### Archetype Q

From A First Course in Linear Algebra
Version 2.90
http://linear.ups.edu/

Summary  Linear transformation with equal-sized domain and codomain, so it has the potential to be invertible, but in this case is not. Neither injective nor surjective. Diagonalizable, though.

A linear transformation: (Definition LT)

 T : {ℂ}^{5} → {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{ −2{x}_{1} + 3{x}_{2} + 3{x}_{3} − 6{x}_{4} + 3{x}_{5} \cr −16{x}_{1} + 9{x}_{2} + 12{x}_{3} − 28{x}_{4} + 28{x}_{5} \cr −19{x}_{1} + 7{x}_{2} + 14{x}_{3} − 32{x}_{4} + 37{x}_{5} \cr −21{x}_{1} + 9{x}_{2} + 15{x}_{3} − 35{x}_{4} + 39{x}_{5} \cr −9{x}_{1} + 5{x}_{2} + 7{x}_{3} − 16{x}_{4} + 16{x}_{5} } \right ]

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

 \left \{\left [\array{ 3\cr 4 \cr 1\cr 3 \cr 3 } \right ]\right \}

Injective: No. (Definition ILT)
Since the kernel is nontrivial Theorem KILT tells us that the linear transformation is not injective. Also, since the rank can not exceed 3, we are guaranteed to have a nullity of at least 2, just from checking dimensions of the domain and the codomain. In particular, verify that

\eqalignno{ T\left (\left [\array{ 1\cr 3 \cr −1\cr 2 \cr 4 } \right ]\right ) & = \left [\array{ 4\cr 55 \cr 72\cr 77 \cr 31 } \right ] &T\left (\left [\array{ 4\cr 7 \cr 0\cr 5 \cr 7 } \right ]\right ) & = \left [\array{ 4\cr 55 \cr 72\cr 77 \cr 31 } \right ] & & & & }

This demonstration that T is not injective is constructed with the observation that

\eqalignno{ \left [\array{ 4\cr 7 \cr 0\cr 5 \cr 7 } \right ] & = \left [\array{ 1\cr 3 \cr −1\cr 2 \cr 4 } \right ] + \left [\array{ 3\cr 4 \cr 1\cr 3 \cr 3 } \right ] & & \text{and} \cr z & = \left [\array{ 3\cr 4 \cr 1\cr 3 \cr 3 } \right ] ∈K\kern -1.95872pt \left (T\right ) & & }

so the vector z effectively “does nothing” in the evaluation of T.

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{ −2\cr −16 \cr −19\cr −21 \cr −9 } \right ],\kern 1.95872pt \left [\array{ 3\cr 9 \cr 7\cr 9 \cr 5 } \right ],\kern 1.95872pt \left [\array{ 3\cr 12 \cr 14\cr 15 \cr 7 } \right ],\kern 1.95872pt \left [\array{ −6\cr −28 \cr −32\cr −35 \cr −16 } \right ],\kern 1.95872pt \left [\array{ 3\cr 28 \cr 37\cr 39 \cr 16 } \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 1 } \right ],\kern 1.95872pt \left [\array{ 0\cr 1 \cr 0\cr 0 \cr −1 } \right ],\kern 1.95872pt \left [\array{ 0\cr 0 \cr 1\cr 0 \cr −1 } \right ],\kern 1.95872pt \left [\array{ 0\cr 0 \cr 0\cr 1 \cr 2 } \right ]\right \}

Surjective: No. (Definition SLT)
The dimension of the range is 4, and the codomain ({ℂ}^{5}) has dimension 5. So ℛ\kern -1.95872pt \left (T\right )\mathrel{≠}{ℂ}^{5} and by Theorem RSLT the transformation is not surjective.

To be more precise, verify that \left [\array{ −1\cr 2 \cr 3\cr −1 \cr 4 } \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{ −1\cr 2 \cr 3\cr −1 \cr 4 } \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: }5 & &\text{Rank: }4 & &\text{Nullity: }1 & & & & & & }

Invertible: No.
Neither injective nor surjective. Notice that since the domain and codomain have the same dimension, either the transformation is both onto and one-to-one (making it invertible) or else it is both not onto and not one-to-one (as in this case) by Theorem RPNDD.

Matrix representation (Theorem MLTCV):

 T : {ℂ}^{5} → {ℂ}^{5},\quad T\left (x\right ) = Ax,\quad A = \left [\array{ −2 &3& 3 & −6 & 3\cr −16 &9 &12 &−28 &28 \cr −19&7&14&−32&37\cr −21 &9 &15 &−35 &39 \cr −9 &5& 7 &−16&16 } \right ]

Eigenvalues and eigenvectors (Definition EELT, Theorem EER):

\eqalignno{ λ & = −1 &{ℰ}_{T }\left (−1\right ) & = \left \langle \left \{\left [\array{ 0\cr 2 \cr 3\cr 3 \cr 1 } \right ]\right \}\right \rangle & & & & \cr λ & = 0 &{ℰ}_{T }\left (0\right ) & = \left \langle \left \{\left [\array{ 3\cr 4 \cr 1\cr 3 \cr 3 } \right ]\right \}\right \rangle & & & & \cr λ & = 1 &{ℰ}_{T }\left (1\right ) & = \left \langle \left \{\left [\array{ 5\cr 3 \cr 0\cr 0 \cr 2 } \right ],\kern 1.95872pt \left [\array{ −3\cr 1 \cr 0\cr 2 \cr 0 } \right ],\kern 1.95872pt \left [\array{ 1\cr −1 \cr 2\cr 0 \cr 0 } \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{ 0\cr 2 \cr 3\cr 3 \cr 1 } \right ],\kern 1.95872pt \left [\array{ 3\cr 4 \cr 1\cr 3 \cr 3 } \right ],\kern 1.95872pt \left [\array{ 5\cr 3 \cr 0\cr 0 \cr 2 } \right ],\kern 1.95872pt \left [\array{ −3\cr 1 \cr 0\cr 2 \cr 0 } \right ],\kern 1.95872pt \left [\array{ 1\cr −1 \cr 2\cr 0 \cr 0 } \right ]\right \} & & \cr {M}_{B,B}^{T } & = \left [\array{ −1&0&0&0&0\cr 0 &0 &0 &0 &0 \cr 0 &0&1&0&0\cr 0 &0 &0 &1 &0 \cr 0 &0&0&0&1 } \right ] & & }