### Archetype R

From A First Course in Linear Algebra
Version 2.00
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/

Summary  Linear transformation with equal-sized domain and codomain. Injective, surjective, invertible, diagonalizable, the works.

A linear transformation: (Definition LT)

 T : {ℂ}^{5}\mathrel{↦}{ℂ}^{5},\quad T\left (\left [\array{ {x}_{1} \cr {x}_{2} \cr {x}_{3} \cr {x}_{4} \cr {x}_{5} } \right ]\right ) = \left [\array{ −65{x}_{1} + 128{x}_{2} + 10{x}_{3} − 262{x}_{4} + 40{x}_{5} \cr 36{x}_{1} − 73{x}_{2} − {x}_{3} + 151{x}_{4} − 16{x}_{5} \cr −44{x}_{1} + 88{x}_{2} + 5{x}_{3} − 180{x}_{4} + 24{x}_{5} \cr 34{x}_{1} − 68{x}_{2} − 3{x}_{3} + 140{x}_{4} − 18{x}_{5} \cr 12{x}_{1} − 24{x}_{2} − {x}_{3} + 49{x}_{4} − 5{x}_{5} } \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{ −65 \cr 36 \cr −44 \cr 34 \cr 12} \right ],\kern 1.95872pt \left [\array{ 128 \cr −73 \cr 88 \cr −68 \cr −24 } \right ],\kern 1.95872pt \left [\array{ 10 \cr −1 \cr 5 \cr −3 \cr −1 } \right ],\kern 1.95872pt \left [\array{ −262 \cr 151 \cr −180 \cr 140 \cr 49 } \right ],\kern 1.95872pt \left [\array{ 40 \cr −16 \cr 24 \cr −18 \cr −5 } \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 0 \cr 0 } \right ],\kern 1.95872pt \left [\array{ 0 \cr 1 \cr 0 \cr 0 \cr 0 } \right ],\kern 1.95872pt \left [\array{ 0 \cr 0 \cr 1 \cr 0 \cr 0 } \right ],\kern 1.95872pt \left [\array{ 0 \cr 0 \cr 0 \cr 1 \cr 0 } \right ],\kern 1.95872pt \left [\array{ 0 \cr 0 \cr 0 \cr 0 \cr 1 } \right ]\right \}

Surjective: Yes. (Definition SLT)
A basis for the range is the standard basis of {ℂ}^{5}, so ℛ\kern -1.95872pt \left (T\right ) = {ℂ}^{5} and Theorem RSLT tells us T is surjective. Or, the dimension of the range is 5, and the codomain ({ℂ}^{5}) has dimension 5. 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: }5 & &\text{Rank: }5 & &\text{Nullity: }0 & & & & & & }

Invertible: Yes.
Both injective and surjective (Theorem ILTIS). Notice that since the domain and codomain have the same dimesion, 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 (Theorem MLTCV):

 T : {ℂ}^{5}\mathrel{↦}{ℂ}^{5},\quad T\left (x\right ) = Ax,\quad A = \left [\array{ −65&128&10&−262& 40 \cr 36 &−73&−1& 151 &−16 \cr −44& 88 & 5 &−180& 24 \cr 34 &−68&−3& 140 &−18 \cr 12 &−24&−1& 49 & −5 } \right ]

The inverse linear transformation (Definition IVLT):

 { T}^{−1} : {ℂ}^{5} → {ℂ}^{5},\quad {T}^{−1}\left (\left [\array{ {x}_{1} \cr {x}_{2} \cr {x}_{3} \cr {x}_{4} \cr {x}_{5} } \right ]\right ) = \left [\array{ −47{x}_{1} + 92{x}_{2} + {x}_{3} − 181{x}_{4} − 14{x}_{5} \cr 27{x}_{1} − 55{x}_{2} + {7\over 2}{x}_{3} + {221\over 4} {x}_{4} + 11{x}_{5} \cr −32{x}_{1} + 64{x}_{2} − {x}_{3} − 126{x}_{4} − 12{x}_{5} \cr 25{x}_{1} − 50{x}_{2} + {3\over 2}{x}_{3} + {199\over 2} {x}_{4} + 9{x}_{5} \cr 9{x}_{1} − 18{x}_{2} + {1\over 2}{x}_{3} + {71\over 2} {x}_{4} + 4{x}_{5} } \right ]

Verify that T\left (T−1\left (x\right )\right ) = x and T\left (T−1\left (x\right )\right ) = x, and notice that the representations of the transformation and its inverse are matrix inverses (Theorem IMR, Definition MI).

Eigenvalues and eigenvectors (Definition EELT, Theorem EER):

\eqalignno{ λ & = −1 &{ℰ}_{T }\left (−1\right ) & = \left \langle \left \{\left [\array{ −57 \cr 0 \cr −18 \cr 14 \cr 5 } \right ],\kern 1.95872pt \left [\array{ 2 \cr 1 \cr 0 \cr 0 \cr 0 } \right ]\right \}\right \rangle & & & & \cr λ & = 1 &{ℰ}_{T }\left (1\right ) & = \left \langle \left \{\left [\array{ −10 \cr −5 \cr −6 \cr 0 \cr 1 } \right ],\kern 1.95872pt \left [\array{ 2 \cr 3 \cr 1 \cr 1 \cr 0 } \right ]\right \}\right \rangle & & & & \cr λ & = 2 &{ℰ}_{T }\left (2\right ) & = \left \langle \left \{\left [\array{ −6 \cr 3 \cr −4 \cr 3 \cr 1 } \right ]\right \}\right \rangle & & & & }

Evaluate the linear transformation with each of these eigenvectors as an interesting check.

A diagonal matrix representation relative to a basis of eigenvectors, B.

\eqalignno{ B & = \left \{\left [\array{ −57 \cr 0 \cr −18 \cr 14 \cr 5 } \right ],\kern 1.95872pt \left [\array{ 2 \cr 1 \cr 0 \cr 0 \cr 0 } \right ],\kern 1.95872pt \left [\array{ −10 \cr −5 \cr −6 \cr 0 \cr 1 } \right ],\kern 1.95872pt \left [\array{ 2 \cr 3 \cr 1 \cr 1 \cr 0 } \right ],\kern 1.95872pt \left [\array{ −6 \cr 3 \cr −4 \cr 3 \cr 1 } \right ]\right \} & & \cr {M}_{B,B}^{T } & = \left [\array{ −1& 0 &0&0&0 \cr 0 &−1&0&0&0 \cr 0 & 0 &1&0&0 \cr 0 & 0 &0&1&0 \cr 0 & 0 &0&0&2 } \right ] & & }