If kT kop = 0, then kT (y)k = 0 for any y with kyk = 1

1. hw 9

(1) Let L(Rⁿ, R^m) be the space of all linear maps from Rⁿ to R^m. For each T ∈ L(Rⁿ, R^m), we define

kT k_op= sup

kxk_Rn=1

kT (x)k_Rⁿ. (a) Prove that k · kop defines a norm on L(Rⁿ, R^m).

Proof. The number kT (x)k ≥ 0 for any x ∈ Rⁿ, kT kop = sup_kxk=1kT (x)k ≥ 0.

If kT kop = 0, then kT (y)k = 0 for any y with kyk = 1. If x = 0, by linearity of T, T (x) = 0. If x 6= 0, take y = x/kxk. Then T (y) = 0 implies that T (x) = 0 by linearity of T. This shows that T (x) = 0 for all x ∈ Rⁿ, i.e. T is the zero map.

For a ∈ R, kaT k_op= sup

kxk=1

kaT (x)k = sup

kxk=1

|a|kT (x)k = |a| sup

kxk=1

kT (x)k = |a|kT k_op. For S, T ∈ L(Rⁿ, R^m) and any x ∈ Rⁿ with kxk = 1,

k(S + T )(x)k = kS(x) + T (x)k ≤ kS(x)k + kT (x)k ≤ kSk_op+ kT k_op

by triangle inequality. Hence kS + T kop≤ kSk_op+ kT kop.  (b) Prove that kT (x)k_R^m ≤ kT k_opkxk_Rⁿ for all x ∈ Rⁿ.

Proof. When x = 0, the inequality is obvious. When x 6= 0, take y = x/kxk.

Then kT (y)k ≤ kT k implies that kT k ≥

T

 x kxk



= kT (x)k kxk .

Multiplying the inequality on the both side by kxk, we obtain the result.

 (c) Let S ∈ L(R^m, R^p). Prove that kS ◦ T kop≤ kSk_opkT k_op.

Proof. For x ∈ Rⁿ with kxk = 1, by exercise (1b), kS ◦ T (x)k ≤ kSkkT (x)k ≤ kSkkT k.

Thus kS ◦ T k ≤ kSkkT k.

 (2) For each A ∈ M_mn(R), we define the matrix norm of A to be

kAk = kL_Ak_op

where LA: Rⁿ→ R^m is the linear map LA(x) = Ax for any x ∈ Rⁿ. In this exercise, we assume m = n.

(a) Use mathematical induction to prove that kA^kk ≤ kAk^k for any k ≥ 1.

Proof. When k = 1, the statement is obvious. Assume that kA^kk ≤ kAk^k for k ∈ N. Exercise (1c) implies that

kA^k+1k = kA ◦ A^kk ≤ kAkkA^kk ≤ kAk · kAk^k= kAk^k+1.

Thus the statement is true for k + 1. By induction, the statement is true for any

natural number k. 

1

Remark. This implies that the sequence of numbers (kA^kk^1/k) is bounded above by kAk. We define the spectral radius of a square matrix A to be

ρ(A) = lim sup

k→∞

kA^kk^1/k. This implies that ρ(A) ≤ kAk.

(b) Let E_ij be the n × n matrix whose ij-th entry is 1 and zero otherwise. Find kE_ijk for all 1 ≤ i, j ≤ n.

Proof. Let {ei} be the standard basis for Rⁿ. For any x =Pn

i=1xiei, we have kE_ijxk = |xj| ≤

q

x²₁+ · · · + x²_n= kxk.

This shows that kE_ijk ≤ 1. On the other hand, kE_ije_jk = 1 ≤ kE_ijk. We find kE_ijk = 1.

 (c) Let λ₁, · · · , λ_n be n real numbers. Suppose that D = Pn

i=1λ_iE_ii, i.e. D is a diagonal matrix. Prove that

kDk = max{|λ₁|, · · · , |λ_n|}.

We denote D by diag(λ₁, · · · , λ_n). Prove that ρ(D) = kDk.

Proof. Denote λ = max{|λ₁|, · · · , |λ_n|}. Let x =Pn

i=1x_ie_iThen Dx =Pn i=1λ_ix_i. We see that

kDxk =p

(λ₁x₁)²+ · · · + (λ_nx_n)² ≤ q

λ²(x²₁+ · · · + x²_n) ≤ λkxk.

Thus kDk ≤ λ. On the other hand,

|λ_j| = kDe_jk ≤ kDk for 1 ≤ j ≤ n.

This implies that λ ≤ kDk. We conclude that kDk = λ.

 (3) Let R > 0. Suppose that the power series f (x) =P∞

k=0a_kx^k is convergent for any x ∈ (−R, R). Here an∈ R for any n ≥ 0. Let A be an n × n real matrix such that kAk ∈ (−R + δ, R − δ) where 0 < δ < R.

(a) Prove that P∞

k=0a_kA^k is convergent in (M_n(R), k · k). In this case, we define f (A) =P∞

k=0akA^k.

Proof. Let ρ = R − δ. Since ρ ∈ (−R, R), P∞

k=0a_kρ^k is convergent in R. By n-th term test, lim

k→∞a_kρ^k = 0. Choose  = 1. Then we can find N ∈ N so that

|a_kρ^k| < 1 whenever k ≥ N. By exercise (2a), ka_kA^kk = |a_k|kA^kk ≤ |a_k|kAk^k≤ kAk

ρ

k

for k ≥ N .

Since kAk ∈ (−ρ, ρ), 0 < (kAk/ρ) < 1 and hence the geometric series

∞

X

k=N

(kAk/ρ)^k

is convergent in R. By comparison test,

∞

X

k=N

ka_kA^kk is convergent in R. Since

(M_n(R), k · k) is a Banach space,

∞

X

k=N

a_kA^k is convergent in (M_n(R), k · k). Thus

∞

X

k=0

akA^k=

N −1

X

k=0

akA^k+

∞

X

k=N

akA^k

is convergent in (M_n(R), k · k). 

(b) If D = diag(λ₁, · · · , λ_n), where λ₁, · · · , λ_n are real numbers in (−R + δ, R − δ).

Prove that f (D) = diag(f (λ1), · · · , f (λn)), i.e. f (D) =Pn

i=1f (λi)Eii.

Proof. Let B = diag(f (λ1), · · · , f (λn)). For each m ≥ 1, we set fm(x) =

m

X

k=0

a_kx^k for x ∈ (−R, R). To prove that f (D) = B, it we have to show that B = lim

m→∞f_m(D). By induction, we have D^m =Pn

i=1λ^m_i E_ii. Hence fm(D) =

n

X

i=1

fm(λi)Eii. By exercise (2c),

kf_m(D) − Bk = max{|f_m(λ_i) − f (λ_i)| : 1 ≤ i ≤ n}.

Since limm→∞fm(x) = f (x) for x ∈ (−R, R), for any  > 0, there exists N_,1, · · · , N_,n∈ N so that

|f_m(λi) − f (λi)| < 

if m ≥ N_,i. Take N_ = max{N_,i : 1 ≤ i ≤ n}. If m ≥ N_, |f_m(λ_i) − f (λ_i)| <  for all 1 ≤ i ≤ n. Thus kf_m(D) − Bk <  if m ≥ N_. We prove that f (D) = B.

 (c) Let A be diagonalizable¹ with A = SDS⁻¹ where D = diag(λ1, · · · , λn). Sup-

pose λ_i ∈ (−R + δ, R − δ) for 1 ≤ i ≤ n. Show that f (A) = Sf (D)S⁻¹.

Proof. Let f_m : (−R, R) → R be the polynomial function as above. By induc- tion, we see that

A^k= SD^kS⁻¹ for any k ≥ 1.

Hence f_m(A) = Sf_m(D)S⁻¹ for any m ≥ 1. Let C = Sf (D)S⁻¹. By exercise (1c),

kf_m(A) − Ck = kSfm(D)S⁻¹− Sf (D)S⁻¹k

= kS(f_m(D) − f (D))S⁻¹k

≤ kSkkS⁻¹kkf_m(D) − f (D)k.

Since lim

m→∞fm(D) = f (D) by the previous exercise, we find

m→∞lim f_m(A) = C = Sf (D)S⁻¹.



1A matrix A ∈ Mn(R) is said to be diagonalizable if there exists an invertible matrix S and a diagonal matrix D in Mn(R) such that S⁻¹AS = D.

Remark. If A is diagonalizable with A = SDS⁻¹, then exp(A) = S⁻¹exp(D)S, and cos A = S⁻¹cos(D)S, and sin A = S⁻¹sin(D)S.

(d) Let λ ∈ (−R, R) and denote J₃(λ) =





λ 1 0 0 λ 1 0 0 λ



.

Compute f (J3(λ)) in terms of f and λ and compute exp(tJ3(λ)) for all t, λ ∈ R.

In general, compute f (J_n(λ)) and compute exp(tJ_n(λ)), where²

J_n(λ) =















λ 1 0 · · · 0 0 λ 1 · · · 0 ... ... ... . .. ...

0 0 0 λ 1

0 0 0 0 λ















n×n

Solution:

f (J₃(λ)) =











f (λ) f⁰(λ) 1!

f⁰⁰(λ) 2!

0 λ f⁰(λ)

0 0 f (λ)1!









 and

f (J_n(λ)) =

























f (λ) f⁰(λ) 1!

f⁰⁰(λ)

2! · · · f⁽ⁿ⁻¹⁾(λ) (n − 1)!

0 f (λ) f⁰(λ)

1! · · · 0

... ... ... . .. ...

0 0 0 f (λ) f⁰(λ)

0 0 0 0 f (λ)1!

























n×n

(4) Let A, B : (a, b) → Mn(R) be matrix valued differentiable functions³. . (a) Prove that

d

dtTr(A(t)) = Tr(A⁰(t)) for any a < t < b.

Proof. Let us prove that Tr : M_n(R) → R is continuous. For any A = [aij], we have

Tr A =

n

X

i=1

a_ii.

2J_nis called a Jordan matrix.

3Let (V, k · k) be a normed space and f : (a, b) → V be a function. (Such a function is called a V -valued function).

Let t0∈ (a, b), we say that f is differentiable at t₀ if

t→tlim0

1 t − t0

(f (t) − f (t0)) exists.

In this case, we denote the limit by f⁰(t0). If f is differentiable at every point of (a, b), we say that f is differentiable.

We can inductively define f^(k)(t) for any k ≥ 1. (We can also define the right derivative and the left derivative of f.)

Using kAk∞≤ kAk_op,

| Tr(A)| ≤

n

X

i=1

|a_ii| ≤

n

X

i=1

kAk_∞= nkAk∞≤ nkAk_op.

By linearity of T, | Tr A − Tr B| = | Tr(A − B)| ≤ nkA − Bk_op. This shows that Tr is a Lipschitz continuous function on M_n(R). Therefore it is continuous.

Choose δ > 0 such that (t − δ, t + δ) ⊂ (a, b). Let h be a real number such that 0 < |h| < δ. By linearity of Tr,

Tr(A(t + h)) − Tr(A(t))

h = TrA(t + h) − A(t)

h .

By continuity of Tr and the differentiability of A(t), d

dtTr(A(t)) = lim

h→∞

Tr(A(t + h)) − Tr(A(t)) h

= Tr



h→∞lim

A(t + h) − A(t) h



= Tr A⁰(t).

 (b) Prove that

(A(t)B(t))⁰ = A⁰(t)B(t) + A(t)B⁰(t) for any a < t < b.

Proof. Let t, h, δ be as above. Then A(t + h)B(t + h) − A(t)B(t)

h = A(t + h) − A(t)

h B(t + h)

+ A(t)B(t + h) − B(t)

h .

Now we need the following lemmas.

Lemma 1.1. Let A : (a − δ, a + δ) → M_n(R). Suppose lim

t→aA(t) = A(a). Then A(t) is bounded in a neighborhood of a.

Proof. Let  = 1. We can find δ⁰ such that kA(t) − A(a)k < 1 for |t − a| < δ⁰. By triangle inequality, kA(t)k ≤ 1 + kA(a)k for |t − a| < δ⁰. Take M = 1 + kA(a)k.

We find kA(t)k ≤ M for any t with |t − a| < δ⁰.  Lemma 1.2. Let A : (a − δ, a + δ) → M_n(R) and B : (a − δ, a + δ) → Mn(R) be functions. Suppose that

limt→aA(t) = A(a), lim

t→aB(t) = B(a).

Then lim

t→aA(t)B(t) = A(a)B(a).

Proof. By norm inequality and the triangle inequality, we have

kA(t)B(t) − A(a)B(a)k = kA(t)B(t) − A(a)B(t)k + kA(a)B(t) − A(a)B(a)k

≤ kA(t) − A(a)kkB(t)k + kA(a)kkB(t) − B(a)k.

By Lemma 1.1, we can choose M > 0 such that kB(t)k ≤ M and kA(t)k ≤ M for any |t − a| < δ⁰ for some δ⁰ > 0. This implies that

kA(t)B(t) − A(a)B(a)k ≤ M (kA(t) − A(a)k + kB(t) − B(a)k).

Using the standard  − δ argument, we find lim

t→aA(t)B(t) = A(a)B(a). 

 (c) Show that A⁰(t) = 0 for t ∈ (a, b) if and only if there exists A ∈ Mn(R) such

that A(t) = A for any a < t < b.

Proof. Let us write A(t) = [aij(t)] where aij : (a, b) → R are functions.

Lemma 1.3. Let A = [a_ij]. Then lim

t→aA(t) = A if and only if lim

t→aa_ij(t) = a_ij. Proof. This can be proved by the inequality:

kAk_∞≤ kAk ≤ nkAk_∞. This gives

max{|aij(t) − aij| : 1 ≤ i, j ≤ n} ≤ kA(t) − Ak ≤ n · max{|a_ij(t) − aij| : 1 ≤ i, j ≤ n}.

By the standard  − δ argument, we prove our result.

 Lemma 1.4. A : (a, b) → Mn(R) is differentiable if and only if aij are all differentiable. In this case,

A⁰(t) = [a⁰_ij(t)].

Proof. By definition,

A(t + h) − A(t)

h = a_ij(t + h) − a_ij(t) h

 . By lemma 1.3,

A⁰(t) = lim

h→0

A(t + h) − A(t)

h =



h→0lim

aij(t + h) − aij(t) h



= [a⁰_ij(t)].

 Let us go back to our main problem. By Lemma 1.4, A⁰(t) = 0 if and only if a⁰_ij(t) = 0. Assume that A⁰(t) = 0. By mean value theorem, a_ij(t) = a_ij for some aij for all t ∈ (a, b). Let A = [aij]. Then A(t) = [aij(t)] = [aij] = A. When A(t) = A, A⁰(t) = 0 is obvious.

 (d) Suppose that A(t) ∈ GL_n(R) for all a < t < b. Prove that

(1.1) d

dt(A(t))⁻¹ = −A(t)⁻¹A⁰(t)A(t)⁻¹ for any a < t < b.

Proof. By definition, A(t)A(t)⁻¹= I. Using the previous results, A⁰(t)A(t)⁻¹+ A(t)d

dtA(t)⁻¹= 0.

This implies that

A(t)d

dtA(t)⁻¹ = −A⁰(t)A(t)⁻¹.

Multiplying the both side of the equation by A(t)⁻¹ to their left, we obtain that d

dtA(t)⁻¹ = −A(t)⁻¹A⁰(t)A(t)⁻¹.

 (e) Let t0 ∈ (a, b). Let C : [a, b] → M_n(R) be a continuous function. Define F :

[a, b] → M_n(R) by

F (t) = Z t

a

C(s)ds, t ∈ [a, b].

Prove that F is differentiable with F⁰ = C.

Lemma 1.5. Let A : [a, b] → Mn(R) be a continuous function. Write A(t) = [a_ij(t)]. Then

Z b a

A(t)dt =

Z b a

aij(t)dt

 .

Proof. Let P = {t_k : 0 ≤ k ≤ m} be a partition of [a, b] and C = {ξ_k : ξ_k ∈ [tk−1, tk]} be a mark of P. Then the Riemann sum of A with respect to (P, C) is given by

R(A, P, C) =

m

X

k=1

(t_k− t_k−1)A(ξ_i)

=

" _m X

k=1

(t_k− t_k−1)aij(ξi)

#

= [R(a_ij, C, P )] .

Let L = [Lij] be a n × n real matrix. By inequality, kAk∞≤ kAk ≤ nkAk∞, we have

max{|R(aij, C, P ) − Lij|} ≤ kR(A, P, C) − Lk ≤ n · max{|R(a_ij, C, P ) − Lij|}.

This shows that

kP k→0lim R(A, C, P ) = L if and only if lim

kP k→0R(aij, C, P ) = Lij

for any 1 ≤ i, j ≤ n. This proves the result. 

If F (t) =Rt

aC(s)ds, then F (t) =h Rt

ac_ij(s)dsi

. Since c_ij are continuous, by the fundamental theorem of calculus, t 7→Rt

ac_ij(s)ds is differentiable with d

dt Z t

a

c_ij(s)ds = c_ij(t).

By Lemma 1.4, F is differentiable and F⁰(t) = [c_ij(t)] = C(t).

(f) Let A(t) = e^tcos t e^tsin t

−e^tsin t e^tcos t



, for t ∈ R. Find A⁰(t), Z t

0

A(s)ds and verify the equation (1.1) using A(t).

(5) Let A =

 2 1 1 2



∈ M₂(R).

(a) Let χA(λ) = det(λI2− A). Find the roots of χ_A(λ).

(b) Let λ1 and λ2 be the roots of χA(λ). Find unit vectors u1 and u2 such that Au_i = λ_iu_i for i = 1, 2 and a matrix S whose i-th column vector is u_i with det S > 0. More precisely, if ui= (xi, yi), then

S =

 x1 x2

y₁ y₂

 .

(c) Prove that AS = SD where D = diag(λ1, λ2) and that A is diagonalizable.

(d) Use the result obtained in exercise (1.3) to compute exp A, cos A and sin A.

(e) Solve for the matrix differential equation

X⁰(t) = AX(t), X(0) = I₂. (f) Let Y (t) = cos(tA) and Z(t) = sin(tA)

A for t ≥ 0. Verify that Y (t) and Z(t) are both solutions to the matrix differential equation

Φ⁰⁰(t) + A²Φ(t) = 0.

Here we use

sin θ

θ =

∞

X

k=0

(−1)^k θ^2k

(2k + 1)! for any θ.