We are now in a position to revisit our first example in this section, Example NIAQ. In that example, we showed that Archetype Q is not injective by constructing two vectors, which when used to evaluate the linear transformation provided the same output, thus violating Definition ILT. Just where did those two vectors come from?
The key is the vector \begin{equation*} \vect{z}=\colvector{3\\4\\1\\3\\3} \end{equation*} which you can check is an element of $\krn{T}$ for Archetype Q. Choose a vector $\vect{x}$ at random, and then compute $\vect{y}=\vect{x}+\vect{z}$ (verify this computation back in Example NIAQ). Then \begin{align*} \lt{T}{\vect{y}}&=\lt{T}{\vect{x}+\vect{z}}\\ &=\lt{T}{\vect{x}}+\lt{T}{\vect{z}}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\\ &=\lt{T}{\vect{x}}+\zerovector&&\vect{z}\in\krn{T}\\ &=\lt{T}{\vect{x}}&&\knowl{./knowls/property.Z.knowl}{\text{Property Z}} \end{align*}
Whenever the kernel of a linear transformation is nontrivial, we can employ this device and conclude that the linear transformation is not injective. This is another way of viewing Theorem KILT. For an injective linear transformation, the kernel is trivial and our only choice for $\vect{z}$ is the zero vector, which will not help us create two different inputs for $T$ that yield identical outputs. For every one of the archetypes that is not injective, there is an example presented of exactly this form.