Archetype S Summary: Domain is column vectors, codomain is matrices. Domain is dimension 3 and codomain is dimension 4. Not injective, not surjective.

◊ A linear transformation: (Definition LT)
\begin{equation*} \ltdefn{T}{\complex{3}}{M_{22}},    \lt{T}{\colvector{a\\b\\c}}= \begin{bmatrix} a-b&2a+2b+c\\ 3a+b+c&-2a-6b-2c \end{bmatrix} \end{equation*}

◊ A basis for the null space of the linear transformation: (Definition KLT)
\begin{equation*} \set{\colvector{-1\\-1\\4}} \end{equation*}

◊ Injective: No. (Definition ILT)
Since the kernel is nontrivial Theorem KILT tells us that the linear transformation is not injective. Also, since the rank can not exceed 3, we are guaranteed to have a nullity of at least 1, just from checking dimensions of the domain and the codomain. In particular, verify that

\begin{align*} \lt{T}{\colvector{2\\1\\3}}&=\begin{bmatrix}1&9\\10&-16\end{bmatrix} & \lt{T}{\colvector{0\\-1\\11}}&=\begin{bmatrix}1&9\\10&-16\end{bmatrix} \end{align*}

This demonstration that $T$ is not injective is constructed with the observation that

\begin{align*} \colvector{0\\-1\\11}&=\colvector{2\\1\\3}+\colvector{-2\\-2\\8} \end{align*} and \begin{align*} \vect{z}&=\colvector{-2\\-2\\8}\in\krn{T} \end{align*}

so the vector $\vect{z}$ effectively "does nothing" in the evaluation of $T$.

◊ 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{ \begin{bmatrix}1&2\\3&-2\end{bmatrix},\, \begin{bmatrix}-1&2\\1&-6\end{bmatrix},\, \begin{bmatrix}0&1\\1&-2\end{bmatrix} } \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{ \begin{bmatrix}1&0\\1&2\end{bmatrix},\, \begin{bmatrix}0&1\\1&-2\end{bmatrix} } \end{equation*}

◊ Surjective: No. (Definition SLT)
The dimension of the range is 2, and the codomain ($M_{22}$) has dimension 4. 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 $\begin{bmatrix}2&-1\\1&3\end{bmatrix}\not\in\rng{T}$, by setting the output of $T$ equal to this matrix and seeing that the resulting system of linear equations has no solution, i.e. is inconsistent. So the preimage, $\preimage{T}{\begin{bmatrix}2&-1\\1&3\end{bmatrix}}$, 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: }2&& \text{Nullity: }1 \end{align*}

◊ Invertible: No.

Not injective (Theorem ILTIS), and the relative dimensions of the domain and codomain prohibit any possibility of being surjective.

◊ Matrix representation (Definition MR):

\begin{align*} B&=\set{ \colvector{1\\0\\0},\, \colvector{0\\1\\0},\, \colvector{0\\0\\1} }\\ &\\ C&=\set{ \begin{bmatrix}1&0\\0&0\end{bmatrix},\, \begin{bmatrix}0&1\\0&0\end{bmatrix},\, \begin{bmatrix}0&0\\1&0\end{bmatrix},\, \begin{bmatrix}0&0\\0&1\end{bmatrix} }\\ &\\ \matrixrep{T}{B}{C}&= \begin{bmatrix} 1 & -1 & 0 \\ 2 & 2 & 1 \\ 3 & 1 & 1 \\ -2 & -6 & -2 \end{bmatrix} \end{align*}