### Archetype V

From A First Course in Linear Algebra
Version 2.99
http://linear.ups.edu/

Summary  Domain is polynomials, codomain is matrices. Domain and codomain both have dimension 4. Injective, surjective, invertible. Square matrix representation, but domain and codomain are unequal, so no eigenvalue information.

A linear transformation: (Definition LT)

 T : {P}_{3} → {M}_{22},\quad T\left (a + bx + c{x}^{2} + d{x}^{3}\right ) = \left [\array{ a + b&a − 2c \cr d & b − d } \right ]

A basis for the null space of the linear transformation: (Definition KLT)

 \left \{\ \right \}

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):

 \left \{\left [\array{ 1&1\cr 0&0 } \right ],\kern 1.95872pt \left [\array{ 1&0\cr 0&1 } \right ],\kern 1.95872pt \left [\array{ 0&−2\cr 0& 0 } \right ],\kern 1.95872pt \left [\array{ 0& 0\cr 1&−1 } \right ]\right \}

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:

 \left \{\left [\array{ 1&0\cr 0&0 } \right ],\kern 1.95872pt \left [\array{ 0&1\cr 0&0 } \right ],\kern 1.95872pt \left [\array{ 0&0\cr 1&0 } \right ],\kern 1.95872pt \left [\array{ 0&0\cr 0&1 } \right ]\right \}

Surjective: Yes. (Definition SLT)
A basis for the range is the standard basis of {M}_{22}, so ℛ\kern -1.95872pt \left (T\right ) = {M}_{22} and Theorem RSLT tells us T is surjective. Or, the dimension of the range is 4, and the codomain ({M}_{22}) has dimension 4. So the transformation is surjective.

Subspace dimensions associated with the linear transformation. Examine parallels with earlier results for matrices. Verify Theorem RPNDD.

\eqalignno{ \text{Domain dimension: }4 & &\text{Rank: }4 & &\text{Nullity: }0 & & & & & & }

Invertible: Yes.
Both injective and surjective (Theorem ILTIS). Notice that since the domain and codomain have the same dimension, either the transformation is both injective and surjective (making it invertible, as in this case) or else it is both not injective and not surjective.

Matrix representation (Definition MR):

\eqalignno{ B & = \left \{1,\kern 1.95872pt x,\kern 1.95872pt {x}^{2},\kern 1.95872pt {x}^{3}\right \} & & \cr C & = \left \{\left [\array{ 1&0\cr 0&0 } \right ],\kern 1.95872pt \left [\array{ 0&1\cr 0&0 } \right ],\kern 1.95872pt \left [\array{ 0&0\cr 1&0 } \right ],\kern 1.95872pt \left [\array{ 0&0\cr 0&1 } \right ]\right \} & & \cr {M}_{B,C}^{T } & = \left [\array{ 1&1& 0 & 0\cr 1&0 &−2 & 0 \cr 0&0& 0 & 1\cr 0&1 & 0 &−1 } \right ] & & }

Since invertible, the inverse linear transformation. (Definition IVLT)

 { T}^{−1}: {M}_{ 22} → {P}_{3},\quad {T}^{−1}\left (\left [\array{ a&b \cr c&d } \right ]\right ) = (a−c−d)+(c+d)x+{1\over 2}(a−b−c−d){x}^{2}+c{x}^{3}