### Archetype X

From A First Course in Linear Algebra
Version 2.90
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/

Summary  Domain and codomain are square matrices. Domain and codomain both have dimension 4. Not injective, not surjective, not invertible, 3 distinct eigenvalues, diagonalizable.

A linear transformation: (Definition LT)

 T : {M}_{22} → {M}_{22},\quad T\left (\left [\array{ a&b\cr c&d } \right ]\right ) = \left [\array{ −2a + 15b + 3c + 27d& 10b + 6c + 18d\cr a − 5b − 9d &−a − 4b − 5c − 8d } \right ]

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

 \left \{\left [\array{ −6&−3\cr 2 & 1 } \right ]\right \}

Injective: No. (Definition ILT)
Since the kernel is nontrivial Theorem KILT tells us that the linear transformation is not injective. In particular, verify that

\eqalignno{ T\left (\left [\array{ −2& 0\cr 1 &−4 } \right ]\right ) & = \left [\array{ 115& 78\cr −38 &−35 } \right ] &T\left (\left [\array{ 4 &3\cr −1 &3 } \right ]\right ) & = \left [\array{ 115& 78\cr −38 &−35 } \right ] & & & & }

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

\eqalignno{ \left [\array{ 4 &3\cr −1 &3 } \right ] & = \left [\array{ −2& 0\cr 1 &−4 } \right ] + \left [\array{ 6 & 3\cr −2 &−1 } \right ] & & \text{and} \cr z & = \left [\array{ 6 & 3\cr −2 &−1 } \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& 0\cr 1 &−1 } \right ],\kern 1.95872pt \left [\array{ 15&10\cr −5 &−4 } \right ],\kern 1.95872pt \left [\array{ 3& 6\cr 0&−5 } \right ],\kern 1.95872pt \left [\array{ 27&18\cr −9 &−8 } \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 −{1\over 2}&0 } \right ],\kern 1.95872pt \left [\array{ 0&1\cr {1\over 4}&0 } \right ],\kern 1.95872pt \left [\array{ 0&0\cr 0&1 } \right ]\right \}

Surjective: No. (Definition SLT)
The dimension of the range is 3, and the codomain ({M}_{22}) has dimension 5. So ℛ\kern -1.95872pt \left (T\right )\mathrel{≠}{M}_{22} and by Theorem RSLT the transformation is not surjective.

To be more precise, verify that \left [\array{ 2&4\cr 3&1 } \right ]∉ℛ\kern -1.95872pt \left (T\right ), by setting the output of T equal to this matrix 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&4 \cr 3&1 } \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: }4 & &\text{Rank: }3 & &\text{Nullity: }1 & & & & & & }

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

Matrix representation (Definition MR):

\eqalignno{ B & = \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 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{ −2&15& 3 &27\cr 0 & 10 & 6 & 18 \cr 1 &−5& 0 &−9\cr −1 &−4 &−5 &−8 } \right ] & & }

Eigenvalues and eigenvectors (Definition EELT, Theorem EER):

\eqalignno{ λ & = 0 &{ℰ}_{T }\left (0\right ) & = \left \langle \left \{\left [\array{ −6&−3\cr 2 & 1 } \right ]\right \}\right \rangle & & & & \cr λ & = 1 &{ℰ}_{T }\left (1\right ) & = \left \langle \left \{\left [\array{ −7&−2\cr 3 & 0 } \right ],\kern 1.95872pt \left [\array{ −1&−2\cr 0 & 1 } \right ]\right \}\right \rangle & & & & \cr λ & = 3 &{ℰ}_{T }\left (3\right ) & = \left \langle \left \{\left [\array{ −3&−2\cr 1 & 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{ −6&−3\cr 2 & 1 } \right ],\kern 1.95872pt \left [\array{ −7&−2\cr 3 & 0 } \right ],\kern 1.95872pt \left [\array{ −1&−2\cr 0 & 1 } \right ],\kern 1.95872pt \left [\array{ −3&−2\cr 1 & 1 } \right ]\right \} & & \cr {M}_{B,B}^{T } & = \left [\array{ 0&0&0&0\cr 0&1 &0 &0 \cr 0&0&3&0\cr 0&0 &0 &3 } \right ] & & }