Consider the linear transformation $\ltdefn{T}{\complex{3}}{\complex{2}}$ that is required to have the following three values, \begin{align*} \lt{T}{\colvector{1\\0\\0}}=\colvector{2\\1}&& \lt{T}{\colvector{0\\1\\0}}=\colvector{-1\\4}&& \lt{T}{\colvector{0\\0\\1}}=\colvector{6\\0} \end{align*}
Because \begin{equation*} B=\set{ \colvector{1\\0\\0},\, \colvector{0\\1\\0},\, \colvector{0\\0\\1} } \end{equation*} is a basis for $\complex{3}$ (Theorem SUVB), Theorem LTDB says there is a unique linear transformation $T$ that behaves this way.
How do we compute other values of $T$? Consider the input \begin{equation*} \vect{w}=\colvector{2\\-3\\1}=(2)\colvector{1\\0\\0}+(-3)\colvector{0\\1\\0}+(1)\colvector{0\\0\\1} \end{equation*}
Then \begin{equation*} \lt{T}{\vect{w}}=(2)\colvector{2\\1}+ (-3)\colvector{-1\\4}+ (1)\colvector{6\\0}=\colvector{13\\-10} \end{equation*}
Doing it again, \begin{equation*} \vect{x}=\colvector{5\\2\\-3}=(5)\colvector{1\\0\\0}+(2)\colvector{0\\1\\0}+(-3)\colvector{0\\0\\1} \end{equation*} so \begin{equation*} \lt{T}{\vect{x}}=(5)\colvector{2\\1}+ (2)\colvector{-1\\4}+ (-3)\colvector{6\\0}=\colvector{-10\\13} \end{equation*}
Any other value of $T$ could be computed in a similar manner. So rather than being given a formula for the outputs of $T$, the requirement that $T$ behave in a certain way for the inputs chosen from a basis of the domain, is as sufficient as a formula for computing any value of the function. You might notice some parallels between this example and Example MOLT or Theorem MLTCV.