Archetype O Summary: Linear transformation with a domain smaller than the codomain, so it is guaranteed to not be onto. Happens to not be one-to-one.
\begin{equation*}
\ltdefn{T}{\complex{3}}{\complex{5}},
\lt{T}{\colvector{x_1\\x_2\\x_3}}=
\colvector{-x_1 + x_2 - 3 x_3\\
-x_1 + 2 x_2 - 4 x_3\\
x_1 + x_2 + x_3\\
2 x_1 + 3 x_2 + x_3\\
x_1 + 2 x_3
}
\end{equation*}
◊ A basis for the null space of the linear transformation: (
Definition KLT)
\begin{equation*}
\set{\colvector{-2\\1\\1}}
\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{5\\-1\\3}}&=\colvector{-15\\-19\\7\\10\\11}&
\lt{T}{\colvector{1\\1\\5}}&=\colvector{-15\\-19\\7\\10\\11}
\end{align*}
This demonstration that $T$ is not injective is constructed with the observation that
\begin{align*}
\colvector{1\\1\\5}&=\colvector{5\\-1\\3}+\colvector{-4\\2\\2}
\end{align*}
and
\begin{align*}
\vect{z}&=\colvector{-4\\2\\2}\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{-1\\-1\\1\\2\\1},\,\colvector{1\\2\\1\\3\\0},\,
\colvector{-3\\-4\\1\\1\\2}}
\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\\-3\\-7\\-2},\,\colvector{0\\1\\2\\5\\1}
}
\end{equation*}
◊ 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*}
The dimension of the range is 2, and the codomain ($\complex{5}$) has dimension 5. So the transformation is not onto. 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 onto.
To be more precise, verify that $\colvector{2\\3\\1\\1\\1}\not\in\rng{T}$, 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, $\preimage{T}{\colvector{2\\3\\1\\1\\1}}$, is empty. This alone is sufficient to see that the linear transformation is not onto.
◊ Invertible: No.
Not injective, and the relative dimensions of the domain and codomain prohibit any possibility of being surjective.
◊ Matrix representation (Theorem MLTCV):
\begin{equation*}
\ltdefn{T}{\complex{3}}{\complex{5}}, \lt{T}{\vect{x}}=A\vect{x}, A=
\begin{bmatrix}
-1&1&-3\\
-1&2&-4\\
1&1&1\\
2&3&1\\
1&0&2
\end{bmatrix}
\end{equation*}
