Archetype Q Summary: Linear transformation with equal-sized domain and codomain, so it has the potential to be invertible, but in this case is not. Neither injective nor surjective. Diagonalizable, though.
\begin{equation*}
\ltdefn{T}{\complex{5}}{\complex{5}},
\lt{T}{\colvector{x_1\\x_2\\x_3\\x_4\\x_5}}=
\colvector{-2 x_1 + 3 x_2 + 3 x_3 - 6 x_4 + 3 x_5\\
-16 x_1 + 9 x_2 + 12 x_3 - 28 x_4 + 28 x_5\\
-19 x_1 + 7 x_2 + 14 x_3 - 32 x_4 + 37 x_5\\
-21 x_1 + 9 x_2 + 15 x_3 - 35 x_4 + 39 x_5\\
-9 x_1 + 5 x_2 + 7 x_3 - 16 x_4 + 16 x_5}
\end{equation*}
◊ A basis for the null space of the linear transformation: (
Definition KLT)
\begin{equation*}
\set{\colvector{3\\4\\1\\3\\3}}
\end{equation*}
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 2, just from checking dimensions of the domain and the codomain. In particular, verify that
\begin{align*}
\lt{T}{\colvector{1\\3\\-1\\2\\4}}&=\colvector{4\\55\\72\\77\\31}&
\lt{T}{\colvector{4\\7\\0\\5\\7}}&=\colvector{4\\55\\72\\77\\31}
\end{align*}
This demonstration that $T$ is not injective is constructed with the observation that
\begin{align*}
\colvector{4\\7\\0\\5\\7}&=\colvector{1\\3\\-1\\2\\4}+\colvector{3\\4\\1\\3\\3}
\end{align*}
and
\begin{align*}
\vect{z}&=\colvector{3\\4\\1\\3\\3}\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{
\colvector{-2\\-16\\-19\\-21\\-9},\,\colvector{3\\9\\7\\9\\5},\,
\colvector{3\\12\\14\\15\\7},\,\colvector{-6\\-28\\-32\\-35\\-16},\,
\colvector{3\\28\\37\\39\\16}
}
\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\\1},\,\colvector{0\\1\\0\\0\\-1},\,
\colvector{0\\0\\1\\0\\-1},\,\colvector{0\\0\\0\\1\\2}
}
\end{equation*}
The dimension of the range is 4, and the codomain ($\complex{5}$) has dimension 5. So $\rng{T}\neq\complex{5}$ and by
Theorem RSLT the transformation is not surjective.
To be more precise, verify that $\colvector{-1\\2\\3\\-1\\4}\not\in\rng{T}$, 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, $\preimage{T}{\colvector{-1\\2\\3\\-1\\4}}$, 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: }5&&
\text{Rank: }4&&
\text{Nullity: }1
\end{align*}
◊ Invertible: No.
Neither injective nor surjective. Notice that since the domain and codomain have the same dimension, either the transformation is both onto and one-to-one (making it invertible) or else it is both not onto and not one-to-one (as in this case) by Theorem RPNDD.
◊ Matrix representation (Theorem MLTCV):
\begin{equation*}
\ltdefn{T}{\complex{5}}{\complex{5}}, \lt{T}{\vect{x}}=A\vect{x}, A=
\begin{bmatrix}
-2&3&3&-6&3\\
-16&9&12&-28&28\\
-19&7&14&-32&37\\
-21&9&15&-35&39\\
-9&5&7&-16&16
\end{bmatrix}
\end{equation*}
◊ Eigenvalues and eigenvectors (Definition EELT, Theorem EER):
\begin{align*}
\eigensystem{T}{-1}{\colvector{0\\2\\3\\3\\1}}\\
\eigensystem{T}{0}{\colvector{3\\4\\1\\3\\3}}\\
\eigensystem{T}{1}{\colvector{5\\3\\0\\0\\2},\,\colvector{-3\\1\\0\\2\\0},\,\colvector{1\\-1\\2\\0\\0}}
\end{align*}
◊ A diagonal matrix representation relative to a basis of eigenvectors, $B$.
\begin{align*}
B&=
\set{
\colvector{0\\2\\3\\3\\1},\,\colvector{3\\4\\1\\3\\3},\,
\colvector{5\\3\\0\\0\\2},\,\colvector{-3\\1\\0\\2\\0},\,
\colvector{1\\-1\\2\\0\\0}}
\\
&\\
\matrixrep{T}{B}{B}&=\begin{bmatrix}
-1&0&0&0&0\\
0&0&0&0&0\\
0&0&1&0&0\\
0&0&0&1&0\\
0&0&0&0&1
\end{bmatrix}
\end{align*}
