7.1 The eigenvalue method

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In this section we apply eigenvalue methods to determine the geometrical

nature of the second degree equation

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0; (7.1)

where not all of a; h; b are zero.

Let A =

·

a h

h b

¸

be the matrix of the quadratic form ax2+2hxy+by2.

We saw in section 6.1, equation 6.2 that A has real eigenvalues ¸1 and ¸2,

given by

¸1 =

a + b ¡

p

(a ¡ b)2 + 4h2

2

; ¸2 =

a + b +

p

(a ¡ b)2 + 4h2

2

:

We show that it is always possible to rotate the x; y axes to x1; x2 axes whose

positive directions are determined by eigenvectors X1 and X2 corresponding

to ¸1and ¸2 in such a way that relative to the x1; y1 axes, equation 7.1 takes

the form

a0x2 + b0y2 + 2g0x + 2f0y + c = 0: (7.2)

Then by completing the square and suitably translating the x1; y1 axes,

to new x2; y2 axes, equation 7.2 can be reduced to one of several standard

forms, each of which is easy to sketch. We need some preliminary de¯nitions.

129

130 CHAPTER 7. IDENTIFYING SECOND DEGREE EQUATIONS

DEFINITION 7.1.1 (Orthogonal matrix) An n £ n real matrix P is

called orthogonal if

PtP = In:

It follows that if P is orthogonal, then det P = §1. For

det (PtP) = det Pt det P = ( det P)2;

so (det P)2 =det In = 1. Hence det P = §1.

If P is an orthogonal matrix with det P = 1, then P is called a proper

orthogonal matrix.

THEOREM 7.1.1 If P is a 2 £ 2 orthogonal matrix with det P = 1, then

P =

·

cos µ ¡sin µ

sin µ cos µ

¸

for some µ.

REMARK 7.1.1 Hence, by the discusssion at the beginning of Chapter

6, if P is a proper orthogonal matrix, the coordinate transformation

·

x

y

¸

= P

·

x1

y1

¸

represents a rotation of the axes, with new x1 and y1 axes given by the

repective columns of P.

Proof. Suppose that PtP = I2, where ¢ =det P = 1. Let

P =

·

a b

c d

¸

:

Then the equation

Pt = P¡1 =

1

¢

adj P

gives ·

a c

b d

¸

=

·

d ¡b

¡c a

¸

Hence a = d; b = ¡c and so

P =

·

a ¡c

c a

¸

;

where a2 + c2 = 1. But then the point (a; c) lies on the unit circle, so

a = cos µ and c = sin µ, where µ is uniquely determined up to multiples of

2¼.

7.1. THE EIGENVALUE METHOD 131

DEFINITION 7.1.2 (Dot product). If X =

·

a

b

¸

and Y =

·

c

d

¸

, then

X ¢ Y , the dot product of X and Y , is de¯ned by

X ¢ Y = ac + bd:

The dot product has the following properties:

(i) X ¢ (Y + Z) = X ¢ Y + X ¢ Z;

(ii) X ¢ Y = Y ¢ X;

(iii) (tX) ¢ Y = t(X ¢ Y );

(iv) X ¢ X = a2 + b2 if X =

·

a

b

¸

;

(v) X ¢ Y = XtY .

The length of X is de¯ned by

jjXjj =

p

a2 + b2 = (X ¢ X)1=2:

We see that jjXjj is the distance between the origin O = (0; 0) and the point

(a; b).

THEOREM 7.1.2 (Geometrical interpretation of the dot product)

Let A = (x1; y1) and B = (x2; y2) be points, each distinct from the origin

O = (0; 0). Then if X =

·

x1

y1

¸

and Y =

·

x2

y2

¸

, we have

X ¢ Y = OA ¢ OB cos µ;

where µ is the angle between the rays OA and OB.

Proof. By the cosine law applied to triangle OAB, we have

AB2 = OA2 + OB2 ¡ 2OA ¢ OB cos µ: (7.3)

Now AB2 = (x2 ¡ x1)2 + (y2 ¡ y1)2; OA2 = x2

1 + y2

1; OB2 = x2

2 + y2

2.

Substituting in equation 7.3 then gives

(x2 ¡ x1)2 + (y2 ¡ y1)2 = (x2

1 + y2

1) + (x2

2 + y2

2) ¡ 2OA ¢ OB cos µ;

132 CHAPTER 7. IDENTIFYING SECOND DEGREE EQUATIONS

which simpli¯es to give

OA ¢ OB cos µ = x1x2 + y1y2 = X ¢ Y:

It follows from theorem 7.1.2 that if A = (x1; y1) and B = (x2; y2) are

points distinct from O = (0; 0) and X =

·

x1

y1

¸

and Y =

·

x2

y2

¸

, then

X ¢ Y = 0 means that the rays OA and OB are perpendicular. This is the

reason for the following de¯nition:

DEFINITION 7.1.3 (Orthogonal vectors) Vectors X and Y are called

orthogonal if

X ¢ Y = 0:

There is also a connection with orthogonal matrices:

THEOREM 7.1.3 Let P be a 2£2 real matrix. Then P is an orthogonal

matrix if and only if the columns of P are orthogonal and have unit length.

Proof. P is orthogonal if and only if PtP = I2. Now if P = [X1jX2], the

matrix PtP is an important matrix called the Gram matrix of the column

vectors X1 and X2. It is easy to prove that

PtP = [Xi ¢ Xj ] =

·

X1 ¢ X1 X1 ¢ X2

X2 ¢ X1 X2 ¢ X2

¸

:

Hence the equation PtP = I2 is equivalent to

·

X1 ¢ X1 X1 ¢ X2

X2 ¢ X1 X2 ¢ X2

¸

=

·

1 0

0 1

¸

;

or, equating corresponding elements of both sides:

X1 ¢ X1 = 1; X1 ¢ X2 = 0; X2 ¢ X2 = 1;

which says that the columns of P are orthogonal and of unit length.

The next theorem describes a fundamental property of real symmetric

matrices and the proof generalizes to symmetric matrices of any size.

THEOREM 7.1.4 If X1 and X2 are eigenvectors corresponding to distinct

eigenvalues ¸1 and ¸2 of a real symmetric matrix A, then X1 and X2 are

orthogonal vectors.

7.1. THE EIGENVALUE METHOD 133

Proof. Suppose

AX1 = ¸1X1; AX2 = ¸2X2; (7.4)

where X1 and X2 are non{zero column vectors, At = A and ¸1 6= ¸2.

We have to prove that Xt

1 X2 = 0. From equation 7.4,

Xt

2AX1 = ¸1Xt

2 X1 (7.5)

and

Xt

1 AX2 = ¸2Xt

1X2: (7.6)

From equation 7.5, taking transposes,

(Xt

2 AX1)t = (¸1Xt

2X1)t

so

Xt

1AtX2 = ¸1Xt

1X2:

Hence

Xt

1 AX2 = ¸1Xt

1X2: (7.7)

Finally, subtracting equation 7.6 from equation 7.7, we have

(¸1 ¡ ¸2)Xt

1X2 = 0

and hence, since ¸1 6= ¸2,

Xt

1 X2 = 0:

THEOREM 7.1.5 Let A be a real 2 £ 2 symmetric matrix with distinct

eigenvalues ¸1 and ¸2. Then a proper orthogonal 2£2 matrix P exists such

that

PtAP = diag (¸1; ¸2):

Also the rotation of axes

·

x

y

¸

= P

·

x1

y1

¸

\diagonalizes" the quadratic form corresponding to A:

XtAX = ¸1x2

1 + ¸2y2

1:

134 CHAPTER 7. IDENTIFYING SECOND DEGREE EQUATIONS

Proof. Let X1 and X2 be eigenvectors corresponding to ¸1 and ¸2. Then

by theorem 7.1.4, X1 and X2 are orthogonal. By dividing X1 and X2 by

their lengths (i.e. normalizing X1 and X2) if necessary, we can assume that

X1 and X2 have unit length. Then by theorem 7.1.1, P = [X1jX2] is an

orthogonal matrix. By replacing X1 by ¡X1, if necessary, we can assume

that det P = 1. Then by theorem 6.2.1, we have

PtAP = P¡1AP =

·

¸1 0

0 ¸2

¸

:

Also under the rotation X = PY ,

XtAX = (PY )tA(PY ) = Y t(PtAP)Y = Y t diag (¸1; ¸2)Y

= ¸1x2

1 + ¸2y2

1:

EXAMPLE 7.1.1 Let A be the symmetric matrix

A =

·

12 ¡6

¡6 7

¸

:

Find a proper orthogonal matrix P such that PtAP is diagonal.

Solution. The characteristic equation of A is ¸2 ¡ 19¸ + 48 = 0, or

(¸ ¡ 16)(¸ ¡ 3) = 0:

Hence A has distinct eigenvalues ¸1 = 16 and ¸2 = 3. We ¯nd corresponding

eigenvectors

X1 =

·

¡3

2

¸

and X2 =

·

2

3

¸

:

Now jjX1jj = jjX2jj = p13. So we take

X1 =

1

p13

·

¡3

2

¸

and X2 =

1

p13

·

2

3

¸

:

Then if P = [X1jX2], the proof of theorem 7.1.5 shows that

PtAP =

·

16 0

0 3

¸

:

However det P = ¡1, so replacing X1 by ¡X1 will give det P = 1.

7.1. THE EIGENVALUE METHOD 135

y

2

x

2

-4 -2 2 4

2

4

-2

-4

x

y

Figure 7.1: 12x2 ¡ 12xy + 7y2 + 60x ¡ 38y + 31 = 0.

REMARK 7.1.2 (A shortcut) Once we have determined one eigenvec-

tor X1 =

·

a

b

¸

, the other can be taken to be

·

¡b

a

¸

, as these these vectors

are always orthogonal. Also P = [X1jX2] will have det P = a2 + b2 > 0.

We now apply the above ideas to determine the geometric nature of

second degree equations in x and y.

EXAMPLE 7.1.2 Sketch the curve determined by the equation

12x2 ¡ 12xy + 7y2 + 60x ¡ 38y + 31 = 0:

Solution. With P taken to be the proper orthogonal matrix de¯ned in the

previous example by

P =

·

3=p13 2=p13

¡2=p13 3=p13

¸

;

then as theorem 7.1.1 predicts, P is a rotation matrix and the transformation

X =

·

x

y

¸

= PY = P

·

x1

y1

¸

136 CHAPTER 7. IDENTIFYING SECOND DEGREE EQUATIONS

or more explicitly

x =

3x1 + 2y1 p13

; y = ¡2x1 + 3y1 p13

; (7.8)

will rotate the x; y axes to positions given by the respective columns of P.

(More generally, we can always arrange for the x1 axis to point either into

the ¯rst or fourth quadrant.)

Now A =

·

12 ¡6

¡6 7

¸

is the matrix of the quadratic form

12x2 ¡ 12xy + 7y2;

so we have, by Theorem 7.1.5

12x2 ¡ 12xy + 7y2 = 16x2

1 + 3y2

1:

Then under the rotation X = PY , our original quadratic equation becomes

16x2

1 + 3y2

1 +

60

p13

(3x1 + 2y1) ¡

38

p13

(¡2x1 + 3y1) + 31 = 0;

or

16x2

1 + 3y2

1 +

256

p13

x1 +

6

p13

y1 + 31 = 0:

Now complete the square in x1 and y1:

16

µ

x2

1 +

16

p13

x1

+ 3

µ

y2

1 +

2

p13

y1

+ 31 = 0;

16

µ

x1 +

8

p13

¶2

+ 3

µ

y1 +

1

p13

¶2

= 16

µ

8

p13

¶2

+ 3

µ

1

p13

¶2

¡ 31

= 48: (7.9)

Then if we perform a translation of axes to the new origin (x1; y1) =

(¡ 8 p13

; ¡ 1 p13

):

x2 = x1 +

8

p13

; y2 = y1 +

1

p13

;

equation 7.9 reduces to

16x2

2 + 3y2

2 = 48;

or

x2

2

3

+

y2

2

16

= 1:

7.1. THE EIGENVALUE METHOD 137

x

y

Figure 7.2:

x2

a2 +

y2

b2 = 1; 0 < b < a: an ellipse.

This equation is now in one of the standard forms listed below as Figure 7.2

and is that of a whose centre is at (x2; y2) = (0; 0) and whose axes of

symmetry lie along the x2; y2 axes. In terms of the original x; y coordinates,

we ¯nd that the centre is (x; y) = (¡2; 1). Also Y = P tX, so equations 7.8

can be solved to give

x1 =

3x1 ¡ 2y1 p13

; y1 =

2x1 + 3y1 p13

:

Hence the y2{axis is given by

0 = x2 = x1 +

8

p13

=

3x ¡ 2y

p13

+

8

p13

;

or 3x ¡ 2y + 8 = 0. Similarly the x2 axis is given by 2x + 3y + 1 = 0.

This ellipse is sketched in Figure 7.1.

Figures 7.2, 7.3, 7.4 and 7.5 are a collection of standard second degree

equations: Figure 7.2 is an ellipse; Figures 7.3 are hyperbolas (in both these

examples, the asymptotes are the lines y = §

b

a

x); Figures 7.4 and 7.5

represent parabolas.

EXAMPLE 7.1.3 Sketch y2 ¡ 4x ¡ 10y ¡ 7 = 0.

138 CHAPTER 7. IDENTIFYING SECOND DEGREE EQUATIONS

x

y

x

y

Figure 7.3: (i)

x2

a2 ¡

y2

b2 = 1; (ii)

x2

a2 ¡

y2

b2 = ¡1; 0 < b; 0 < a.

x

y

x

y

Figure 7.4: (i) y2 = 4ax; a > 0; (ii) y2 = 4ax; a < 0.

7.1. THE EIGENVALUE METHOD 139

x

y

x

y

Figure 7.5: (iii) x2 = 4ay; a > 0; (iv) x2 = 4ay; a < 0.

Solution. Complete the square:

y2 ¡ 10y + 25 ¡ 4x ¡ 32 = 0

(y ¡ 5)2 = 4x + 32 = 4(x + 8);

or y2

1 = 4x1, under the translation of axes x1 = x+8; y1 = y ¡5. Hence we

get a parabola with vertex at the new origin (x1; y1) = (0; 0), i.e. (x; y) =

(¡8; 5).

The parabola is sketched in Figure 7.6.

EXAMPLE 7.1.4 Sketch the curve x2 ¡ 4xy + 4y2 + 5y ¡ 9 = 0.

Solution. We have x2 ¡ 4xy + 4y2 = XtAX, where

A =

·

1 ¡2

¡2 4

¸

:

The characteristic equation of A is ¸2¡5¸ = 0, so A has distinct eigenvalues

¸1 = 5 and ¸2 = 0. We ¯nd corresponding unit length eigenvectors

X1 =

1

p5

·

1

¡2

¸

; X2 =

1

p5

·

2

1

¸

:

Then P = [X1jX2] is a proper orthogonal matrix and under the rotation of

axes X = PY , or

x =

x1 + 2y1 p5

y = ¡2x1 + y1 p5

;

140 CHAPTER 7. IDENTIFYING SECOND DEGREE EQUATIONS

x

1

y

1

-8 -4 4 8 12

4

8

12

-4

-8

x

y

Figure 7.6: y2 ¡ 4x ¡ 10y ¡ 7 = 0.

we have

x2 ¡ 4xy + 4y2 = ¸1x2

1 + ¸2y2

1 = 5x2

1:

The original quadratic equation becomes

5x2

1 +

p5

p5

(¡2x1 + y1) ¡ 9 = 0

5(x2

1 ¡

2

p5

x1) + p5y1 ¡ 9 = 0

5(x1 ¡

1

p5

)2 = 10 ¡

p5y1 = p5(y1 ¡ 2p5);

or 5x2

2 = ¡ 1 p5

y2, where the x1; y1 axes have been translated to x2; y2 axes

using the transformation

x2 = x1 ¡

1

p5

; y2 = y1 ¡ 2p5:

Hence the vertex of the parabola is at (x2; y2) = (0; 0), i.e. (x1; y1) =

( 1 p5

; 2p5), or (x; y) = ( 21

5 ; 8

5 ). The axis of symmetry of the parabola is the

line x2 = 0, i.e. x1 = 1=p5. Using the rotation equations in the form

x1 =

x ¡ 2y

p5

7.2. A CLASSIFICATION ALGORITHM 141

x

2

y

2

-4 -2 2 4

2

4

-2

-4

x

y

Figure 7.7: x2 ¡ 4xy + 4y2 + 5y ¡ 9 = 0.

y1 =

2x + y

p5

;

we have

x ¡ 2y

p5

=

1

p5

; or x ¡ 2y = 1:

The parabola is sketched in Figure 7.7.

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