$\square$  Summary: Domain and codomain are square matrices. Domain and codomain both have dimension 4. Not injective, not surjective, not invertible, 3 distinct eigenvalues, diagonalizable.

$\square$  A linear transformation (Definition LT).\begin{equation*}\ltdefn{T}{M_{22}}{M_{22}},\quad \lt{T}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}= \begin{bmatrix} -2a+15b+3c+27d & 10b+6c+18d \\ a-5b -9d & -a-4b-5c-8d \end{bmatrix} \end{equation*}

$\square$  A basis for the kernel of the linear transformation (Definition KLT).\begin{align*}\set{ \begin{bmatrix} -6 & -3 \\ 2 & 1 \end{bmatrix} } \end{align*}

$\square$  Is the linear transformation injective (Definition ILT)?No.

Since the kernel is nontrivial Theorem KILT tells us that the linear transformation is not injective. In particular, verify that \begin{align*} \lt{T}{\begin{bmatrix}-2&0\\1&-4\end{bmatrix}}&=\begin{bmatrix}115&78\\-38&-35\end{bmatrix} & \lt{T}{\begin{bmatrix}4&3\\-1&3\end{bmatrix}}&=\begin{bmatrix}115&78\\-38&-35\end{bmatrix} \end{align*} This demonstration that $T$ is not injective is constructed with the observation that \begin{align*} \begin{bmatrix}4&3\\-1&3\end{bmatrix} &=\begin{bmatrix}-2&0\\1&-4\end{bmatrix}+\begin{bmatrix}6&3\\-2&-1\end{bmatrix} \intertext{and} \vect{z}&=\begin{bmatrix}6&3\\-2&-1\end{bmatrix}\in\krn{T} \end{align*} so the vector $\vect{z}$ effectively “does nothing” in the evaluation of $T$.

$\square$  A spanning set for the range of a linear transformation (Definition RLT)can be constructed easily by evaluating the linear transformation on a standard basis (Theorem SSRLT).\begin{align*}\set{\begin{bmatrix}-2&0\\1&-1\end{bmatrix},\, \begin{bmatrix}15&10\\-5&-4\end{bmatrix},\, \begin{bmatrix}3&6\\0&-5\end{bmatrix},\, \begin{bmatrix}27&18\\-9&-8\end{bmatrix}} \end{align*}

$\square$  A basis for the range of the linear transformation (Definition RLT). If the linear transformation is injective, then the spanning set just constructed is guaranteed to be linearly independent (Theorem ILTLI) and is therefore a basis of the range with no changes. Injective or not, this spanning set can be converted to a “nice” linearly independent spanning set by making the vectors the rows of a matrix (perhaps after using a vector representation), row-reducing, and retaining the nonzero rows (Theorem BRS), and perhaps un-coordinatizing.\begin{align*}\set{ \begin{bmatrix}1 & 0 \\ -\frac{1}{2} & 0\end{bmatrix},\, \begin{bmatrix}0 & 1 \\ \frac{1}{4} & 0\end{bmatrix},\, \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix} } \end{align*}

$\square$  Is the linear transformation surjective (Definition SLT)?No.

The dimension of the range is 3, and the codomain ($M_{22}$) has dimension 5. So $\rng{T}\neq M_{22}$ and by Theorem RSLT the transformation is not surjective.

To be more precise, verify that $\begin{bmatrix}2 & 4\\ 3 & 1\end{bmatrix}\not\in\rng{T}$, by setting the output of $T$ equal to this matrix and seeing that the resulting system of linear equations has no solution, <ie /> is inconsistent. So the preimage, $\preimage{T}{\begin{bmatrix}2 & 4\\ 3 & 1\end{bmatrix}}$, is empty. This alone is sufficient to see that the linear transformation is not onto.

$\square$  Subspace dimensions associated with the linear transformation (Definition ROLT, Definition NOLT). Verify Theorem RPNDD, and examine parallels with earlier results for matrices.\begin{align*}\text{Rank: }3&&\text{Nullity: }1&&\text{Domain dimension: }4&\end{align*}

$\square$  Is the linear transformation invertible (Definition IVLT, and examine parallels with the existence of matrix inverses.)?No.

Neither injective nor surjective (Theorem ILTIS). Notice that since the domain and codomain have the same dimension, either the transformation is both injective and surjective or else it is both not injective and not surjective (making it not invertible, as in this case).

$\square$  Matrix representation of the linear transformation, as given by Definition MR and explained by Theorem FTMR. \begin{align*} B&=\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}} &&\text{(Domain basis)}\\ 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}} &&\text{(Codomain basis)}\\ \matrixrep{T}{B}{C}&=\begin{bmatrix} -2 & 15 & 3 & 27 \\ 0 & 10 & 6 & 18 \\ 1 & -5 & 0 & -9 \\ -1 & -4 & -5 & -8 \end{bmatrix} \end{align*}

$\square$  Eigenvalues, and bases for eigenspaces (Definition EELT, Theorem EER). Evaluate the linear transformation with each eigenvector as an interesting check.\begin{align*}\eigensystem{T}{0}{\begin{bmatrix}-6 & -3 \\ 2 & 1\end{bmatrix}}\\ \eigensystem{T}{1}{ \begin{bmatrix}-7 & -2\\3 & 0\end{bmatrix},\, \begin{bmatrix}-1 & -2 \\ 0 & 1\end{bmatrix} }\\ \eigensystem{T}{3}{\begin{bmatrix}-3 & -2 \\ 1 & 1\end{bmatrix}} \end{align*}

$\square$  A diagonal matrix representation relative to a basis of eigenvectors. \begin{align*} B&=\set{\begin{bmatrix}-6 & -3 \\ 2 & 1\end{bmatrix},\, \begin{bmatrix}-7 & -2 \\ 3 & 0\end{bmatrix},\, \begin{bmatrix}-1 & -2 \\ 0 & 1\end{bmatrix},\, \begin{bmatrix}-3 & -2 \\ 1 & 1\end{bmatrix}} &&\text{(Domain, codomain basis)}\\ \matrixrep{T}{B}{B}&=\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix} \end{align*}