We are now in a position to revisit our first example in this section, Example NSAQ. In that example, we showed that Archetype Q is not surjective by constructing a vector in the codomain where no element of the domain could be used to evaluate the linear transformation to create the output, thus violating Definition SLT. Just where did this vector come from?
The short answer is that the vector \begin{equation*} \vect{v}=\colvector{-1\\2\\3\\-1\\4} \end{equation*} was constructed to lie outside of the range of $T$. How was this accomplished? First, the range of $T$ is given by \begin{equation*} \rng{T}=\spn{\set{ \colvector{1\\0\\0\\0\\1},\,\colvector{0\\1\\0\\0\\-1},\, \colvector{0\\0\\1\\0\\-1},\,\colvector{0\\0\\0\\1\\2} } }\end{equation*}
Suppose an element of the range $\vect{v^*}$ has its first 4 components equal to $-1$, $2$, $3$, $-1$, in that order. Then to be an element of $\rng{T}$, we would have \begin{equation*} \vect{v^*}=(-1)\colvector{1\\0\\0\\0\\1}+(2)\colvector{0\\1\\0\\0\\-1}+(3) \colvector{0\\0\\1\\0\\-1}+(-1)\colvector{0\\0\\0\\1\\2} =\colvector{-1\\2\\3\\-1\\-8} \end{equation*}
So the only vector in the range with these first four components specified, must have $-8$ in the fifth component. To set the fifth component to any other value (say, 4) will result in a vector ($\vect{v}$ in Example NSAQ) outside of the range. Any attempt to find an input for $T$ that will produce $\vect{v}$ as an output will be doomed to failure.
Whenever the range of a linear transformation is not the whole codomain, we can employ this device and conclude that the linear transformation is not surjective. This is another way of viewing Theorem RSLT. For a surjective linear transformation, the range is all of the codomain and there is no choice for a vector $\vect{v}$ that lies in $V$, yet not in the range. For every one of the archetypes that is not surjective, there is an example presented of exactly this form.