ONE-TO-ONE LINEAR TRANSFORMATIONS

Linear Algebra Project

Objectives:

Use 'linalg' package of Maple

1. To find whether a linear transformation is one-to-one.

2. To find the inverse transformation whenever possible.

Required Information:

If A is an n x n square matrix and T[A] : R^n -> R^n is multiplication by A

Then

T[A] is one-to-one if the range of T[A] is R^n

This also means A is invertible.

If T[A] is one-to-one then the inverse transformation is defined by T[A^(-1)]

Solved Example 1:

Let A = matrix([[1, -2, 2], [2, 1, 1], [1, 1, 0]])

The transformation

T[A] is given by w[1] = x[1] - 2 x[2] + 2 x[3]

w[2] = 2 x[1] + x[2] + x[3]

w[3] = x[1] + x[2]

a) Determine whether T[A] is one-to-one.

b) If T[A] is one-to-one find T[A^(-1)]

Solution:

a) First we check if A is invertible using the inverse command.

A = matrix([[1, -2, 2], [2, 1, 1], [1, 1, 0]])

> with(linalg):

> A := matrix([[1, -2, 2], [2, 1, 1], [1, 1, 0]]);

A := matrix([[1, -2, 2], [2, 1, 1], [1, 1, 0]])

> Ainverse:=inverse(A);

Ainverse := matrix([[1, -2, 4], [-1, 2, -3], [-1, 3...

b) The existence of the inverse of A shows that T is one-to-one, and the range of T is R^3 . In fact for any vector

w = matrix([[w1], [w2], [w3]]) in R^3 there is a pre-image matrix([[x1], [x2], [x3]]) given by A^(-1) w.

> w := matrix(3,1,[w1, w2, w3]);

w := matrix([[w1], [w2], [w3]])

> x= evalm(Ainverse&*(w));

x = matrix([[w1-2*w2+4*w3], [-w1+2*w2-3*w3], [-w1+3...

The vector x is given by matrix([[x1], [x2], [x3]]) = x = matrix([[w1-2*w2+4*w3], [-w1+2*w2-3*w3], [-w1+3...

Solved Example 2:

Let A = matrix([[1, 2, 1], [-2, 1, 4], [7, 4, -5]])

T[A] is given by w[1] = x[1] + 4 x[2] - x[3]

w[2] = 2 x[1] +7 x[2] + x[3]

w[3] = x[1] +3 x[3]

a) Show that T[A] is not one-to-one.

b) Find all vectors x in R^3 which are mapped onto the 0 vector.

(Note that: For a one-to-one transformation T[A] only the 0- vector is mapped onto 0 vector.

In other words Ax = 0 or A matrix([[x[1]], [x[2]], [x[3]]]) = matrix([[0], [0], [0]]) has only the trivial solution

or the solution space of the system of corresponding linear equations consists of x[1] = 0, x[2] = 0, x[3] = 0

Solution:

(a) We check whether matrix A = matrix([[1, 2, 1], [-2, 1, 4], [7, 4, -5]]) is invertible.

> A:= matrix(3,3, [1,2,1,-2,1,4,7,4,-5]);

A := matrix([[1, 2, 1], [-2, 1, 4], [7, 4, -5]])

> inverse(A);

Error, (in inverse) singular matrix

This means matrix A is not invertible.

(b) Now we find all vector w in R^3 which map onto the 0 vector.

We do this by finding the rref of the system Ax = 0 using the augmented matrix.

Ax = 0

or matrix([[1, 2, 1], [-2, 1, 4], [7, 4, -5]]) matrix([[x1], [x2], [x3]]) = matrix([[0], [0], [0]])

> w := matrix(3,1,[0,0,0]);

w := matrix([[0], [0], [0]])

> Aw := augment(A,w);

Aw := matrix([[1, 2, 1, 0], [-2, 1, 4, 0], [7, 4, -...

> Awreduced:= rref(Aw);

Awreduced := matrix([[1, 0, -7/5, 0], [0, 1, 6/5, 0...

This means that the solution space consists of vectors matrix([[x[1]], [x[2]], [x[3]]]) where x[1] = 7*x[3]/5 , x[2] = 6*x[3]/5

or all vectors lambda matrix([[7/5], [6/5], [1]]) where lambda is a scalar.

__________________________________________________________________________

Assignment

In each of the problems for the matrix A

a) define T[A] .

b) determine whether T[A] is one-to-one.

c) if T[A] is one-to-one find the inverse transformation. If T[A] is not one-to-one find all the vectors which map onto the 0 vector.

Problem 1:

A = matrix([[1, 3, -4], [2, 7, 1], [1, 3, 0]])

Problem 2:

A = matrix([[2, 0, -4], [2, 5, -1], [-1, 3, 0]])

_____________________________________________________________________________

MSIP Grant #P120A80089-98: "Three Urban Calculus Reform programs: Adopting the Best" 1998-2001, MSEIP Grant #P120A010031: "Four Colleges: Calculus + Enhancements" 2001-04