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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*}
‚¨ú¬†¬†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 ‚Äúnice‚ÄĚ 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*}