Example LTDB2 Linear transformation defined on a basis

Consider the linear transformation $\ltdefn{R}{\complex{3}}{\complex{2}}$ with the three values, \begin{align*} \lt{R}{\colvector{1\\2\\1}}=\colvector{5\\-1}&& \lt{R}{\colvector{-1\\5\\1}}=\colvector{0\\4}&& \lt{R}{\colvector{3\\1\\4}}=\colvector{2\\3} \end{align*}

You can check that \begin{equation*} D=\set{ \colvector{1\\2\\1},\, \colvector{-1\\5\\1},\, \colvector{3\\1\\4} } \end{equation*} is a basis for $\complex{3}$ (make the vectors the columns of a square matrix and check that the matrix is nonsingular, Theorem CNMB). By Theorem LTDB we know there is a unique linear transformation $R$ with the three specified outputs. However, we have to work just a bit harder to take an input vector and express it as a linear combination of the vectors in $D$.

For example, consider, \begin{equation*} \vect{y}=\colvector{8\\-3\\5} \end{equation*}

Then we must first write $\vect{y}$ as a linear combination of the vectors in $D$ and solve for the unknown scalars, to arrive at \begin{equation*} \vect{y}=\colvector{8\\-3\\5}= (3)\colvector{1\\2\\1}+ (-2)\colvector{-1\\5\\1}+ (1)\colvector{3\\1\\4} \end{equation*}

Then the proof of Theorem LTDB gives us \begin{equation*} \lt{R}{\vect{y}}=(3)\colvector{5\\-1}+ (-2)\colvector{0\\4}+ (1)\colvector{2\\3}= \colvector{17\\-8} \end{equation*}

Any other value of $R$ could be computed in a similar manner.