Archetype
Archetype Archetype Dark Mode Archetype Archetype Archetype Scratch ActiveCode Profile \(\newcommand{\orderof}[1]{\sim #1}
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Archetype T Archetype T
⬜
⬜ Summary Domain and codomain are polynomials. Domain has dimension 5, while codomain has dimension 6. Is injective, can not be surjective.
⬜
Definition A linear transformation (
Definition LT ).
\begin{equation*}
\ltdefn{T}{P_4}{P_5},\quad\lteval{T}{p(x)}=(x-2)p(x)
\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 the kernel is trivial
Theorem KILT tells us that the linear transformation 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{
x-2,\,
x^2-2x,\,
x^3-2x^2,\,
x^4-2x^3,
x^5-2x^4,
x^6-2x^5}
\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{
-\frac{1}{32}x^5+1,\,
-\frac{1}{16}x^5+x,\,
-\frac{1}{8}x^5+x^2,\,
-\frac{1}{4}x^5+x^3,\,
-\frac{1}{2}x^5+x^4
}
\end{equation*}
⬜
Surjective? Is the linear transformation surjective (
Definition SLT )? No.
The dimension of the range is 5, and the codomain (
\(P_5\) ) has dimension 6. So the transformation is not surjective. Notice too that since the domain
\(P_4\) has dimension 5, it is impossible for the range to have a dimension greater than 5, and no matter what the actual definition of the function, it cannot possibly be surjective in this situation.
To be more precise, verify that
\(1+x+x^2+x^3+x^4\not\in\rng{T}\text{,}\) 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,
\(\preimage{T}{1+x+x^2+x^3+x^4}\text{,}\) is nonempty. 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}&=5&\text{nullity}&=0&\text{domain}&=5
\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 given by
Definition MR and explained by
Theorem FTMR .
\begin{equation*}
\text{domain basis}=\set{1,\,x,\,x^2,\,x^3,\,x^4}
\end{equation*}
\begin{equation*}
\text{codomain basis}=\set{1,\,x,\,x^2,\,x^3,\,x^4,\,x^5}
\end{equation*}
\begin{equation*}
\text{matrix representation}=\begin{bmatrix}
-2 & 0 & 0 & 0 & 0 \\
1 & -2 & 0 & 0 & 0 \\
0 & 1 & -2 & 0 & 0 \\
0 & 0 & 1 & -2 & 0 \\
0 & 0 & 0 & 1 & -2 \\
0 & 0 & 0 & 0 & 1
\end{bmatrix}
\end{equation*}
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