Archetype R   

From A First Course in Linear Algebra
Version 2.20
© 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 : 5 5,T x1 x2 x3 x4 x5 = 65x1 + 128x2 + 10x3 262x4 + 40x5 36x1 73x2 x3 + 151x4 16x5 44x1 + 88x2 + 5x3 180x4 + 24x5 34x1 68x2 3x3 + 140x4 18x5 12x1 24x2 x3 + 49x4 5x5

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

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):

65 36 44 34 12 , 12873 88 68 24 , 101 5 3 1 , 262 151 180 140 49 , 40 16 24 18 5

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:

1 0 0 0 0 , 0 1 0 0 0 , 0 0 1 0 0 , 0 0 0 1 0 , 0 0 0 0 1

Surjective: Yes. (Definition SLT)
A basis for the range is the standard basis of 5, so T = 5 and Theorem RSLT tells us T is surjective. Or, the dimension of the range is 5, and the codomain (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.

 Domain dimension:  5  Rank:  5  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 (Theorem MLTCV):

T : 5 5,T x = Ax,A = 6512810262 40 36 73 1 151 16 44 88 5 180 24 34 68 3 140 18 12 241 49 5

The inverse linear transformation (Definition IVLT):

T1 : 5 5,T1 x1 x2 x3 x4 x5 = 47x1 + 92x2 + x3 181x4 14x5 27x1 55x2 + 7 2x3 + 221 2 x4 + 11x5 32x1 + 64x2 x3 126x4 12x5 25x1 50x2 + 3 2x3 + 199 2 x4 + 9x5 9x1 18x2 + 1 2x3 + 71 2 x4 + 4x5

Verify that T T1 x = x and T T1 x = 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):

λ = 1 T 1 = 57 0 18 14 5 , 2 1 0 0 0 λ = 1 T 1 = 10 5 6 0 1 , 2 3 1 1 0 λ = 2 T 2 = 6 3 4 3 1

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

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

B = 57 0 18 14 5 , 2 1 0 0 0 , 10 5 6 0 1 , 2 3 1 1 0 , 6 3 4 3 1 MB,BT = 1 0 000 0 1 0 0 0 0 0 100 0 0 0 1 0 0 0 002