Archetype P Archetype P
⬜ Summary Linear transformation with a domain smaller that its codomain, so it is guaranteed to not be surjective. Happens to be injective.
⬜
Definition A linear transformation (
Definition LT).
\begin{equation*}
\ltdefn{T}{\complex{3}}{\complex{5}},\quad
\lteval{T}{\colvector{x_1\\x_2\\x_3}}=
\colvector{-x_1 + x_2 + x_3\\
-x_1 + 2 x_2 + 2 x_3\\
x_1 + x_2 + 3 x_3\\
2 x_1 + 3 x_2 + x_3\\
-2 x_1 + x_2 + 3 x_3}
\end{equation*}
⬜
Kernel A basis for the kernel of the linear transformation (
Definition KLT).
\begin{equation*}
\set{\ }
\end{equation*}
⬜
Injective? Is the linear transformation injective (
Definition ILT)? Yes.
Since \(\krn{T}=\set{\zerovector}\text{,}\) Theorem KILT tells us that \(T\) is injective.
⬜
Spanning Set for Range 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{equation*}
\set{\colvector{-1\\-1\\1\\2\\-2},\,\colvector{1\\2\\1\\3\\1},\,\colvector{1\\2\\3\\1\\3}}
\end{equation*}
⬜
Range 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{equation*}
\set{\colvector{1\\0\\0\\-10\\6},\,\colvector{0\\1\\0\\7\\-3},\,\colvector{0\\0\\1\\-1\\1}}
\end{equation*}
⬜
Surjective? Is the linear transformation surjective (
Definition SLT)? No.
The dimension of the range is 3, and the codomain (\(\complex{5}\)) has dimension 5. So the transformation is not surjective. 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 surjective in this situation.
To be more precise, verify that \(\colvector{2\\1\\-3\\2\\6}\not\in\rng{T}\text{,}\) 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, \(\preimage{T}{\colvector{2\\1\\-3\\2\\6}}\text{,}\) is empty. This alone is sufficient to see that the linear transformation is not onto.
⬜
Subspace Dimensions 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}&=0&\text{domain}&=3
\end{align*}
⬜
Invertible? Is the linear transformation invertible (
Definition IVLT, and examine parallels with the existence of matrix inverses.)? No.
The relative dimensions of the domain and codomain prohibit any possibility of being surjective, so apply Theorem ILTIS.
⬜
Matrix Representation Matrix representation of the linear transformation, as described in
Theorem MLTCV. (See also
Example MOLT.) If \(A\) is the matrix below, then \(\lteval{T}{\vect{x}} = A\vect{x}\text{.}\) 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{equation*}
\begin{bmatrix}
-1&1&1\\
-1&2&2\\
1&1&3\\
2&3&1\\
-2&1&3
\end{bmatrix}
\end{equation*}
