### Archetype P

From A First Course in Linear Algebra
Version 2.11
http://linear.ups.edu/

Summary  Linear transformation with a domain smaller that its codomain, so it is guaranteed to not be surjective. Happens to be injective.

A linear transformation: (Definition LT)

 T : {ℂ}^{3}\mathrel{↦}{ℂ}^{5},\quad T\left (\left [\array{ {x}_{1}\cr {x}_{2} \cr {x}_{3} } \right ]\right ) = \left [\array{ −{x}_{1} + {x}_{2} + {x}_{3} \cr −{x}_{1} + 2{x}_{2} + 2{x}_{3} \cr {x}_{1} + {x}_{2} + 3{x}_{3} \cr 2{x}_{1} + 3{x}_{2} + {x}_{3} \cr −2{x}_{1} + {x}_{2} + 3{x}_{3} } \right ]

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

 \left \{\ \right \}

Injective: Yes. (Definition ILT)
Since K\kern -1.95872pt \left (T\right ) = \left \{0\right \}, Theorem KILT tells us that T 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\cr −1 \cr 1\cr 2 \cr −2 } \right ],\kern 1.95872pt \left [\array{ 1\cr 2 \cr 1\cr 3 \cr 1 } \right ],\kern 1.95872pt \left [\array{ 1\cr 2 \cr 3\cr 1 \cr 3 } \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\cr 0 \cr 0\cr −10 \cr 6 } \right ],\kern 1.95872pt \left [\array{ 0\cr 1 \cr 0\cr 7 \cr −3 } \right ],\kern 1.95872pt \left [\array{ 0\cr 0 \cr 1\cr −1 \cr 1 } \right ]\right \}

Surjective: No. (Definition SLT)
The dimension of the range is 3, and the codomain ({ℂ}^{5}) has dimension 5. So the transformation is not surjective. Notice too that since the domain {ℂ}^{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 \left [\array{ 2\cr 1 \cr −3\cr 2 \cr 6 } \right ]∉ℛ\kern -1.95872pt \left (T\right ), 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, {T}^{−1}\left (\left [\array{ 2\cr 1 \cr −3\cr 2 \cr 6 } \right ]\right ), is empty. 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.

\eqalignno{ \text{Domain dimension: }3 & &\text{Rank: }3 & &\text{Nullity: }0 & & & & & & }

Invertible: No.
The relative dimensions of the domain and codomain prohibit any possibility of being surjective, so apply Theorem ILTIS.

Matrix representation (Theorem MLTCV):

 T : {ℂ}^{3}\mathrel{↦}{ℂ}^{5},\quad T\left (x\right ) = Ax,\quad A = \left [\array{ −1&1&1\cr −1 &2 &2 \cr 1 &1&3\cr 2 &3 &1 \cr −2&1&3 } \right ]