1&3&-7 \end{bmatrix} \end{equation*} and define a function $\ltdefn{P}{\complex{4}}{\complex{3}}$ by \begin{equation*} \lt{P}{\vect{x}}=A\vect{x} \end{equation*}
So we are using an old friend, the matrix-vector product (Definition MVP) as a way to convert a vector with 4 components into a vector with 3 components. Applying Definition MVP allows us to write the defining formula for $P$ in a slightly different form, \begin{equation*} \lt{P}{\vect{x}}=A\vect{x}= \begin{bmatrix} 3&-1&8&1\\ 2&0&5&-2\\ 1&1&3&-7 \end{bmatrix} \colvector{x_1\\x_2\\x_3\\x_4} = x_1\colvector{3\\2\\1}+ x_2\colvector{-1\\0\\1}+ x_3\colvector{8\\5\\3}+ x_4\colvector{1\\-2\\-7} \end{equation*}
So we recognize the action of the function $P$ as using the components of the vector ($x_1,\,x_2,\,x_3,\,x_4$) as scalars to form the output of $P$ as a linear combination of the four columns of the matrix $A$, which are all members of $\complex{3}$, so the result is a vector in $\complex{3}$. We can rearrange this expression further, using our definitions of operations in $\complex{3}$ (Section VO). \begin{align*} \lt{P}{\vect{x}} &=A\vect{x}&&\text{Definition of $P$}\\ &= x_1\colvector{3\\2\\1}+ x_2\colvector{-1\\0\\1}+ x_3\colvector{8\\5\\3}+ x_4\colvector{1\\-2\\-7}&&\knowl{./knowls/definition.MVP.knowl}{\text{Definition MVP}}\\ &= \colvector{3x_1\\2x_1\\x_1}+ \colvector{-x_2\\0\\x_2}+ \colvector{8x_3\\5x_3\\3x_3}+ \colvector{x_4\\-2x_4\\-7x_4}&&\knowl{./knowls/definition.CVSM.knowl}{\text{Definition CVSM}}\\ &=\colvector{3x_1-x_2+8x_3+x_4\\2x_1+5x_3-2x_4\\x_1+x_2+3x_3-7x_4}&&\knowl{./knowls/definition.CVA.knowl}{\text{Definition CVA}} \end{align*}
You might recognize this final expression as being similar in style to some previous examples (Example ALT) and some linear transformations defined in the archetypes (Archetype M through Archetype R). But the expression that says the output of this linear transformation is a linear combination of the columns of $A$ is probably the most powerful way of thinking about examples of this type.
Almost forgot — we should verify that $P$ is indeed a linear transformation. This is easy with two matrix properties from Section MM. \begin{align*} \lt{P}{\vect{x}+\vect{y}} &=A\left(\vect{x}+\vect{y}\right)&&\text{Definition of $P$}\\ &=A\vect{x}+A\vect{y}&&\knowl{./knowls/theorem.MMDAA.knowl}{\text{Theorem MMDAA}}\\ &=\lt{P}{\vect{x}}+\lt{P}{\vect{y}}&&\text{Definition of $P$} \end{align*} and \begin{align*} \lt{P}{\alpha\vect{x}} &=A\left(\alpha\vect{x}\right)&&\text{Definition of $P$}\\ &=\alpha\left(A\vect{x}\right)&&\knowl{./knowls/theorem.MMSMM.knowl}{\text{Theorem MMSMM}}\\ &=\alpha\lt{P}{\vect{x}}&&\text{Definition of $P$} \end{align*}
So by Definition LT, $P$ is a linear transformation.