Archetype T Summary: Domain and codomain are polynomials. Domain has dimension 5, while codomain has dimension 6. Is injective, can't be surjective.

◊ A linear transformation: (Definition LT)
\begin{equation*} \ltdefn{T}{P_4}{P_5},    \lt{T}{p(x)}=(x-2)p(x) \end{equation*}

◊ A basis for the null space of the linear transformation: (Definition KLT)
\begin{equation*} \set{\ } \end{equation*}

◊ Injective: Yes. (Definition ILT)
Since the kernel is trivial Theorem KILT tells us that the linear transformation is injective.

◊ 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{ x-2,\, x^2-2x,\, x^3-2x^2,\, x^4-2x^3, x^5-2x^4, x^6-2x^5 } \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{ -\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: No. (Definition SLT)
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 $1+x+x^2+x^3+x^4\not\in\rng{T}$, 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, $\preimage{T}{1+x+x^2+x^3+x^4}$, is nonempty. 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: }5&& \text{Nullity: }0 \end{align*}

◊ Invertible: No.

The relative dimensions of the domain and codomain prohibit any possibility of being surjective, so apply Theorem ILTIS.

◊ Matrix representation (Definition MR):

\begin{align*} B&=\set{1,\,x,\,x^2,\,x^3,\,x^4}\\ &\\ C&=\set{1,\,x,\,x^2,\,x^3,\,x^4,\,x^5}\\ &\\ \matrixrep{T}{B}{C}&= \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{align*}