$\square$  Summary: Linear transformation with equal-sized domain and codomain. Injective, surjective, invertible, diagonalizable, the works.

$\square$  A linear transformation (Definition LT).\begin{equation*}\ltdefn{T}{\complex{5}}{\complex{5}},\quad \lt{T}{\colvector{x_1\\x_2\\x_3\\x_4\\x_5}}= \colvector{-65 x_1 + 128 x_2 + 10 x_3 - 262 x_4 + 40 x_5\\ 36 x_1 - 73 x_2 - x_3 + 151 x_4 - 16 x_5\\ -44 x_1 + 88 x_2 + 5 x_3 - 180 x_4 + 24 x_5\\ 34 x_1 - 68 x_2 - 3 x_3 + 140 x_4 - 18 x_5\\ 12 x_1 - 24 x_2 - x_3 + 49 x_4 - 5 x_5} \end{equation*}

$\square$  A basis for the kernel of the linear transformation (Definition KLT).\begin{align*}\set{\ } \end{align*}

$\square$  Is the linear transformation injective (Definition ILT)?Yes.

Since the kernel is trivial Theorem KILT tells us that the linear transformation is injective.

$\square$  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{ \colvector{-65\\36\\-44\\34\\12},\, \colvector{128\\-73\\88\\-68\\-24},\, \colvector{10\\-1\\5\\-3\\-1},\, \colvector{-262\\151\\-180\\140\\49},\, \colvector{40\\-16\\24\\-18\\-5}} \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{ \colvector{1\\0\\0\\0\\0},\,\colvector{0\\1\\0\\0\\0},\, \colvector{0\\0\\1\\0\\0},\,\colvector{0\\0\\0\\1\\0},\,\colvector{0\\0\\0\\0\\1} } \end{align*}

$\square$  Is the linear transformation surjective (Definition SLT)?Yes.

A basis for the range is the standard basis of $\complex{5}$, so $\rng{T}=\complex{5}$ and Theorem RSLT tells us $T$ is surjective. Or, the dimension of the range is 5, and the codomain ($\complex{5}$) has dimension 5. So the transformation is surjective.

$\square$  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: }5&&\text{Nullity: }0&&\text{Domain dimension: }5&\end{align*}

$\square$  Is the linear transformation invertible (Definition IVLT, and examine parallels with the existence of matrix inverses.)?Yes.

Both injective and surjective (Theorem ILTIS). Notice that since the domain and codomain have the same dimension, either the transformation is both injective and surjective (making it invertible, as in this case) or else it is both not injective and not surjective.

$\square$  Matrix representation of the linear transformation, as described in Theorem MLTCV. (See also Example MOLT.) If $A$ is the matrix below, then $\lt{T}{\vect{x}} = A\vect{x}$. This computation may also be viewed as an application of Definition MR and Theorem FTMR from Section MR, where the bases are chosen to be the standard bases of $\complex{m}$ (Definition SUV).\begin{bmatrix} -65&128&10&-262&40\\ 36&-73&-1&151&-16\\ -44&88&5&-180&24\\ 34&-68&-3&140&-18\\ 12&-24&-1&49&-5 \end{bmatrix}

$\square$  Eigenvalues, and bases for eigenspaces (Definition EELT, Theorem EER). Evaluate the linear transformation with each eigenvector as an interesting check.\begin{align*}\eigensystem{T}{-1}{\colvector{-57\\0\\-18\\14\\5},\,\colvector{2\\1\\0\\0\\0}}\\ \eigensystem{T}{1}{\colvector{-10\\-5\\-6\\0\\1},\,\colvector{2\\3\\1\\1\\0}}\\ \eigensystem{T}{2}{\colvector{-6\\3\\-4\\3\\1}} \end{align*}

$\square$  A diagonal matrix representation relative to a basis of eigenvectors. \begin{align*} B&=\set{\colvector{-57\\0\\-18\\14\\5},\,\colvector{2\\1\\0\\0\\0},\, \colvector{-10\\-5\\-6\\0\\1},\,\colvector{2\\3\\1\\1\\0},\,\colvector{-6\\3\\-4\\3\\1}} &&\text{(Domain, codomain basis)}\\ \matrixrep{T}{B}{B}&=\begin{bmatrix} -1&0&0&0&0\\ 0&-1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&2\end{bmatrix} \end{align*}