Eigencircles of 2 2 matrices Graham Farr Faculty of IT Monash - - PowerPoint PPT Presentation

Eigencircles of 2 × 2 matrices

Graham Farr

Faculty of IT Monash University Graham.Farr@infotech.monash.edu.au

18 July 2007 Joint work with Michael Englefield (School of Mathematical Sciences, Monash)

Eigenvalues and eigenpairs

Eigenvalue of a 2 × 2 matrix: number λ such that a b c d x y

• =

λ x y

• with x, y not both 0.

To start with, λ ∈ R.

Eigenvalues and eigenpairs

Eigenvalue of a 2 × 2 matrix: number λ such that a b c d x y

• =

λ x y

• with x, y not both 0.

To start with, λ ∈ R. Field isomorphism: λ ← → λ λ

Eigenvalues and eigenpairs

Eigenvalue of a 2 × 2 matrix: number λ such that a b c d x y

• =

λ λ x y

• with x, y not both 0.

To start with, λ ∈ R. Field isomorphism: λ ← → λ λ

Eigenvalues and eigenpairs

Eigenvalue of a 2 × 2 matrix: number λ such that a b c d x y

• =

λ λ x y

• with x, y not both 0.

To start with, λ ∈ R. Field isomorphism: λ ← → λ λ

• Extend using field isomorphism:

λ + µi ← →

• λ

µ −µ λ

Eigenvalues and eigenpairs

Eigenvalue of a 2 × 2 matrix: number λ such that a b c d x y

• =

λ λ x y

• with x, y not both 0.

To start with, λ ∈ R. Field isomorphism: λ ← → λ λ

• Extend using field isomorphism:

λ + µi ← →

• λ

µ −µ λ

• Eigenpair of a 2 × 2 matrix: (λ, µ) ∈ R2 such that

a b c d x y

• =
• λ

µ −µ λ x y

• with x, y not both 0.

The Eigencircle

a b c d x y

• =
• λ

µ −µ λ x y

The Eigencircle

a b c d x y

• =
• λ

µ −µ λ x y

• Eigenpairs must satisfy
• a − λ

b − µ c + µ d − λ

• = 0

The Eigencircle

a b c d x y

• =
• λ

µ −µ λ x y

• Eigenpairs must satisfy
• a − λ

b − µ c + µ d − λ

• = 0

Some eigenpairs: (a, b), (a, −c), (d, b), (d, −c).

The Eigencircle

a b c d x y

• =
• λ

µ −µ λ x y

• Eigenpairs must satisfy
• a − λ

b − µ c + µ d − λ

• = 0

Some eigenpairs: (a, b), (a, −c), (d, b), (d, −c). Eigenpairs form a circle, the eigencircle:

• λ − a + d

2 2 +

• µ − b − c

2 2 = a + d 2 2 + b − c 2 2 −(ad−bc) ( λ− f )2 + ( µ− g )2 = f 2 + g2 − det A

The Eigencircle

λ µ O a d b −c

The Eigencircle

λ µ O a d b −c E F G H

The Eigencircle

λ µ O a d b −c E F G H C

The Eigencircle

λ µ O a d b −c E F G H C f g

The Eigencircle

λ µ O a d b −c C f g

The Eigencircle

λ µ O a d b −c C f g R

The Eigencircle

λ µ O a d b −c C R

The Eigencircle

λ µ O a d b −c C R ρ √ det A

The Eigencircle

λ µ O a d b −c C R ρ √ det A . . . provided det A > 0

The Eigencircle: det A < 0

The Eigencircle: det A < 0

λ µ O C

The Eigencircle: det A < 0

λ µ O C R

The Eigencircle: det A < 0

λ µ O C R ρ √ − det A

The Eigencircle: det A = 0

The Eigencircle: det A = 0

λ µ O C

The Eigencircle: det A = 0

λ µ O C R = ρ

The Eigencircle

Determinant Origin is   

• utside
• n

inside    eigencircle ⇐ ⇒ det A    > 0 = 0 < 0

The Eigencircle

Determinant Origin is   

• utside
• n

inside    eigencircle ⇐ ⇒ det A    > 0 = 0 < 0 Real eigenvalues Eigencircle meets λ-axis ⇐ ⇒ eigenvalues are real

Eigenvectors

Eigenvectors

Given (real) eigenvalue λ, a b c d x y

• =

λ λ x y

• get eigenvectors:

x y

• = any multiple of
• d

−c

• −

λ

• .

Eigenvectors

Given (real) eigenvalue λ, a b c d x y

• =

λ λ x y

• get eigenvectors:

x y

• = any multiple of
• d

−c

• −

λ

• .

λ µ O C (λ1, 0) (λ2, 0) (d, −c)

Eigenvectors

For a real symmetric 2 × 2 matrix, distinct real eigenvalues have perpendicular eigenvectors.

Eigenvectors

For a real symmetric 2 × 2 matrix, distinct real eigenvalues have perpendicular eigenvectors. Proof without words: λ µ O C (λ1, 0) (λ2, 0) (d, −c)

(λ, µ)-eigenvectors

A (λ, µ)-eigenvector is a nonzero

„ x y «

corresponding to the eigenpair (λ, µ).

(λ, µ)-eigenvectors

A (λ, µ)-eigenvector is a nonzero

„ x y «

corresponding to the eigenpair (λ, µ). λ µ O C (λ, µ) (d, −c)

(λ, µ)-eigenvectors

A (λ, µ)-eigenvector is a nonzero

„ x y «

corresponding to the eigenpair (λ, µ). λ µ O C (λ, µ) (d, −c) Diametrically opposite eigenpairs have perpendicular (λ, µ)-eigenvectors

(λ, µ)-eigenvectors

A (λ, µ)-eigenvector is a nonzero

„ x y «

corresponding to the eigenpair (λ, µ). λ µ O C (λ, µ) (d, −c) (λ′, −µ′) Diametrically opposite eigenpairs have perpendicular (λ, µ)-eigenvectors

Power and determinant

P C Euclid’s Elements III.35–36: Power of point = PQ · PR, independent of direction of line Q R

Power and determinant

P C Euclid’s Elements III.35–36: Power of point = PQ · PR, independent of direction of line Q R

Power and determinant

P C Euclid’s Elements III.35–36: Power of point = PQ · PR, independent of direction of line Q R

Power and determinant

P C Euclid’s Elements III.35–36: Power of point = PQ · PR, independent of direction of line Q R

Power and determinant

P C Euclid’s Elements III.35–36: Power of point = PQ · PR, independent of direction of line Q = R

Power and determinant

P C Euclid’s Elements III.35–36: Power of point = PQ · PR, independent of direction of line Q = R eigencircle

Power and determinant

O = P C Euclid’s Elements III.35–36: Power of point = PQ · PR, independent of direction of line Q = R eigencircle λ µ

Power and determinant

O = P C Euclid’s Elements III.35–36: Power of point = PQ · PR, independent of direction of line Q = R eigencircle λ µ √ det A

Power and determinant

O = P C Euclid’s Elements III.35–36: Power of point = PQ · PR, independent of direction of line Q = R eigencircle λ µ √ det A det A = power of origin w.r.t. eigencircle

Power and determinant

Power and determinant

Power of point = PQ · PR; lengths are signed, so now < 0 P C Q R

Power and determinant

Power of point = PQ · PR; lengths are signed, so now < 0 P C Q R

Power and determinant

Power of point = PQ · PR; lengths are signed, so now < 0 P C Q R

Power and determinant

Power of point = PQ · PR; lengths are signed, so now < 0 P C Q R

Power and determinant

Power of point = PQ · PR; lengths are signed, so now < 0 P C Q R eigencircle

Power and determinant

Power of point = PQ · PR; lengths are signed, so now < 0 O C Q R eigencircle λ µ

Power and determinant

Power of point = PQ · PR; lengths are signed, so now < 0 O C Q R eigencircle λ µ √ − det A

Power and determinant

Power of point = PQ · PR; lengths are signed, so now < 0 O C Q R eigencircle λ µ √ − det A det A = power of origin w.r.t. eigencircle

Power and discriminant

Power and discriminant

Characteristic equation of A: λ2 − (a + d)λ + (ad − bc) = 0

Power and discriminant

Characteristic equation of A: λ2 − (a + d)λ + (ad − bc) = 0 Real eigenvalues:

Power and discriminant

Characteristic equation of A: λ2 − (a + d)λ + (ad − bc) = 0 Real eigenvalues: λ µ O C M N Y L1 (λ1, 0) L2 (λ2, 0) Discriminant: ∆ = (a + d)2 − 4 det A = 4(f 2 − det A) = 4(ρ2 − g2) = −4(g − ρ)(g + ρ) = −4 · YM · YN = −4 · (power of Y )

Power and discriminant

Characteristic equation of A: λ2 − (a + d)λ + (ad − bc) = 0 Real eigenvalues: λ µ O C M N Y L1 (λ1, 0) L2 (λ2, 0) Discriminant: ∆ = (a + d)2 − 4 det A = 4(f 2 − det A) = 4(ρ2 − g2) = −4(g − ρ)(g + ρ) = −4 · YM · YN = −4 · (power of Y ) = 4 · (YLi)2 ;

Power and discriminant

Characteristic equation of A: λ2 − (a + d)λ + (ad − bc) = 0 Real eigenvalues: λ µ O C M N Y L1 (λ1, 0) L2 (λ2, 0) Discriminant: ∆ = (a + d)2 − 4 det A = 4(f 2 − det A) = 4(ρ2 − g2) = −4(g − ρ)(g + ρ) = −4 · YM · YN = −4 · (power of Y ) = 4 · (YLi)2 ; λ1, λ2 = OY ± YLi

Power and discriminant

Characteristic equation of A: λ2 − (a + d)λ + (ad − bc) = 0 Real eigenvalues: λ µ O C M N Y L1 (λ1, 0) L2 (λ2, 0) Discriminant: ∆ = (a + d)2 − 4 det A = 4(f 2 − det A) = 4(ρ2 − g2) = −4(g − ρ)(g + ρ) = −4 · YM · YN = −4 · (power of Y )

Power and discriminant

Characteristic equation of A: λ2 − (a + d)λ + (ad − bc) = 0 Complex eigenvalues: λ µ O Discriminant: ∆ = (a + d)2 − 4 det A = 4(f 2 − det A) = 4(ρ2 − g2) = −4(g − ρ)(g + ρ) = −4 · YM · YN = −4 · (power of Y )

Power and discriminant

Characteristic equation of A: λ2 − (a + d)λ + (ad − bc) = 0 Complex eigenvalues: λ µ O C M N Y V Discriminant: ∆ = (a + d)2 − 4 det A = 4(f 2 − det A) = 4(ρ2 − g2) = −4(g − ρ)(g + ρ) = −4 · YM · YN = −4 · (power of Y )

Power and discriminant

Characteristic equation of A: λ2 − (a + d)λ + (ad − bc) = 0 Complex eigenvalues: λ µ O C M N Y V Discriminant: ∆ = (a + d)2 − 4 det A = 4(f 2 − det A) = 4(ρ2 − g2) = −4(g − ρ)(g + ρ) = −4 · YM · YN = −4 · (power of Y ) = −4 · (YV )2 ;

Power and discriminant

Characteristic equation of A: λ2 − (a + d)λ + (ad − bc) = 0 Complex eigenvalues: λ µ O C M N Y V Discriminant: ∆ = (a + d)2 − 4 det A = 4(f 2 − det A) = 4(ρ2 − g2) = −4(g − ρ)(g + ρ) = −4 · YM · YN = −4 · (power of Y ) = −4 · (YV )2 ; λ1, λ2 = OY ± i YV

Multiparameter eigenvalue problems

Multiparameter eigenvalue problems

Typical system: find k-tuple (λ1, . . . , λk) such that

k

• j=1

λjAijxi = 0, where:

◮ each Aij is an mi × ni matrix, ◮ xi is a nonzero ni-element vector.

Work includes:

◮ R D Carmichael, Amer. J. Math., 1921; ◮ F V Atkinson, Bull. Amer. Math. Soc., 1968; ◮ P Binding and P J Browne, Linear Algebra Appl., 1989; ◮ B D Sleeman, Mulitparameter Spectral Theory in Hilbert

Space, Pitman, 1978;

◮ H Volkmer, Multiparameter eigenvalue problems and

expansion theorems, Springer, 1988.

Other results

Can use eigencircle to illustrate:

◮ geometric proof that product of eigenvalues is determinant ◮ expression for angle between eigenvectors ◮ set of all matrices with a given pair of eigenvalues ◮ matrices are linear combinations of rotation and reflection

matrices

◮ complex eigenpairs: can illustrate using hyperbola in a 3rd

dimension, and a hyperboloid in 4D

◮ quaternions: can illustrate using an eigensphere

Note: all this is for 2 × 2 matrices.

Future work

◮ Pick your favourite property of matrices.

See if the 2 × 2 case can be illustrated using the eigencircle.

◮ Extend to larger matrices: 3 × 3, 4 × 4, n × n. ◮ Develop further for quaternions (eigensphere).

Further information

◮ M J Englefield and G E Farr, Eigencircles of 2x2 matrices,

Mathematics Magazine 79 (October 2006) 281–289.

◮ M J Englefield and G E Farr, Eigencircles and associated

surfaces, preprint, 2006.