Archetype R Summary: Linear transformation with equal-sized domain and codomain. Injective, surjective, invertible, diagonalizable, the works.
\begin{equation*}
\ltdefn{T}{\complex{5}}{\complex{5}},
\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*}
◊ A basis for the null space of the linear transformation: (
Definition KLT)
\begin{equation*}
\set{\ }
\end{equation*}
Since the kernel is trivial
Theorem KILT tells us that the linear transformation 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{-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{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\\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{equation*}
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.
◊ Subspace dimensions associated with
the linear transformation. Examine parallels with earlier results for matrices. Verify
Theorem RPNDD.
\begin{align*}
\text{Domain dimension: }5&&
\text{Rank: }5&&
\text{Nullity: }0
\end{align*}
◊ Invertible: 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.
◊ Matrix representation (Theorem MLTCV):
\begin{equation*}
\ltdefn{T}{\complex{5}}{\complex{5}}, \lt{T}{\vect{x}}=A\vect{x}, A=
\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}
\end{equation*}
◊ The inverse linear transformation (Definition IVLT):
\begin{equation*}
T^{-1}:\complex{5}\rightarrow\complex{5},
T^{-1}\left(\colvector{x_1\\x_2\\x_3\\x_4\\x_5}\right)=
\colvector{-47 x_1 + 92 x_2 + x_3 - 181 x_4 - 14 x_5\\
27 x_1 - 55 x_2 + \frac{7}{2} x_3 + \frac{221}{2} x_4 + 11 x_5\\
-32 x_1 + 64 x_2 - x_3 - 126 x_4 - 12 x_5\\
25 x_1 - 50 x_2 + \frac{3}{2} x_3 + \frac{199}{2} x_4 + 9 x_5\\
9 x_1 - 18 x_2 + \frac{1}{2} x_3 + \frac{71}{2} x_4 + 4 x_5}
\end{equation*}
◊ Eigenvalues and eigenvectors (Definition EELT, Theorem EER):
\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*}
◊ A diagonal matrix representation relative to a basis of eigenvectors, $B$.
\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}}
\\
&\\
\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*}
