Exercises for Section 16.1

Question 16.6

Determine if the following functions \(T{:}\ \mathbb R^2 \rightarrow \mathbb R^2\) are one-to-one and/or onto.

  • (a) \(T(x,y) = (2x,y)\)
  • (b) \(T(x,y) = (x^2,y)\)
  • (c) \(T(x,y) = (\sqrt[3]{x},\sqrt[3]{y})\)
  • (d) \(T(x,y) = (\sin x, \cos y)\)

Question 16.7

Determine if the following functions \(T{:}\ \mathbb R^2 \rightarrow \mathbb R^2\) are one-to-one and/or onto.

  • (a) \(T(x,y,z) = (2x+y+3z,3y-4z,5x)\)
  • (b) \(T(x,y,z) = (y \sin x, z \cos y, xy)\)
  • (c) \(T(x,y,z) = (xy, yz, xz)\)
  • (d) \(T(x,y,z) = (e^x, e^y, e^z)\)

Question 16.8

Let \(D\) be a square with vertices \((0,0),(1,1),(2,0),(1,-1)\) and \(D^*\) be a parallelogram with vertices \((0,0),(1,2),(2,1),(1,-1)\). Find a linear map \(T\) taking \(D^*\) onto \(D\).

Question 16.9

Let \(D\) be a parallelogram with vertices \((0,0),(-1,3),(-2,0),(-1,-3)\). Let \(D^* = [0,1] \times [0,1]\). Find a linear map \(T\) such that \(T(D^*)=D\).

Question 16.10

Let \(S^* = (0,1] \times [0, 2 \pi)\) and define \(T (r, \theta) = ( r \cos \theta, r \sin \theta)\). Determine the image set \(S\). Show that \(T\) is one-to-one on \(S^*\).

Question 16.11

Define \[ T ( x^* , y^*) = \left( \frac{x^* - y^*}{\sqrt{2}}, \frac{x^* + y^*}{\sqrt{2}} \right). \] Show that \(T\) rotates the unit square, \(D^* = [0,1] \times [0,1]\).

Question 16.12

Let \(D^* = [0,1] \times [0,1]\) and define \(T\) on \(D^*\) by \(T (u, v) = ( - u^2 + 4u,v)\). Find the image \(D\). Is \(T\) one-to-one?

Question 16.13

Let \(D^*\) be the parallelogram bounded by the lines \(y = 3x - 4, y = 3x, y = {\textstyle \frac{1}{2}}x\), and \(y= {\textstyle \frac{1}{2}} (x+4)\). Let \(D = [0,1] \times [0,1]\). Find a \(T\) such that \(D\) is the image of \(D^*\) under \(T\).

Question 16.14

Let \(D^* = [0,1] \times [0,1]\) and define \(T\) on \(D^*\) by \(T (x^*, y^*) = (x^*y^*, x^*)\). Determine the image set \(D\). Is \(T\) one-to-one? If not, can we eliminate some subset of \(D^*\) so that on the remainder \(T\) is one-to-one?

Question 16.15

Let \(D^*\) be the parallelogram with vertices at \((-1,3), (0,0), (2,-1)\), and \((1,2)\), and \(D\) be the rectangle \(D= [0,1] \times [0,1]\). Find a \(T\) such that \(D\) is the image set of \(D^*\) under \(T\).

Question 16.16

Let \(T \colon\, {\mathbb R}^3 \to {\mathbb R}^3\) be the spherical coordinate mapping defined by \((\rho, \phi, \theta) \mapsto (x,y,z)\), where \[ x = \rho \sin \phi \cos \theta, y = \rho \sin \phi \sin \theta, z = \rho \cos \phi. \]

314

Let \(D^*\) be the set of points \((\rho, \phi, \theta)\) such that \(\phi \in [0, \pi], \theta \in [0, 2 \pi], \rho \in [0,1]\). Find \(D = T ( D^*)\). Is \(T\) one-to-one? If not, can we eliminate some subset of \(D^*\) so that, on the remainder, \(T\) will be one-to-one?

In the next two exercises, let \(T ( {\bf x}) = A {\bf x}\), where A is a \(2 \times 2\) matrix.

Question 16.17

Show that \(T\) is one-to-one if and only if the determinant of \(A\) is not zero.

Question 16.18

Show that det \(A \ne 0\) if and only if \(T\) is onto.

Question 16.19

Suppose \(T \colon\, {\mathbb R}^2 \to {\mathbb R}^2\) is linear and is given by \(T ({\bf x}) = A {\bf x}\), where \(A\) is a \(2 \times 2\) matrix. Show that if det \(A \ne 0\), then \(T\) takes parallelograms onto parallelograms. [HINT: The general parallelogram in \({\mathbb R}^2\) can be described by the set of points \({\bf q} = {\bf p} + \lambda {\bf v}+ \mu {\bf w}\) for \(\lambda, \mu \in (0,1)\) where \({\bf p}, {\bf v}, {\bf w}\) are vectors in \({\mathbb R}^2\) with \({\bf v}\) not a scalar multiple of \({\bf w}\).]

Question 16.20

A map \(T: \mathbb R^2 \rightarrow \mathbb R^2\) is called affine if \(T(\bf{x}) = A\bf{x} + \bf{v},\) where \(A\) is a \(2 \times 2\) matrix, and \(\bf{v}\) is a fixed vector in \(\mathbb R^2\). Show that Exercises 12, 13, and 14 hold for \(T\).

Question 16.21

Suppose \(T \colon\, {\mathbb R}^2 \to {\mathbb R}^2\) is as in Exercise 14 and that \(T ( P^*) = P\) is a parallelogram. Show that \(P^*\) is a parallelogram.

Question 16.22

Show that \(T\) is not one-to-one.