6.1 Motivation

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We motivate the chapter on eigenvalues by discussing the equation

ax2 + 2hxy + by2 = c;

where not all of a; h; b are zero. The expression ax2 + 2hxy + by2 is called

a quadratic form in x and y and we have the identity

ax2 + 2hxy + by2 =

£

x y

¤ ·

a h

h b

¸ ·

x

y

¸

= XtAX;

where X =

·

x

y

¸

and A =

·

a h

h b

¸

. A is called the matrix of the quadratic

form.

We now rotate the x; y axes anticlockwise through µ radians to new

x1; y1 axes. The equations describing the rotation of axes are derived as

follows:

Let P have coordinates (x; y) relative to the x; y axes and coordinates

(x1; y1) relative to the x1; y1 axes. Then referring to Figure 6.1:

115

116 CHAPTER 6. EIGENVALUES AND EIGENVECTORS

¾ -

6

?

@

@

@

@

@

@

@

@I

¡

¡

¡

¡

¡

¡

¡

¡µ

@

@

@

@

@

@

@

@R

¯

¯

¯

¯

¯

¯

¯

¯

¯

¯@

@

@

x

y

y1 x1

P

Q

R

O

®

µ

Figure 6.1: Rotating the axes.

x = OQ = OP cos (µ + ®)

= OP(cos µ cos ® ¡ sin µ sin ®)

= (OP cos ®) cos µ ¡ (OP sin ®) sin µ

= ORcos µ ¡ PRsin µ

= x1 cos µ ¡ y1 sin µ:

Similarly y = x1 sin µ + y1 cos µ.

We can combine these transformation equations into the single matrix

equation: ·

x

y

¸

=

·

cos µ ¡sin µ

sin µ cos µ

¸ ·

x1

y1

¸

;

or X = PY , where X =

·

x

y

¸

; Y =

·

x1

y1

¸

and P =

·

cos µ ¡sin µ

sin µ cos µ

¸

.

We note that the columns of P give the directions of the positive x1 and y1

axes. Also P is an orthogonal matrix { we have PP t = I2 and so P¡1 = Pt.

The matrix P has the special property that det P = 1.

A matrix of the type P =

·

cos µ ¡sin µ

sin µ cos µ

¸

is called a rotation matrix.

We shall show soon that any 2£2 real orthogonal matrix with determinant

6.1. MOTIVATION 117

equal to 1 is a rotation matrix.

We can also solve for the new coordinates in terms of the old ones: ·

x1

y1

¸

= Y = PtX =

·

cos µ sin µ

¡sin µ cos µ

¸ ·

x

y

¸

;

so x1 = x cos µ + y sin µ and y1 = ¡x sin µ + y cos µ. Then

XtAX = (PY )tA(PY ) = Y t(PtAP)Y:

Now suppose, as we later show, that it is possible to choose an angle µ so

that PtAP is a diagonal matrix, say diag(¸1; ¸2). Then

XtAX =

£

x1 y1

¤ ·

¸1 0

0 ¸2

¸ ·

x1

y1

¸

= ¸1x2

1 + ¸2y2

1 (6.1)

and relative to the new axes, the equation ax2 + 2hxy + by2 = c becomes

¸1x2

1 + ¸2y2

1 = c, which is quite easy to sketch. This curve is symmetrical

about the x1 and y1 axes, with P1 and P2, the respective columns of P,

giving the directions of the axes of symmetry.

Also it can be veri¯ed that P1 and P2 satisfy the equations

AP1 = ¸1P1 and AP2 = ¸2P2:

These equations force a restriction on ¸1 and ¸2. For if P1 =

·

u1

v1

¸

, the

¯rst equation becomes

·

a h

h b

¸ ·

u1

v1

¸

= ¸1

·

u1

v1

¸

or

·

a ¡ ¸1 h

h b ¡ ¸1

¸ ·

u1

v1

¸

=

·

0

0

¸

:

Hence we are dealing with a homogeneous system of two linear equations in

two unknowns, having a non{trivial solution (u1; v1). Hence

¯¯¯¯

a ¡ ¸1 h

h b ¡ ¸1

¯¯¯¯

= 0:

Similarly, ¸2 satis¯es the same equation. In expanded form, ¸1 and ¸2

satisfy

¸2 ¡ (a + b)¸ + ab ¡ h2 = 0:

This equation has real roots

¸ =

a + b §

p

(a + b)2 ¡ 4(ab ¡ h2)

2

=

a + b §

p

(a ¡ b)2 + 4h2

2

(6.2)

(The roots are distinct if a 6= b or h 6= 0. The case a = b and h = 0 needs

no investigation, as it gives an equation of a circle.)

The equation ¸2¡(a+b)¸+ab¡h2 = 0 is called the eigenvalue equation

of the matrix A.

118 CHAPTER 6. EIGENVALUES AND EIGENVECTORS

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