Linear mapping of the plane into itself.

Linear mapping of the plane into itself.


	Linear mapping of the plane into itself.

Consider a linear mapping

$\begin{displaymath} A:\;{\rm R}^2 \; \to \; {\rm R}^2 \end{displaymath}$

After fixing a basis in

${\rm R}^2$

one can assign a matrix to the linear operator A,

$\begin{displaymath} A= \left( \begin{array}{cc} a_{11} & a_{12} \ a_{21} & a_{22} \end{array} \right) \end{displaymath}$

Given a vector

$x\in {\rm R}^2$

its image

$y \in {\rm R}^2$

after applying

is defined as

$\begin{displaymath} y = A x. \end{displaymath}$

After change of coordinates

$\begin{displaymath} x_{\mbox{old}} = C x_{\mbox{new}} \end{displaymath}$

the matrices of the operator in "old" and "new" coordinates are related to each other in the following way

$\begin{displaymath} A_{\mbox{new}} = C^{-1} A_{\mbox{old}}C, \end{displaymath}$

where

$C^{-1}$

denotes the inverse of C.

Our goal is to find C so that the matrix

$A_{\mbox{new}}$

has the simplest possible form (canonical form). There is a standard procedure for calculating C.

Canonical From.

1.

Write the characteristic polynomial for A.

$\begin{displaymath} \det(A- \lambda E)=\det\left( \begin{array}{cc} a_{11}-\lambda & a_{12} \ a_{21} & a_{22}-\lambda \end{array} \right)=0 \end{displaymath}$

2.

Calculate the roots of the characteristic polynomial (eigenvalues of A)

3.

The canonical form for A is dictated by the structure of its eigenvalues (Fig.1.1).

**Figure 1.1:** Canonical forms of linear operators on the plane
$\begin{figure}\begin{center} \psfig{file=images/canonicalForm2.eps,scale=0.7} \end{center}\end{figure}$

Consider an example of calculating a canonical form.

Example 1 Our goal is to calculate the canonical form and the corresponding change of coordinates for

$\begin{displaymath} \left( \begin{array}{cc} 1 & 8 \ 2 & 1 \end{array}\right) \end{displaymath}$

The characteristic polynomial is

$\begin{displaymath} \det\left( \begin{array}{cc} 1-\lambda & 8 \ 2 & 1-\lambda \end{array}\right)= (1-\lambda)^2 - 16 = 0. \end{displaymath}$

Hence, the eigenvalues are

$\begin{displaymath} \lambda_1 = -3,\;\; \lambda_2=5. \end{displaymath}$

The corresponding eigenvectors are

$\begin{displaymath} \xi_1 = \left( \begin{array}{c} 2 \ 1 \end{array} \righ... ...xi_2 = \left( \begin{array}{c} 2 \ -1 \end{array} \right) \end{displaymath}$

Thus,

$\begin{displaymath} \left( \begin{array}{cc} 5 & 0 \ 0 & -3 \end{array}\right) \end{displaymath}$

is the canonical form and

$\begin{displaymath} (\xi_1\;\; \xi_2) = \left( \begin{array}{cc} 2 & 2\ 1 &-1 \end{array} \right) \end{displaymath}$

is the matrix for the corresponding change of coordinates.

Let us derive the canonical forms presented in Fig. 1.1.

Complex eigenvalues .

If A has complex conjugate eigenvalues

$\begin{displaymath} \lambda_1 = \alpha + i \beta,\;\; \lambda_2 = \alpha - i \beta \end{displaymath}$

then after calculating the eigenvector

$\begin{displaymath} A\xi_1 = \lambda_1 \xi_1, \;\;\xi_1 = a + i b \end{displaymath}$

we have

$\begin{displaymath} A(a + i b) = Aa + i Ab=(\alpha + i \beta)(a + i b)= \alpha a - \beta b + i(\beta a + \alpha b) \end{displaymath}$

Hence,

$\begin{displaymath} Aa = \alpha a - \beta b\;\;\mbox{ and } Ab = \beta a + \alpha b. \end{displaymath}$

After the change of coordinates

$\begin{displaymath} x_{\mbox{old}} = C x_{\mbox{new}}, \mbox{ where } C=(a\;\;b) \end{displaymath}$

the new matrix of the operator A is calculated as follows

$\begin{displaymath} \bar A = C^{-1} A C. \end{displaymath}$

$C^{-1}$

is the inverse for

That means

$\begin{displaymath} C^{-1}(a \; b)= (C^{-1}a\; C^{-1}b)= \left( \begin{array}{cc} 1 & 0\ 0 & 1 \end{array}\right) \end{displaymath}$

Thus,

$\begin{displaymath} \bar A = C^{-1} A C = C^{-1} A (a \; b) = C^{-1}(Aa\; Ab)=(\alpha a - \beta b \;\; \; \beta a + \alpha b)= \end{displaymath}$

$\begin{displaymath} (\alpha C^{-1}a - \beta C^{-1}b \;\;\;\; \beta C^{-1}a + \a... ...{cc} \alpha & \beta \ -\beta & \alpha \end{array} \right) \end{displaymath}$

We proved that after changing the coordinates we obtain the canonical form

$\begin{displaymath} \bar A = \left( \begin{array}{cc} \alpha & \beta \ -\beta & \alpha \end{array} \right) \end{displaymath}$

Real and different eigenvalues.

If A has real and different eigenvalues then

$\begin{displaymath} A\xi_1 = \lambda_1 \xi_1,\;\;\; A\xi_2 = \lambda_2 \xi_2 \end{displaymath}$

and the change of coordinates is defined by the matrix

$\begin{displaymath} C= ( \xi_1 \;\; \xi_2). \end{displaymath}$

Thus,

$\begin{displaymath} C^{-1} C = ( C^{-1} \xi_1\;\; C^{-1} \xi_2 ) = \left( \begin{array}{cc} 1 & 0\ 0 & 1 \end{array}\right) \end{displaymath}$

Hence,

$\begin{displaymath} \bar A = C^{-1} A C = C^{-1} A ( \xi_1 \;\; \xi_2 ) = C^{-1} (A \xi_1 \;\; A \xi_2 )= \end{displaymath}$

$\begin{displaymath} C^{-1} ( \lambda_1 \xi_1 \;\; \lambda_2 \xi_2 ) = ( \lambda... ...rray}{cc} \lambda_1 & 0\ 0 & \lambda_2 \end{array}\right) \end{displaymath}$

The eigenvalue of multiplicity 2

$\lambda_1 = \lambda_2 = \mu$

and there are two linearly independent eigenvectors

$\xi_1, \;\; \xi_2$

corresponding to

$\mu$

then the canonical form is the diagonal matrix

$\begin{displaymath} \bar A = \left( \begin{array}{cc} \mu & 0\ 0 & \mu \end{array}\right) \end{displaymath}$

and

$\begin{displaymath} C= ( \xi_1 \;\; \xi_2). \end{displaymath}$

If there is only one eigenvector

$\xi_1$

for

$\mu$

then we need to construct an adjoint vector

$\xi_{12}$

which is the solution of the system

$\begin{displaymath} (A - \mu E) \xi_{12} = \xi_1. \end{displaymath}$

The matrix of the coordinate change becomes

$\begin{displaymath} C= (\xi_1 \;\; \xi_{12} ). \end{displaymath}$

The canonical form is the Jordan block of the 2nd order,

$\begin{displaymath} \bar A = C^{-1} A C = C^{-1} A (\xi_1 \;\; \xi_{12} ) = C^{-1} ( A \xi_1 \;\; A \xi_{12} )= \end{displaymath}$

$\begin{displaymath} C^{-1}(\mu \xi_1 \;\; \xi_1 + \mu \xi_{12} )= ( \mu C^{-1} ... ...t( \begin{array}{cc} \mu & 1\ 0 & \mu \end{array}\right). \end{displaymath}$

This completes the justification of the diagram presented in Fig.1.1.

Next:Canonical formsUp:Linear mapping of the planePrevious:Linear mapping of the plane

Sergey Nikitin 2004-10-22