Archetype P Summary: Linear transformation with a domain smaller that its codomain, so it is guaranteed to not be surjective. Happens to be injective.

◊ A linear transformation: (Definition LT)
\begin{equation*} \ltdefn{T}{\complex{3}}{\complex{5}},    \lt{T}{\colvector{x_1\\x_2\\x_3}}= \colvector{-x_1 + x_2 + x_3\\ -x_1 + 2 x_2 + 2 x_3\\ x_1 + x_2 + 3 x_3\\ 2 x_1 + 3 x_2 + x_3\\ -2 x_1 + x_2 + 3 x_3} \end{equation*}

◊ A basis for the null space of the linear transformation: (Definition KLT)
\begin{equation*} \set{\ } \end{equation*}

◊ Injective: Yes. (Definition ILT)
Since $\krn{T}=\set{\zerovector}$, Theorem KILT tells us that $T$ is injective.

◊ A basis for the range of the linear transformation: (Definition RLT)
Evaluate the linear transformation on a standard basis to get a spanning set for the range (Theorem SSRLT):
\begin{equation*} \set{\colvector{-1\\-1\\1\\2\\-2},\,\colvector{1\\2\\1\\3\\1},\, \colvector{1\\2\\3\\1\\3}} \end{equation*}
If the linear transformation is injective, then the set above is guaranteed to be linearly independent (Theorem ILTLI). This spanning set may be converted to a nice'' basis, by making the vectors the rows of a matrix (perhaps after using a vector reperesentation), row-reducing, and retaining the nonzero rows (Theorem BRS), and perhaps un-coordinatizing. A basis for the range is:
\begin{equation*} \set{ \colvector{1\\0\\0\\-10\\6},\,\colvector{0\\1\\0\\7\\-3},\,\colvector{0\\0\\1\\-1\\1} } \end{equation*}

◊ Surjective: No. (Definition SLT)
The dimension of the range is 3, and the codomain ($\complex{5}$) has dimension 5. So the transformation is not surjective. Notice too that since the domain $\complex{3}$ has dimension 3, it is impossible for the range to have a dimension greater than 3, and no matter what the actual definition of the function, it cannot possibly be surjective in this situation.

To be more precise, verify that $\colvector{2\\1\\-3\\2\\6}\not\in\rng{T}$, by setting the output equal to this vector and seeing that the resulting system of linear equations has no solution, i.e. is inconsistent. So the preimage, $\preimage{T}{\colvector{2\\1\\-3\\2\\6}}$, is empty. This alone is sufficient to see that the linear transformation is not onto.

◊ Subspace dimensions associated with the linear transformation. Examine parallels with earlier results for matrices. Verify Theorem RPNDD. \begin{align*} \text{Domain dimension: }3&& \text{Rank: }3&& \text{Nullity: }0 \end{align*}

◊ Invertible: No.

The relative dimensions of the domain and codomain prohibit any possibility of being surjective, so apply Theorem ILTIS.

◊ Matrix representation (Theorem MLTCV):

\begin{equation*} \ltdefn{T}{\complex{3}}{\complex{5}},   \lt{T}{\vect{x}}=A\vect{x},    A= \begin{bmatrix} -1&1&1\\ -1&2&2\\ 1&1&3\\ 2&3&1\\ -2&1&3 \end{bmatrix} \end{equation*}