lgebra Linear e Aplicaes EIGENVALUES AND EIGENVECTORS Motivation #1 - - PowerPoint PPT Presentation

Álgebra Linear e Aplicações

EIGENVALUES AND EIGENVECTORS

Motivation #1

• We have so far been focusing on systems of linear

algebraic equations

• We now move on to study systems of linear

differential equations

• Solutions to are of the form
• This motivates us to seek solutions in the same form

du1 dt = 7u1 − 4u2 du2 dt = 5u1 − 2u2 ✓u0

1

u0

2

◆ = ✓7 −4 5 −2 ◆ ✓u1 u2 ◆ u0 = Au

u0 = λu u = αeλt

Motivation #2

• Substituting and we get
• In other words,
• Each pair , with nonzero x, gives us a solution

to the differential equation in the form we sought

u1 = α1eλt u2 = α2eλt

du1 dt = 7u1 − 4u2 du2 dt = 5u1 − 2u2 λα1eλt = 7α1eλt − 4α2eλt λα2eλt = 5α1eλt − 2α2eλt λα2 = 5α1 − 2α2 λα1 = 7α1 − 4α2 ⇒ ⇒ ✓7 −4 5 −2 ◆ ✓α1 α2 ◆ = λ ✓α1 α2 ◆ ⇒ Ax = λx

(λ, x)

x = ✓α1 α2 ◆

with

Eigenvectors and Eigenvalues

• For an n×n matrix A, scalars and vectors xn×1 ≠ 0

such that are called eigenvalues and eigenvectors of A, respectively

• Each pair is called an eigenpair
• The set of distinct eigenvalues, denoted by , is

called the spectrum of A

• is singular
• , the set of eigenvectors

associated to , is an eigenspace for A

• Nonzero row vectors such that are

called left-hand eigenvectors for A

λ Ax = λx (λ, x) σ(A)

λ ∈ σ(A) ⇔ A − λI ⇔ det(A − λI) = 0 {x 6= 0 | x 2 N(A λI)}

λ

y∗(A − λI) = 0 y∗

Geometrically

• Eigenvectors experience only change in magnitude or

sign under transformation by A

Ax x

generic vector x

x

eigenvector x

Ax = λx

Back to Motivation #1

• Let us find the eigenvalues of matrix A in our differential equation
• must be singular, so
• Eigenvalues and make singular
• To find the eigenvectors, find and

✓u0

1

u0

2

◆ = ✓7 −4 5 −2 ◆ ✓u1 u2 ◆ det(A − λI) = 0 ⇒

• 7 − λ

−4 5 −2 − λ

• = 0 ⇒ p(λ) = λ2 − 5λ + 6 = 0

A − λI u0 = Au N(A − 2I) N(A − 3I) λ = 2 λ = 3 A − λI

(A − 2I)x = 0 ⇒ ✓5 −4 5 −4 ◆ x = 0 (A − 3I)x = 0 ⇒ ✓4 −4 5 −5 ◆ x = 0 x = α ✓4 5 ◆ x = β ✓1 1 ◆

Back to Motivation #2

• Each eigenpair leads to an independent solution
• Linear combinations of u1 and u2 are also solutions
• In fact, all solutions are linear combinations of u1 and u2

u1 = eλ1tx1 = e2t ✓ 4 5 ◆ u2 = eλ2tx2 = e3t ✓ 1 1 ◆ du1 dt = 7u1 − 4u2 du2 dt = 5u1 − 2u2 u1 = α14e2t + α2e3t u2 = α15e2t + α2e3t ⇒

Characteristic Polynomial and Equation

• The characteristic polynomial of An×n is
• The degree is n and the leading term is
• The characteristic equation for A is
• The eigenvalues of A are the solutions of the characteristic

equation, i.e., the roots of the characteristic polynomial

• Altogether, A has n eigenvalues, but some may be complex

numbers (even if A is real) and some may be repeated

• If A is real, then its complex eigenvalues must appear in

conjugate pairs. I.e., if then

p(λ) = det(A − λI) (−1)nλn

p(λ) = 0

λ ∈ σ(A) ¯ λ ∈ σ(A)

The Companion Matrix

• Consider the following matrix
• It’s monic characteristic polynomial is
• Proof: expand by minors on the last column

Ap =        · · · −a0 1 · · · −a1 1 · · · −a2 . . . . . . ... . . . . . . · · · 1 −an−1       

det(xI − Ap) = p(x) = a0 + a1x + a2x2 + · · · + an−1xn−1 + xn

Poor man’s root finder

• Assume there is some magical finite

eigenvalue-finding algorithm

• We can take any polynomial, build its

companion matrix, and feed to it

• This would give us the roots of any polynomial
• But there is no magical root-finding algorithm
• So there can’t be a magical eigenvalue-finding

algorithm either

DIAGONALIZATION BY SIMILARITY TRANSFORMATIONS

Example of Diagonalization

• A good choice of coordinate system (basis) can

significantly simplify an equation or problem

• Consider the oblique ellipse
• No cross-term in new coordinates:

13x2 + 10xy + 13y2 = 72

x y u v x y ✓13 5 5 13 ◆ ✓x y ◆ = 72 ✓x y ◆ = 1 √ 2 ✓1 1 1 −1 ◆ ✓u v ◆ u v ✓1 1 1 −1 ◆ 1 √ 2 ✓13 5 5 13 ◆ 1 √ 2 ✓1 1 1 −1 ◆ ✓u v ◆ = 72 u v ✓18 8 ◆ ✓u v ◆ = 72

u2 4 + v2 9 = 1

Similarities revisited

• Matrices An×n and Bn×n are said to be similar

matrices whenever there exists a nonsingular matrix P such that P–1AP = B. The product P–1AP is called a similarity transformation on A

• Similarities preserve the characteristic polynomial
• Similar matrices share the same eigenvalues
• Eigenvalues are intrinsic to the linear operator
• Eigenvectors depend on coordinate system
• Row reductions do not preserve eigenvalues

det(B − λI) = det(P−1AP − λI) = det

• P−1(A − λI)P
• = det(A − λI)

= det(P−1) det(A − λI) det(P)

A Fundamental Problem

• Given a square matrix A, reduce it to the

simplest possible form by means of a similarity

• Diagonal matrices are the simplest
• Are all matrices similar to a diagonal matrix?
• So no, certainly not for nilpotent matrices
• Can we at least triangularize?

A = ✓0 1 ◆ A2 = 0 ⇒ 0 = D2 ⇒ A = 0 P−1AP = D ⇒ P−1A2P = D2 ⇒ P−1APP−1AP = D2

Schur’s Triangularization Theorem

• Every square matrix is unitarily similar to an

upper-triangular matrix

• I.e., for each An×n there is an unitary matrix U and

an upper-triangular matrix T such that U–1AU = T

• Proof by induction
• Basis: If a matrix is 1×1, nothing needs to be done
• Assumption: Every n–1 × n–1 matrix can be triangularized
• Induction step: Prove n × n matrix can be triangularized

Ax = λx U = x X U∗AU = ✓x∗Ax x∗AX X∗Ax X∗AX ◆ B = Q∗ ˜ TQ V = ✓1 Q ◆ = T = ✓λ a∗Q ˜ T ◆ = ✓λ a∗ Bn−1×n−1 ◆ (UV)∗AUV = ✓λ a∗Q Q∗BQ ◆

Can we apply Scur’s theorem?

• Not if we need knowledge of eigenvalues
• Can we use Householder reduction instead?
• So, no

PA =      ∗ ∗ · · · ∗ ∗ · · · ∗ . . . . . . ... . . . ∗ · · · ∗     

PAP−1 =      ∗ ∗ · · · ∗ ∗ ∗ · · · ∗ . . . . . . ... . . . ∗ ∗ · · · ∗     

Hessenberg form

• But can eliminate all entries below subdiagonal!

P1 = ✓1 Q1 ◆ P1A =          ∗ ∗ ∗ · · · ∗ ∗ ∗ ∗ · · · ∗ ∗ ∗ · · · ∗ ∗ ∗ · · · ∗ . . . . . . . . . ... . . . ∗ ∗ · · · ∗          (P1A)P−1

1

=          ∗ ∗ ∗ · · · ∗ ∗ ∗ ∗ · · · ∗ ∗ ∗ · · · ∗ ∗ ∗ · · · ∗ . . . . . . . . . ... . . . ∗ ∗ · · · ∗         

P2P1AP−1

1

=          ∗ ∗ ∗ · · · ∗ ∗ ∗ ∗ · · · ∗ ∗ ∗ · · · ∗ ∗ · · · ∗ . . . . . . . . . ... . . . ∗ · · · ∗         

Hessenberg form

• But can eliminate all entries below subdiagonal!

P2 =   1 1 Q2   (P2P1AP−1

1 )P−1 2

=          ∗ ∗ ∗ · · · ∗ ∗ ∗ ∗ · · · ∗ ∗ ∗ · · · ∗ ∗ · · · ∗ . . . . . . . . . ... . . . ∗ · · · ∗          (P1A)P−1

1

=          ∗ ∗ ∗ · · · ∗ ∗ ∗ ∗ · · · ∗ ∗ ∗ · · · ∗ ∗ ∗ · · · ∗ . . . . . . . . . ... . . . ∗ ∗ · · · ∗         

Diagonalization and Eigenpairs

• If A can be diagonalized, then rewrite equation
• Looking column by column, we get ,

so is an eigenpair for A

• So each column of P is an eigenvector for A, and since P is

non-singular, the columns are linearly independent

• Conversely, columns of P are linearly independent

eigenvectors of A, and if D has the corresponding eigenvalues, then P–1AP = D

P−1An×nP = D =      λ1 · · · λ2 · · · . . . . . . ... . . . · · · λn      AP = P      λ1 · · · λ2 · · · . . . . . . ... . . . · · · λn     

as

(λi, P∗i) AP∗i = λiP∗i

Summary of Diagonalizability

• A square matrix A is said to be diagonalizable

whenever A is similar to a diagonal matrix

• A complete set of eigenvectors for A is any set of

linearly independent eigenvectors for A

• An×n is diagonalizable iff A possesses a complete

set of eigenvectors. Not all matrices do.

• iff the columns of P

constitute a complete set of eigenvectors for A and are the eigenpairs

• Otherwise, A is defective, or deficient

(λi, P∗i) P−1AP = diag(λ1, λ2, . . . λn)

Multiplicity

• For we adopt the

following definitions

• The algebraic multiplicity of is the number of times it is

repeated as root of the characteristic polynomial

• When , we say is a simple eigenvalue
• The geometric multiplicity of is , the

maximum number of l.i. eigenvectors associated to

• Eigenvalues for which

are called semisimple eigenvalues

λ ∈ σ(A) = {λ1, λ2, . . . , λs}

λ p(x) = (x − λ1)a1(x − λ2)a2 · · · (x − λs)as alg multA(λi) = ai alg multA(λ) = 1 λ λ dim N(A − λI) λ alg multA(λ) = geo multA(λ)

Multiplicity Inequality

• For every matrix A in Cn×n and for each
• Proof

λ ∈ σ(A) geo multA(λ) ≤ alg multA(λ)

U = X Y N(A − λiI) = span{xi1, xi2, · · · , xigi} geo multA(λi) = gi X = xi1 xi2 · · · xigi

• p(λ) = det(A − λI) = det(U∗AU − λI)

U∗AU = ✓X∗ Y∗ ◆ A X Y = ✓X∗AX X∗AY Y∗AX Y∗AY ◆ U∗AU − λI = ✓(λi − λ)I X∗AY Y∗AY − λI ◆ = (λi − λ)gi det(Y∗AY − λI) = ✓λiIgi×gi X∗AY Y∗AY ◆ = det

• (λi − λ)I
• det(Y∗AY − λI)

p(λ) = det(U∗AU − λI)

Independent Eigenvectors

• Eigenvectors associated to distinct eigenvalues are l.i.
• In other words, if

then when i ≠ j

• Proof
• Let

be a basis for

• Corollary: is an l.i. set

σ(A) = {λ1, λ2, . . . , λk} N(A − λiI) ∩ N(A − λjI) = {0}

N(A − λiI)

Bi = {xi1, xi2, . . . , xigi}

=

gj

X

k=1

αjkAxjk − αjkλixjk =

gj

X

k=1

αjkλjxjk − αjkλixjk

gi

X

k=1

αikxik =

gj

X

k=1

αjkxjk ⇒ (A − λiI)

gi

X

k=1

αikxik = (A − λiI)

gj

X

k=1

αjkxjk ⇒ 0 = (A − λiI)

gj

X

k=1

αjkxjk = (λj − λi)

gj

X

k=1

αjkxjk ⇒

gj

X

k=1

αjkxjk = 0

B1 ∪ B2 ∪ · · · ∪ Bk

Diagonalizability and Multiplicities

• A matrix An×n is diagonalizable iff for each
• Proof
• Proof

geo multA(λ) = alg multA(λ) λ ∈ σ(A) (⇐)

⇒ B is a basis of eigenvectors

A is diagonalizable

⇒ AP = P      λ1Im1 · · · λ2Im2 · · · . . . . . . ... . . . · · · λkImk      det(A − λI) = det(P−1AP − λI) =

k

Y

i=1

(λi − λ)mi alg multA(λi) = mi geo multA(λi) = mi

(⇒)

P = X1 X2 · · · Xk

• A[Xi]∗j = λi[Xi]∗j

|B| =

k

X

i=1

geo multA(λi) =

k

X

i=1

alg multA(λi) = n

n

is l.i.

B = B1 ∪ B2 ∪ · · · ∪ Bk

Spectral Theorem for Diagonalizable Matrices #1

• A matrix An×n with spectrum

is diagonalizable iff there exist matrices such that and the Gi’s have the following properties

• Gi is the projector onto along
• GiGj = 0 whenever i ≠ j
• This expansion is called the spectral decomposition of A.

The Gi are the spectral projectors associated with A.

σ(A) = {λ1, λ2 . . . , λk} {G1, G2, . . . , Gk} A = λ1G1 + λ2G2 + · · · + λkGk

N(A − λiI) R(A − λiI) G1 + G2 + · · · + Gk = I

• Proof

Spectral Theorem for Diagonalizable Matrices #2

A = PDP−1 = λ1X1YT

1 + λ2X2YT 2 + · · · + λkXkYT k

= λ1G1 + λ2G2 + · · · + λkGk =

• X1

X2 · · · Xk

• B

B B @ λ1I · · · λ2I · · · . . . . . . ... . . . · · · λkI 1 C C C A B B B @ YT

1

YT

2

. . . YT

k

1 C C C A PP−1 = I ⇒

k

X

i=1

Gi = I P−1P = I ) YT

i Xj =

( I, i = j 0, i 6= j ) ( G2

i = Gi

GiGj = 0, i 6= j R(Gi) = R(XiYT

i ) ⊆ R(Xi) = R(XiYT i Xi) = R(GiXi) ⊆ R(Gi)

⇒ R(Gi) = R(Xi) = N(A − λiI) Gi(A − λiI) = Gi

k

X

j=1

λjGj − λiGi = 0 ⇒ R(A − λiI) ⊆ N(Gi) dim R(A − λiI) = n − dim N(A − λiI) = n − dim R(Gi) ⇒ N(Gi) = R(A − λiI) = dim N(Gi)

(⇒)

Spectral Theorem for Diagonalizable Matrices #3

• Proof (⇐)

n = dim R(I) = dim R(G1 + G2 + · · · + Gk)

G1 + G2 + · · · + Gk = I

= dim ⇥ R(G1) + R(G2) + · · · + R(Gk) ⇤

⇒ G2

i x = GiGjy ⇒ Gix = 0 ⇒ R(Gi) ∩ R(Gj) = {0}

Gix = Gjy

= dim R(G1) + dim R(G2) + · · · + dim R(Gk) = dim N(A − λ1I) + dim N(A − λ2I) + · · · + dim N(A − λkI) = geo multA(λ1) + geo multA(λ2) + · · · + geo multA(λk)

geo multA(λi) ≤ alg multA(λi)

= alg multA(λ1) + alg multA(λ2) + · · · + alg multA(λk) ⇒ geo multA(λi) = alg multA(λi) ⇒ A is diagonalizable

FUNCTIONS OF DIAGONALIZABLE MATRICES

Intuition #1

• For square matrices A, what could we possibly mean

by sin A, eA, or ln A?

• Does not respect identities expected of sin and cos
• An alternative is to use infinite series

sin ✓a11 a12 a21 a22 ◆

?

= ✓sin a11 sin a12 sin a21 sin a22 ◆ sin2 A + cos2 A 6= I sin x =

∞

X

i=0

(−1)i x2i+1 (2i + 1)! = x − x3 3! + x5 5! + · · ·

Intuition #2

• Using the same series with matrix A
• Properties satisfied, but convergence questionable
• Simpler when A is diagonalizable

sin A =

∞

X

i=0

(−1)i A2i+1 (2i + 1)! A = PDP−1 ⇒ Ak = PDkP−1

sin A = A − A3 3! + A5 5! + · · ·

=

∞

X

i=0

(−1)i

• PDP−12i+1

(2i + 1)! = P ∞ X

i=0

(−1)i D2i+1 (2i + 1)! ! P−1 D = diag(λ1, λ2, . . . , λn) = P diag(sin λ1, sin λ2, . . . , sin λn) P−1

Intuition #3

• Given a function f and a diagonalizable matrix A,

we can avoid convergence issues defining f(A) as

• But diagonalization is not unique, so is f(A) unique?
• Use spectral decomposition
• Spectral projectors depend only on A

f(A) = P f(D) P−1 = P diag

• f(λ1), f(λ2), . . . , f(λn)
• P−1

f(A) = P f(D) P−1 = X1 X2 · · · Xk

• B

B B @ f(λ1)I · · · f(λ2)I · · · . . . . . . ... . . . · · · f(λk)I 1 C C C A B B @ YT

1

YT

2

· · · YT

k

1 C C A =

k

X

i=1

f(λi) XiYT

i = k

X

i=1

f(λi) Gi

Functions of Diagonalizable Matrices

• Let A = PDP-1 be a diagonalizable matrix where

the eigenvalues in are grouped by repetition

• Let f be a function defined at each
• Define
• Each Gi is the spectral projector associated to

D = diag(λ1I, λ2I, . . . , λkI) λi ∈ σ(A)

f(A) = P f(D) P−1 = P      f(λ1)I · · · f(λ2)I · · · . . . . . . ... . . . · · · f(λk)I      P−1 = f(λ1)G1 + f(λ2)G2 + · · · + f(λk)Gk

λi

Matrix Polynomials

• Every function of An×n can be expressed as a polynomial in A
• If f(A) exists, there is a polynomial p(z) such that p(A) = f(A)
• Simply use the Lagrange interpolating polynomials
• We have just produced formulas for each spectral projector Gi

p(A) =

k

X

i=1

p(λi)Gi =

k

X

i=1

f(λi)Gi = f(A) p(λi) = f(λi) p(z) =

k

X

i=1

f(λi) Qk

j6=i(z − λj)

Qk

j6=i(λi − λj)

! gi(z) = ( 1, z = λi 0, z 6= λi gi(A) =

k

X

i=1

gi(λi)Gi = Gi ⇒ f(A) = p(A) =

k

X

i=1

f(λi) Qk

j6=i(A − λjI)

Qk

j6=i(λi − λj)

! gi(A) =

k

X

i=1

gi(λi) Qk

j6=i(A − λjI)

Qk

j6=i(λi − λj)

! = Qk

j6=i(A − λjI)

Qk

j6=i(λi − λj)

= Qk

j6=i(A − λjI)

Qk

j6=i(λi − λj)

Gi

Spectral Projectors

• If A is diagonalizable with ,

then the spectral projector onto along is given by

• Consequently, if f(z) is defined on , then

is a polynomial in A of degree at most k – 1

σ(A) = {λ1, λ2, . . . , λk} N(A − λiI) R(A − λiI) Gi = Qk

j6=i(A − λj)

Qk

j6=i(λi − λj)

σ(A) f(A) =

k

X

i=1

f(λi)Gi

SYSTEM OF DIFFERENTIAL EQUATIONS

Systems of Differential Equations #1

• Back to our motivating example, we now want to solve
• In matrix form, with
• The solution of the scalar equation

with is

• Is a solution to the matrix equation?
• Is it the only solution?

u0

1 = a11u1 + a12u2 + · · · + a1nun

u0

2 = a21u1 + a22u2 + · · · + a2nun

. . . u0

n = an1u1 + an2u2 + · · · + annun

u0

1(0) = c1

u0

2(0) = c2

. . . u0

n(0) = cn

with

u0 = Au u(0) = c u = eAtc u(t) = eαtc u0 = αu u(0) = c

Relevant Properties of the Matrix Exponential

• If A is diagonalizable with
• Relevant properties

σ(A) = {λ1, λ2, . . . , λk}

deAt dt =

k

X

i=1

λieλitGi = k X

i=1

λiGi ! k X

i=1

eλitGi ! = AeAt eAt = eλ1tG1 + eλ2tG2 + · · · + eλktGk eA = eλ1G1 + eλ2G2 + · · · + eλkGk eAte−At = e−AteAt = I = e0 G2

i = Gi

GiGj = 0, i 6= j Af(A) = f(A)A AeAt = eAtA

Systems of Differential Equations #2

• So is certainly a solution
• To show it is the only solution, consider a different

solution v and show it equals u

• We know that
• To prove , differentiate
• So it is constant. Evaluate at t = 0 to find value
• So

u = eAtc e−Atu = c e−Atv = c

deAtv dt = eAtv0 − AeAtv = eAtv0 − eAtAv = eAtv0 − eAtv0 = 0 eA0v(0) = Ic = c

e−Atv = c ⇒ v = eAtc = u

Summary of Differential Equations

• If A is diagonalizable with

then the unique solution of , is where vi is the eigenvector vi = Gi c, and Gi is the spectral projector associated to

σ(A) = {λ1, λ2, . . . , λk} u0 = Au u(0) = c u = eAtc = eλ1tv1 + eλ2tv2 + · · · + eλltvk λi

Long-term behavior

• What is the behavior of the solution as ?
• Since vi are constant, it must be controlled by eigenvalues
• Term due to imaginary part eiyt oscillates around unit circle
• Norm is controlled by real part of eigenvalues
• For example, if for all i, then as

and for all c, i.e., it evolves to 0

u = eAtc = eλ1tv1 + eλ2tv2 + · · · + eλltvk

|eλt| = |exteiyt| = |ext| |eiyt| = |ext| eλt = e(x+iy)t = exteiyt eiyt = cos yt + i sin yt

Re(λi) < 0 t → ∞ eAt → 0 u(t) → 0 t → ∞

Summary of Stability

• Let and , and let A be diagonalizable

with eigenvalues

• If for each i, then and

for every c. The system and matrix are said to be stable

• If for some i, components of u(t) can become

unbounded as . The system and matrix are unstable

• If , components of u(t) remain bounded, but some

can oscillate indefinitely. The system and matrix are semistable

u0 = Au u(0) = c λi

Re(λi) < 0 lim

t→∞ eAt = 0

lim

t→∞ u(t) = 0

Re(λi) > 0 t → ∞ Re(λi) ≤ 0

Predator–Prey Application #1

• Assume populations u1(t) of predators and u2(t) of

prey, and assume the populations are governed by

• Find the size of the populations at each time and

determine if and when either will become extinct

• Solution

with

u1(0) = u2(0) = 100 u0

1 =

u1 + u2 u0

2 = −u1 + u2

u0 = Au ⇒ u = eAtc c = ✓100 100 ◆ A = ✓ 1 1 −1 1 ◆

Predator–Prey Application #2

• Find eigenvalues for A
• Use spectral projectors to compute matrix power

p(λ) = λ2 − 2λ + 2 = 0 ⇒ ( λ1 = 1 + i λ2 = 1 − i u(t) = eAtc = eλ1tG1c + eλ2tG2c G1 + G2 = I = eλ1tv1 + eλ2tv2 ⇒ v1 + v2 = c G1 = A − λ2I λ1 − λ2 = 1 2 ✓1 −i i 1 ◆ ⇒ v1 = 50 ✓1 − i 1 + i ◆ ⇒ v2 = 50 ✓1 + i 1 − i ◆ = 100et(cos t + sin t) = 100et(cos t − sin t) u2(t) = 50

• (1 + i)e(1+i)t + (1 − i)e(1−i)t

u1(t) = 50

• (1 − i)e(1+i)t + (1 + i)e(1−i)t

Predator–Prey Application #3

• Plotting the solution

u1(t) = 100et(cos t + sin t) u2(t) = 100et(cos t − sin t) u2(t) u1(t) u2(t) = 0 ⇒ cos t = sin t ⇒ t = π 4

Example of Difference Equation

• Fibonacci sequence
• 0, 1, 1, 2, 3, 5, 8, 13, 21, …
• Recurrence relation
• Appears in many places in Nature

f0 = 0 f1 = 1 fi = fi−1 + fi−2

Fibonacci Sequence

• Find the general term of the Fibonacci Sequence
• First, rewrite equation in matrix form
• Then use diagonalization to compute matrix power

f0 = 0 f1 = 1 fi = fi−1 + fi−2 ✓ fi fi−1 ◆ = ✓1 1 1 ◆ ✓fi−1 fi−2 ◆ fi = ✓ fi fi−1 ◆ A = ✓1 1 1 ◆ fi = A fi−1 = Ai−1 f1 p(λ) = λ2 − λ − 1 = 0 λ1 = 1 + √ 5 2 λ2 = 1 − √ 5 2 x1 = ✓λ1 1 ◆ x2 = ✓λ2 1 ◆ A = PΛP−1 ⇒ Ai = PΛiP−1 ⇒ fi = PΛi−1P−1 f1 P = ✓λ1 λ2 1 1 ◆ fi = [fi]1 P−1 = 1 √ 5 ✓ 1 −λ2 −1 λ1 ◆ = 1 √ 5 @ 1 + √ 5 2 !i − 1 − √ 5 2 !i1 A = [PΛi−1P−1 f0]1 = 1 √ 5

• λi

1 − λi 2

Invertibility and Diagonalizability

• Invertibility is concerned with eigenvalues
• A is singular iff

is an eigenvalue for A

• Diagonalization is concerned with eigenvectors
• n independent eigenvectors
• No connection between two concepts
• Some invertible matrices can be diagonalized
• Some diagonalizable matrices are singular

λ = 0 x 6= 0 2 N(A) , Ax = 0x det(A − 0I) = det(A) = 0

DISCRETE FOURIER TRANSFORM

(AND THE FFT)

NORMAL MATRICES

Unitary diagonalization

• Vectors associated to different eigenvalues are always l.i.
• When all eigenvalues are distinct, the matrix is diagonalizable
• Repeated eigenvalues not necessarily a problem: identity matrix
• We can always pick an orthonormal basis for the

eigenspace of each eigenvalue

• But what about vectors associated to distinct eivenvalues?

Can we make them mutually orthogonal?

• If so, we can diagonalize using orthogonal matrices
• Does not work for all matrices
• Nevertheless, a large class of matrices is unitarily diagonalizable

Hermitian or Symmetric Matrices

• Mutually orthogonal eigenspaces
• Proof in two parts
• Eigenvalues of Hermitian or symmetric matrices are real
• Eigenvectors from distinct eigenvalues are orthogonal

Ax = λx

• x∗Ax

∗ = ¯ c = x∗A∗x = c x∗Ax = λx∗x = λkxk2 = c ⇒ c ∈ R ⇒ λ ∈ R = x∗Ax Ax1 = λ1x1 Ax2 = λ2x2 (λ1x1)∗x2 = (Ax1)∗x2 = x∗

1Ax2 = x∗ 1λ2x2

⇒ λ1x∗

1x2 = λ2x∗ 1x2 ⇒ x∗ 1x2 = 0

Hermitian or Symmetric Matrices

• We have shown that Hermitian or Symmetric matrices

have real eigenvalues and orthogonal eigenvectors

• We now show such matrices are unitarily diagonalizable,

i.e., unitarily similar to a diagonal matrix

• Proof immediate from Schur’s triangularization

A = U∗TU A∗ = U∗T∗U = D A = A∗ ⇒ U∗TU = U∗T∗U ⇒ TU = T∗U ⇒ T = T∗

Lanczos Tridiagonalization

• Schur’s theorem is impractical
• It needs the eigenvalues
• So triangulation is impractical
• In practice, instead of triangulation, we can

reduce to Hessenberg form

• If the matrix is symmetric, the result is a

tridiagonal matrix

• This leads to a massive improvement in

performance

Unitary Diagonalization #1

• If matrix A is unitarily diagonalizable, then
• A is symmetric iff D = D*
• All are eigenvalues real
• A is anti-symmetric iff D = –D*
• All eigenvalues imaginary
• If A itself is unitary, then
• A is unitarily diagonalizable too
• All eigenvalues on unit circle in the complex plane

A∗ = U∗D∗U A = U∗DU

and

A = Q∗TQ A∗A = I ⇒ Q∗T∗TQ = I ⇒ T∗T = I ⇒ T = D ⇒ Q∗T∗QQ∗TQ = I

n

Unitary Diagonalization #2

• If matrix A is unitarily diagonalizable, then
• Matrix A is called normal if it commutes with A*
• Every unitarily diagonalizable matrix is normal
• Symmetric, anti-symmetric, and unitary matrices are all

examples of normal matrices

• Is every normal matrix unitarily diagonalizable?

A∗A =

• U∗D∗U
• U∗DU
• = U∗D∗DU

= U∗DD∗U=

• U∗DU
• U∗D∗U

= AA∗

Unitary Diagonalization #3

• Yes, every normal matrix is unitarily diagonalizable
• Proof in two parts
• So T is triangular and normal. Is it diagonal?

A = Q∗TQ A∗ = Q∗T∗Q A∗A = AA∗ Q∗T∗QQ∗TQ = Q∗TQQ∗T∗Q kTxk2 = x∗T∗Tx = x∗TT∗x = kT∗xk2 x = ei ) kT∗ik = kT∗

∗ik (true of every normal matrix)

kT∗1k2 = |t11|2 kT∗2k2 = |t22|2 kT∗

∗2k2 = |t22|2 + n

X

j=3

|t2j|2 ⇒ t1j = 0, i ∈ {2, . . . , n}

n n

⇒ t2j = 0, i ∈ {3, . . . , n} kT∗

∗1k2 = |t11|2 + n

X

j=2

|t1j|2 ⇒ Q∗T∗TQ = Q∗TT∗Q ⇒ T∗T = TT∗

Summary of Normal Matrices

• A matrix A is called normal iff it commutes with A*
• A*A = AA*
• A matrix A is unitarily diagonalizable iff it is normal
• If A is a normal matrix with
• A is RPN, i.e.,
• is also normal, hence also RPN
• when
• The spectral projectors Gi are orthogonal projectors
• Examples of normal matrices
• Symmetric, skew-symmetric, Hermitian, skew-Hermitian,
• rthogonal, and unitary matrices

σ(A) = {λ1, λ2, . . . , λk}

A − λiI N(A − λiI) ⊥ N(A − λjI) λi 6= λj R(A) ⊥ N(A)

POSITIVE DEFINITE MATRICES

Positive Definite Matrices

• When a matrix is symmetric or Hermitian, all eigenvalues are real
• What additional property will make them positive?
• Non-negative eigenvalues iff A can be factored as A = B*B
• Positive eigenvalues iff additionally B is non-singular
• Alternative characterization

A = UDU∗ λi ≥ 0

n

⇒ A = UD

1 2 D 1 2 U∗ = B∗B

B = D

1 2 U∗

A = B∗B ⇒ λi = x∗

i Axi

x∗

i xi

= x∗

i B∗Bxi

x∗

i xi

= kBxk2 kxik2 ≥ 0 A = B∗B ⇒ x∗Ax = x∗B∗Bx = kBxk2 0 ⇒ x∗

i Axi

x∗

i xi

= λi ≥ 0 x∗Ax ≥ 0 x∗

i Axi ≥ 0

⇒

Positive Definite Matrices

• For any matrix A, the following statements are equivalent
• A is symmetric positive definite, denoted
• for every non-zero x
• All eigenvalues of A are positive
• A = B*B for some non-singular B
• Although B is not unique, there is only one upper triangular R with

positive diagonals such that A = R*R, the Cholesky factorization of A

• A has an LU (LDU) factorization where every pivot is positive
• This factorization is such that LDL* = R*R, with R = D

1 2 L∗

x∗Ax > 0

A 0

Polar decomposition

• Every complex square matrix A accepts a

factorization , with Q unitary and H Hermitian positive semidefinite

• Geometrically, a scale along perpendicular

directions followed by a rotation (maybe reflection)

• Left polar decomposition also exists
• If A is non-singular, decompositions are unique

A = QH

A = HQ0

Polar decomposition

• Proof from SVD

A = UΣV∗= U(V∗V)ΣV∗= (UV∗)(VΣV∗) Q = UV∗ H = VΣV∗ (z = ρeiθ) = √ A∗A

Diagonalizable matrices

• Matrix A is diagonalizable iff

with and

• i.e., N normal and H symmetric positive definite
• Proof
• Proof

A = H−1NH N∗N = NN∗ H 0

(⇐) A = PDP−1 P−1 = QH A = H−1Q∗DQH = H−1NH N = Q∗DQ N∗N = NN∗ (⇐) N = PDP−1 A = H−1NH = H−1PDP−1H = SDS−1 S = H−1P

A map of matrix types

RPN Normal Hermitian Real-Symmetric Non-singular Diagonalizable

THE JORDAN NORMAL FORM

Cyclic subspaces

• A subspace is cyclic w.r.t

if

• is nilpotent
• E is an invariant subspace of A
• is a basis for E
• Construction
• Let and take

A : V → V

A\E index(A\E) = k v 2 E, Ak−1v 6= 0

⇒ α1 = 0 α1 = α2 = · · · = αk = 0 s = 0 ⇒ Ak−1s = α1Ak−1v = 0 Ak−2s = α1Ak−2v + α2Ak−1v = α2Ak−1v = 0 ⇒ α2 = 0

E ⊂ V

l.i invariant

E = {Ak−1v, . . . , Av, v}

s =

k

X

i=1

αiAi−1v As =

k−1

X

i=1

αiAiv ∈ E

Cyclic Jordan structure

• Let be nilpotent with index k and E a cyclic

subspace with basis

• What is structure of ?

=       1 ... ... ... 1      

A\E E = {Ak−1v, . . . , Av, v} A\E

A\E = ⇥ A(Ak−1v)E A(Ak−2v)E · · · A(v)E ⇤

Jordan chains

• Let be nilpotent with index k and E a cyclic

subspace with basis

• Define so
• v1 is an eigenvector with eigenvalue
• Furthermore,

A\E E = {Ak−1v, . . . , Av, v} λ = 0 E = {v1, . . . , vk} vi = Ak−iv

Av1 = 0 = λv1 Av2 = v1 = λv2 + v1 Av3 = v2 = λv3 + v2 Avk = vk−1 = λvk + vk−1 · · ·

Nilpotent cyclic decomposition #1

• If is nilpotent, then V can be

decomposed into a direct sum where Ei are all cyclic subspaces of A

• Proof by induction on index k of A
• Base case

E1 ⊕ E2 ⊕ · · · ⊕ Em k = 1 V = span{v1, v2, . . . , vn} Ei = span{vi}

A : V → V

A = 0 AEi = 0 ∈ Ei

Nilpotent cyclic decomposition #2

• Proof by induction on index k of A
• Induction step: Assume A has index
• Consider
• , so by inductive hypothesis

with

• Since , there is wi such that
• Define

k + 1 Awi = vi vi ∈ R(A) A0 = A\R(A) index(A0) = k R(A) = E0

1 ⊕ E0 2 ⊕ · · · ⊕ E0 m

E0

i = span{Aki1vi, Aki2vi, . . . , vi}

= span{Akiwi, Aki−1wi, . . . , wi} Ei = span{Aki−1vi, Aki−2vi, . . . , vi, wi}

Nilpotent cyclic decomposition #3

• Proof by induction on index k of A
• Define
• Note that
• Complete basis so that
• Define
• We now prove that

Em+i = span{zi} V = E1 ⊕ E2 ⊕ · · · ⊕ Em+p Ei = span{Aki−1vi, Aki−2vi, . . . , vi, wi} R(A) ∩ N(A) = span{Ak1−1v1, . . . , Akm−1vm} N(A) = span{Ak1−1v1, . . . , Akm−1vm, z1, z2, . . . , zp}

Nilpotent cyclic decomposition #4

Em+i = span{zi} R(A) = E0

1 ⊕ E0 2 ⊕ · · · ⊕ E0 m

Awi = vi

multiply by A basis for !

R(A) 0 =

m

X

i=1

@γivi +

kj−1

X

j=1

βijAjvi 1 A 0 = p X

i=1

αizi ! +

m

X

i=1

βi,kiAki−1vi 0 = p X

i=1

αizi ! +

m

X

i=1

@γiwi +

ki

X

j=1

βijAj−1vi 1 A

basis for ! N(A)

αi = 0, i ∈ {1, . . . , p} βiki = 0, i ∈ {1, . . . , m} γi = 0 βij = 0 j ∈ {1, . . . , ki − 1} i ∈ {1, . . . , m}

is a direct sum (l.i)

E1 + E2 + · · · + Em+p Ei = span{Aki−1vi, Aki−2vi, . . . , vi, wi} E0

i = span{Aki1vi, Aki2vi, . . . , vi}

N(A) = span{Ak1−1v1, . . . , Akm−1vm, z1, z2, . . . , zp}

R(A) ∩ N(A) = span{Ak1−1v1, . . . , Akm−1vm}

Nilpotent cyclic decomposition #5

Em+i = span{zi} R(A) = E0

1 ⊕ E0 2 ⊕ · · · ⊕ E0 m

dim(E1 ⊕ E2 ⊕ · · · ⊕ Em+p) =

m

X

i=1

dim(Ei) +

p

X

i=1

dim(Em+p) = m +

m

X

i=1

dim(E0

i)

! + dim

• N(A)
• − m

= dim

• R(A)
• + dim
• N(A)
• = dim(V )

Awi = vi Ei = span{Aki−1vi, Aki−2vi, . . . , vi, wi} E0

i = span{Aki1vi, Aki2vi, . . . , vi}

dim(E1 ⊕ E2 ⊕ · · · ⊕ Em+p) = dim(V ) N(A) = span{Ak1−1v1, . . . , Akm−1vm, z1, z2, . . . , zp} R(A) ∩ N(A) = span{Ak1−1v1, . . . , Akm−1vm}

Another visualization

• Assume A nilpotent and

N(A) N(A) ∩ R(A) N(A) ∩ R(A2) U0 V1 X2 A2U2 AV2 AU1 N(A) ∩ R(A3) = R(A3)

index(A) = 4

Y3 AX3 R(A3) : ⇥Y3 ⇤ A2V3 A3U3 V : ⇥U3 AU3 A2U3 A3U3 U2 AU2 A2U2 U1 AU1 U0 ⇤ R(A2) : ⇥X3 AX3 X2 ⇤ R(A) : ⇥V3 AV3 A2V3 V2 AV2 V1 ⇤ dim

• N(A) ∩ R(Ai)
• = rank(Ai) − rank(Ai+1) = ri − ri+1

cols(Ui) = dim

• N(A) ∩ R(Ai)
• − dim
• N(A) ∩ R(Ai+1)
• = ri − 2ri+1 + ri+2

Nilpotent Jordan form #1

• Every nilpotent matrix of index k

is similar to a block diagonal matrix in the form

• , where each Ei contains

the vectors from Ei, the basis of cyclic subspace Ei from the nilpotent cyclic decomposition of V

Ni =       1 ... ... ... 1      

with

P−1LP = N =      N1 · · · N2 · · · . . . ... . . . · · · Nt     

P = ⇥E1 E2 · · · Et ⇤

L : V → V

Nilpotent Jordan form #2

• Every nilpotent matrix of index k

is similar to a block diagonal matrix in the form

• The number of blocks is
• The size of the largest block is
• The number of blocks is

where

• Since ri depends only on L, the structure is unique

Ni =       1 ... ... ... 1      

with

t = dim(N(L)) k × k

P−1LP = N =      N1 · · · N2 · · · . . . ... . . . · · · Nt     

L : V → V

i × i νi = ri−1 − 2ri + ri+1 ri = rank(L

i)

Derivation of Jordan form

• Let and
• From the core-nilpotent factorization of
• From the Nilpotent Jordan form
• Using block multiplication
• Proceed by blocks and by induction with A1

A ∈ Cn×n σ(A) = {λ1, λ2, . . . , λs}

index(L1) = index(A − λ1I)

C1 non-singular, L1 nilpotent

Y−1

1 L1Y1 = N(λ1)

X−1

1 (A − λ1I)X1 =

L1 C1

• Q1 = X1

Y1 I

• Q−1

1 (A − λ1I)Q1 =

N(λ1) C1

• J(λ1) = N(λ1) + λ1I

A1 = C1 + λ1I σ(A1) = {λ2, λ3, . . . , λs} Q−1

1 AQ1 =

N(λ1) + λ1I C1 + λ1I

• =

J(λ1) A1

• A − λ1I

The Jordan form

• For every matrix with spectrum

, there is a matrix P s.t.

• J has one Jordan segment

for each

• Each segment is made up of

Jordan blocks

A ∈ Cn×n σ(A) = {λ1, λ2, . . . , λs}

P−1AP = J =      J(λ1) · · · J(λ2) · · · . . . ... . . . · · · J(λs)     

J(λi) λi ∈ σ(A) ti = dim

• N(A − λiI)
• J(λi) =

     J1(λi) · · · J2(λi) · · · . . . ... . . . · · · Jti(λi)      Jj(λi) =       λi 1 ... ... ... 1 λi      

Jordan chains again

• Focus on a single Jordan block
• Let be the

section of P corresponding to block

Jj(λi) Pj(λi) = ⇥x1 x2 · · · xpji ⇤ Jj(λi)

A ⇥ x1 x2 · · · xpji ⇤ = ⇥ x1 x2 · · · xpji ⇤ 2 6 6 6 6 4 λi 1 ... ... ... 1 λi 3 7 7 7 7 5

pji×pji

Ax1 = λix1 Ax2 = λix2 + x1 Ax3 = λix3 + x2 Axpji = λixpji + xpji−1 . . . (A − λiI)x1 = 0 (A − λiI)x2 = x1 (A − λiI)x3 = x2 (A − λiI)xpji = xpji−1 . . .

eigenvector generalized eigenvectors

. . . (A − λiI)x1 = 0 (A − λiI)2x2 = 0 (A − λiI)3x3 = 0 (A − λiI)pjixpji = 0

Constructing Jordan chains

• For each , set

for ,

• Sequentially extend basis with

such that is a basis for

• Build a Jordan chain on top of each vector
• Solve
• When all are put in a matrix P, then P is non-

singular and

λ ∈ σ(An×n) Mi = R

• (A − λI)i

∩ N(A − λI) k = index(A − λI) i ∈ {0, 1, . . . , k − 1} Sk−1 Sk−2, . . . , S0 Sk−1 ∪ · · · ∪ Sk−i Mi bij ∈ Si

(A − λI)ixij = bij Pij = ⇥(A − λI)ixij (A − λI)ixij · · · (A − λI)ixij xij ⇤

Pij PAP−1 = J

Functions of non-diagonalizable matrices

• Direct generalization of the diagonalizable case
• If , then
• Same issues as before
• must make sense
• must be unique even though P isn’t
• First, let’s make sure it makes sense
• Then, let’s find a definition based on projectors

to ensure we have an uniquely defined matrix

A = PJP−1 f(A) = Pf(J)P−1 f(J) Pf(J)P−1

Functions of a Jordan block #1

• Problem reduces to finding

J(λi) =      J1(λi) · · · J2(λi) · · · . . . ... . . . · · · Jti(λi)      Jj(λi) =       λi 1 ... ... ... 1 λi       J =      J(λ1) · · · J(λ2) · · · . . . ... . . . · · · J(λs)     

f

• Jj(λi)
• f(J) =

2 6 6 6 4 f

• J(λ1)
• · · ·

f

• J(λ2)
• · · ·

. . . ... . . . · · · f

• J(λs)
• 3

7 7 7 5 f

• J(λi)
• =

2 6 6 6 4 f

• J1(λi)
• · · ·

f

• J2(λi)
• · · ·

. . . ... . . . · · · f

• Jti(λi)
• 3

7 7 7 5 f

• Jj(λi)
• = ?

Functions of a Jordan block #2

• Consider for a moment that has a

Taylor expansion about

• Obvious analogy is
• But is nilpotent! This sum is finite!

f : C → C λ

f(z) = f(λ) + f 0(λ)(z − λ) + f 00(λ) 2! (z − λ)2 + f 000(λ) 3! (z − λ)3 + · · · |z − λ| < r

for

f

• Jj(λi)
• = f(λi)I + f 0(λi)
• Jj(λi) − λiI
• + f 00(λi)

2!

• Jj(λi) − λiI

2 + · · · f

• Jj(λi)
• =

kij−1

• p=0

f (p) p!

• Jj(λi) − λiI

p

Jj(λi) − λiI

Jj(λi) − λiI =       1 ... ... ... 1      

kij×kij

Functions of Jordan block #3

• For a Jordan block and a function

for which exist,

Jj(λi) f(z) f(λi), f 0(λi), . . . , f (kij1)(λi)

f

• Jj(λi)
• =

kij−1

X

p=0

f (p) p!

• Jj(λi) − λiI

p =                   f(λi) f 0(λi)

f 00(λi) 2!

· · ·

f (kij 1)(λi) (kij1)!

f(λi) f 0(λi) ... . . . ... ...

f 00(λi) 2!

f(λi) f 0(λi) f(λi)                  

Is uniquely defined? #1

• Start partitioning J into its s Jordan segments and

partition P and P-1 conformably

• Define spectral projectors
• Gi projects onto along
• Perfectly consistent with diagonalizable case

f(A)

P = ⇥ P1 · · · Ps ⇤ P−1 =    Q1 . . . Qs    J =    J(λ1) ... J(λs)   

R

• (A − λiI)ki

N

• (A − λiI)ki

Gi = PiQi

A − λiI = P(J − λiI)P−1 = P

        J(λ1) − λiI ... J(λi) − λiI ... J(λs) − λiI         P−1 nilpotent with index ki non-singular core-nilpotent decomposition!

Is uniquely defined? #2

• Continuing with the development of
• But
• So
• We now show this can be simplified further to

f(A)

f(A)

f(A) = Pf(J)P−1 =

s

X

i=1

Pif

• J(λi)
• Qi

f

• J(λi)
• =

ki−1

X

j=0

f (j)(λi) j!

• J(λi) − λiI

j f(A) =

s

X

i=1

Pi @

ki−1

X

j=0

f (j)(λi) j!

• J(λi) − λiI

j 1 A Qi =

s

X

i=1 ki−1

X

j=0

f (j)(λi) j! Pi

• J(λi) − λiI

jQi f(A) =

s

X

i=1 ki−1

X

j=0

f (j)(λi) j! (A − λiI)jGi

Is uniquely defined? #3

• Proof

f(A)

f(A) =

s

X

i=1 ki−1

X

j=0

f (j)(λi) j! Pi

• J(λi) − λiI

jQi =

s

X

i=1 ki−1

X

j=0

f (j)(λi) j! (A − λiI)jGi (A − λiI)jGi = P(J − λiI)jP−1Gi = h P1

• J(λ1) − λiI

j · · · Pi

• J(λi) − λiI

j · · · Ps

• J(λs) − λiI

ji 2 6 6 6 6 6 6 4 . . . Qi . . . 3 7 7 7 7 7 7 5

= ⇥P1 · · · Pi · · · Ps ⇤ 2 6 6 6 6 6 6 6 4

• J(λ1) − λiI

j ...

• J(λi) − λiI

j ...

• J(λs) − λiI

j 3 7 7 7 7 7 7 7 5 2 6 6 6 6 6 6 4 Q1 . . . Qi . . . Qs 3 7 7 7 7 7 7 5 PiQi

= Pi

• J(λi) − λiI

jQi

QjPi = δjiI

Spectral resolution of .

• For with such that

and for a function such that exist for each , the value of is

• is the projector onto the generalized eigenspace

along

• when
• is nilpotent of index ki

f(A)

A ∈ Cn×n σ(A) = {λ1, . . . , λs} ki = index(A − λiI) f : C → C f(λi), f 0(λi), . . . , f (ki1)(λi) λi ∈ σ(A) f(A)

f(A) =

s

X

i=1 ki−1

X

j=0

f (j)(λi) j! (A − λiI)jGi

N

• (A − λiI)ki

R

• (A − λiI)ki

Gi G1 + G2 + · · · + Gs = I GiGj = 0 i 6= j

• A − λiI
• Gi = Gi
• A − λiI
• = Pi
• J(λi) − λiI
• Qi

Back to motivation #1

• Consider the following system of linear ordinary

differential equations

• Want to compute
• Now what?

u0

2 = 4(u2 − u1)

u0

1 = u2

u0

3 = u1 − u2 + 3u3

A =   1 −4 4 1 −1 3   u0 = Au det(A − λI) = −λ3 + 7λ2 − 16λ + 12 = (3 − λ)

• −λ

1 −4 4 − λ

• = (3 − λ)(2 − λ)2

N(A − 3I) = α   1   N(A − 2I) = β   1 2 1   u1(0) = 2 u2(0) = 1 u3(0) = 3

u = eAtc

u(0) = c c =   2 1 3   P =   1 −2 2 −3 1 1   P−1AP = J =   2 1 2 3   (A − 2I) v =   1 2 1   v =   −2 −3  

Back to motivation #1

• From the Jordan form
• We have

P−1AP = J =   2 1 2 3   eAt = P eJt P−1 = P   e2t te2t e2t e3t   P−1 =   (1 − 2t)e2t te2t −4te2t (2t + 1)e2t 3e3t − (2t + 3)e2t (t + 2)e2t − 2e3t e3t   eAtc =   (2 − 3t)e2t (1 − 6t)e2t 7e3t − (3t + 4)e2t  