MA 554: Homework 1

MA 554: Homework 1

Yingwei Wang ∗

Department of Mathematics, Purdue University, West Lafayette, IN, USA

1 The induction of Japanese empire succession

Theorem 1.1. Let S =∐

∞

i=1 Si, where∐

means disjoint union. Suppose the cardinality¯Si < ∞, ∀i. Let

fi : Si → Si−1, ∀i = 2, 3, · · · .

If ¯Si > 0, ∀i, then ∃ai ∈ Si such that fi(ai) = ai−1.

Proof. 1. Step I.

Consider the image of S2 in S1 under the map f2:

f2 : S2 → f2(S2) ⊂ S1.

Choose a1 ∈ f2(S2) and a2 ∈ f−12 (a1) ∈ S2.

2. Step II.

Consider the image of S3 in S2 under the map f3:

f3 : S3 → f3(S3) ⊂ S2.

If a2 ∈ f3(S3), then it is fine. We can choose a3 ∈ S3 such that a3 ∈ f−13 (a2).

If not, we have to choose another a′1 ∈ f2(S2) and a′2 ∈ f−12 (a′1) ∈ S2 in Step I. Since

0 < ¯Si < ∞, ∀i, we can always find proper a1 and a2 such that a2 ∈ f3(S3).Keep doing the Step I and Step II, we can get a sequence {ai}

∞

i=1 such that

fi(ai) = ai−1,

where ai ∈ Si.

∗E-mail address : [email protected]; Tel : 765 237 7149

1

Yingwei Wang Linear Algebra

2 Notherian Ring

Definition 2.1. Let R be a ring. If every ideal can be generated by finitely many elements,

then say R is notherian.

Theorem 2.1. If R is noethrian, then every element r ∈ R can be written as

r = Πni=1qi,

where q′is are irreducible.

Proof. First, claim that in the Definition 2.1, “finitely generated”⇔ “ascending chaincondition”.

(⇐) : For any ideal I ⊂ R, let us choose a1 ∈ I such that (a1) ⊂ I; If (a1) 6= I, thenchoose a2 ∈ I \ (a1); If (a1, a2) & I, then we can choose a3 ∈ I \ (a1, a2). Keep dong this,we can get an ascending chain:

(a1) ⊂ (a1, a2) ⊂ (a1, a2, a3) ⊂ · · · ⊂ I.

This chain should satisfy the “ascending chain condition”. Thus, there exists j such that

(a1, a2, · · · , aj) = (a1, a2, · · · , aj, aj+1) = · · · = I.

It follows that I is finitely generated.(⇒) : Consider any ascending chain

I1 & I2 & · · · Ii & Ii+1 · · ·

Suppose

I1 = (a1, · · · , ak1),

I2 = (a1, · · · , ak1, · · · , ak2),

· · ·

Ii = (a1, · · · , ak1, · · · , ak2, · · · , aki),

· · · .

Since for ∀i, Ii is finitely generated, there exists j such that

Ij = Ij+1 = · · · .

It follows that R satisfy the “ascending chain condition”.Second, let us prove this theorem.Let r ∈ R, a 6= 0. If r is irreducible, then we are done.If r is reducible, then r = r1r2, where r1 and r2 are non-unit. If both r1, r2 are

irreducible, then we are done. If not, let us suppose r2 is reducible, then we have r2 = r3r4.

2


Keep doing this, by Theorem 1.1, we can get a sequence {r2k}∞

k=1, such that

· · · r2k | r2(k−1) | · · · r4 | r2 | r.

Now we have an ascending chain

(r) ⊂ (r2) ⊂ (r4) · · · .

Since R is notherian, then there exists j such that

(r2j) = (r2(j+1)) = · · · .

It implies that r2j is irreducible.Besides, if r1 is reducible at the beginning, then we can also get an ascending chain,

and get the same conclusion.It follows that finally, r can be written as

r = Πni=1qi,

where q′is are irreducible.

3

MA 554: Homework 2

Yingwei Wang ∗


1 The Chinese Remainder Theorem

Lemma 1.1. Suppose q1, q2, · · · , qn be pairwise coprime positive integers, i.e.

(qi, qj) = 1, ∀i 6= j, i, j = 1, 2, · · · , n.

Let

di =∏

j 6=i

qj.

Then

(d1, d2, · · · , dn) = 1. (1.1)

Proof. First, claim that

(di, qi) = 1, ∀i = 1, 2, · · · , n. (1.2)

Without loss of generality, assume there exists a prime p1 such that

(d1, q1) = p1 6= 1.

Then on one hand,

p1 | d1 = q2q3 · · · qn,

⇒ p1 | qk, for some k = 2, 3, · · ·n. (1.3)


1


On the other hand,p1 | q1. (1.4)

By Eqs.(1.3)-(1.4), we can get

(p1, pk) = p1 6= 1,

which contradicts with the fact that p1 and pk are coprime.Now we have proved that (1.2) is true.Second, claim that (1.1) is true.Assume there exists a prime p such that

(d1, d2, · · · , dn) = p 6= 1.

Then we havep | di, ∀i = 1, 2, · · · , n. (1.5)

In addition, by (1.2), we have

p ∤ qi, ∀i = 1, 2, · · · , n. (1.6)

It is obvious that (1.5) contradicts with (1.6) by the definition of di and the factthat p is a prime hence irreducible.

2 RSA algorithm

The keys for the RSA algorithm are generated the following way:

1. Choose two distinct large prime numbers p and q;

2. Compute n = pq;

3. Compute ϕ(n) = (p− 1)(q − 1), where ϕ(n) is Euler’s totient function;

4. Choose an integer e such that 1 < e < ϕ(n) and greatest common denominatorof (e, ϕ(n)) = 1, i.e. e and ϕ(n) are coprime;

5. Determine de = 1 mod ϕ(n), i.e. d is the multiplicative inverse of e mod ϕ(n).

How to use the RSA algorithm?

2


1. Encryption:m = me, mod n. (2.1)

2. Decryption:m = (m)d, mod n. (2.2)

Let p = 11, q = 13. Take e = 7. Find d and decrypt 123.

Solution:

In this case, n = 143, ϕ(n) = 120. Find d such that

7d = 120k + 1.

It is easy to know thatd = 103, k = 6.

Now by (2.2), we need to figure out m such that

m = (123)103, mod 143.

We know that

123103 = (143− 20)103 = (−20)103, mod 143.

Furthermore,

(−20)103 =(

(−20)10)10

(−20)3,

= 5610 × 8,

= 100× 8,

= 85, mod 143.

It follows thatm = 85.

By Eq.(2.1), it is easy to check that

123 = 857, mod 143.

3

MA 554: Homework 3

Yingwei Wang ∗


1 Maximal ideal

Definition 1.1. An ideal I in a ring R is called maximal if I 6= R and the onlyideal strictly containing I is R.

Lemma 1.1. Let R be PID. If p ∈ R is irreducible (prime), then the ideal (p)is maximal.

Proof. It is obvious that (p) 6= R since p is not unit.Choose α ∈ R but α /∈ (p). Since R is PID, we can find β ∈ R such that

(β) = (p, α).

It follows that β | p. The fact that p is irreducible implies that β is unit or β = p.If β is unit, then (p, α) = (1) = R, which means the ideal (p) is maximal.If β = p, then

(p, α) = (p) ⇒ α ∈ (p),

which contradicts with the fact that α /∈ (p).In conclusion, if p ∈ R is irreducible (prime), then the ideal (p) is maximal.


1


2 Finite abelian group

Question: Suppose

Z/20Z⊕ Z/12Z ≃s∑

i

⊕Z/ciZ,

where c1 | c2 | · · · | cs.

Solution: Since

20 = 22 × 5,

12 = 22 × 3,

we can getc1 = 22 = 4, c2 = 22 × 3× 5 = 60.

It follows thatZ/20Z⊕ Z/12Z ≃ Z/4Z⊕ Z/60Z.

2

MA 554: Homework 4

Yingwei Wang ∗

Department of Mathematics, Purdue University

This homework is the application of these two theorems.

Theorem 0.1. Let R be a free module of finite rank over D,

R ≃

n∑

⊕D,

where D is a PID.

Let M be a submodule of R, then

M ≃

s∑

i=1

⊕(ciD),

where c1 | c2 | · · · | cs.

Theorem 0.2. Let M be a finite generated D-module, where D is a PID. Then

M ≃n

∑

i=1

D/(ci),

where c1 | c2 | · · · | cn.


1


1 Free module of rank 2

Prove that2Z⊕ 3Z ≃ Z⊕ 6Z. (1.1)

Proof. Let us choose the basis

e1 =

(

10

)

, e2 =

(

01

)

.

Then ∀m ∈ 2Z⊕ 3Z, we have

m = 2xe1 + 3ye2 = (e1, e2)

(

2x3y

)

. (1.2)

Since (2, 3) = 1, we have2a+ 3b = 1,

where a, b ∈ Z.Now we choose the matrix

A =

(

3 −2a b

)

.

It is easy to check that det(A) = 1 and

A−1 =

(

b 2−a 3

)

.

By Eq.(1.2), we have

m = (e1, e2)

(

2x3y

)

,

= (e1, e2)A−1A

(

2x3y

)

,

= (be1 − ae2, 2e1 + 3e2)

(

6(x− y)2ax+ 3by

)

,

= (e′1, e′

2)

(

6(x− y)x+ 3b(y − x)

)

, (1.3)

2


where

e′1= be1 − ae2,

e′2= 2e1 + 3e2.

Since 6(x − y) ∈ 6Z, x + 3b(y − x) ∈ Z, we know that under the new basis{e′

1, e′

2}, the element m can be written in the form (1.3). It follows that m ∈

Z⊕ 6Z.Similarly, it also can be shown that if n ∈ Z⊕ 6Z, then n ∈ 2Z⊕ 3Z.In conclusion, we can get Eq.(1.1)

2 Free module of rank 3

Prove that2Z⊕ 3Z⊕ 10Z ≃ Z⊕ 2Z⊕ 30Z. (2.1)

Proof. By the previous conclusion Eq.(1.1), we can get

2Z⊕ 3Z⊕ 10Z ≃ Z⊕ 6Z⊕ 10Z. (2.2)

It follows that in order to prove Eq.(2.1), we just need to show that

6Z⊕ 10Z ≃ Z⊕ 30Z. (2.3)

Let us choose the basis

e1 =

(

10

)

, e2 =

(

01

)

.

Then ∀m ∈ 6Z⊕ 10Z, we have

m = 6xe1 + 10ye2 = (e1, e2)

(

6x10y

)

. (2.4)

Since (6, 10) = 2, we have

6a + 10b = 2,

⇒ 3a + 5b = 1.

3


where a, b ∈ Z.Now we choose the matrix

A =

(

5 −3a b

)

.

It is easy to check that det(A) = 1 and

A−1 =

(

b 3−a 5

)

.

By Eq.(2.4), we have

m = (e1, e2)

(

6x10y

)

,

= (e1, e2)A−1A

(

6x10y

)

,

= (be1 − ae2, 3e1 + 5e2)

(

30(x− y)6ax+ 10by

)

,

= (e′1, e′

2)

(

30(x− y)2(x+ 5b(y − x))

)

, (2.5)

where

e′1= be1 − ae2,

e′2= 3e1 + 5e2.

Since 30(x − y) ∈ 30Z, 2(x + 5b(y − x)) ∈ 2Z, we know that under the newbasis {e′

1, e′

2}, the element m can be written in the form (2.5). It follows that

m ∈ 2Z⊕ 30Z.Similarly, it also can be shown that if n ∈ 2Z⊕ 30Z, then n ∈ 6Z⊕ 10Z.In conclusion, we can get Eq.(2.3). Further, we can get Eq.(2.1).

4


3 Jordan form

Let V ≃ R[x]/(f(x))n, A ≃ x, where

f(x) = (x− a)2 + b2, b 6= 0.

Prove that

A ≃

E 0 0 · · · 0 0I E 0 · · · 0 00 I E · · · 0 0

· · · · · ·0 0 0 · · · E 00 0 0 · · · I E

2n×2n

, (3.1)

where

E =

(

a −bb a

)

, I =

(

1 00 1

)

.

Proof. If n = 1, thenV ≃ R[x]/((x− a)2 + b2).

Let us choose the basise1 = b, e2 = (x− a).

It follows thatx(e1, e2) = (e1, e2)A1,

where the matrix A1 is(

a −bb a

)

,

If n = 2, thenV ≃ R[x]/((x− a)2 + b2)2.

Let us choose the basis

e1 = 2b(x−a), e2 = (x−a)2−b2, e3 = b(x−a)2+b3, e4 = (x−a)3+b2(x−a).

It follows thatx(e1, e2, e3, e4) = (e1, e2, e3, e4)A2,

5


where the matrix A2 is(

E 0I E

)

.

By induction, for n ∈ N, A should be in the form (3.1).

4 Minimal polynomial

Let matrix A be in this form

A ≃

a 0 0 0 0 01 a 0 0 0 00 0 a 0 0 00 0 0 b 0 00 0 0 1 b 00 0 0 0 0 b

. (4.1)

Find the minimal polynomial of A.

Solution: Let A ≃ x, then

V ≃

(

k[x]

(x− a)2

)

⊕

(

k[x]

(x− a)

)

⊕

(

k[x]

(x− b)2

)

⊕

(

k[x]

(x− b)

)

,

≃

(

k[x]

(x− a)(x− b)

)

⊕

(

k[x]

(x− a)2(x− b)2

)

.

It follows that the minimal polynomial of A is

p(x) = (x− a)2(x− b)2.

6

MA 554: Homework 5

Yingwei Wang ∗


1 Reflection

Definition 1.1. A matrix A is said to be a reflection matrix iff

AAT = ATA = I, (1.1)

det(A) = −1. (1.2)

Theorem 1.1. Let A be a 3 × 3 reflection matrix. Then −1 is one of the

eigenvalues.

Proof. Consider the characteristic polynomial

χA(λ) = det(A− λI),

= −λ3 + a2λ2 + a1λ+ det(A),

= −λ3 + a2λ2 + a1λ− 1. (1.3)

On one hand, it is easy to know that χA(0) = −1 and χA(−∞) = ∞, whichimplies that χA(λ) = 0 must have at least one root in (−∞, 0).

On the other hand, due to the property Eq.(1.1), all of the eigenvalues of Ashould satisfy

|λ| = 1 ⇒ λ = ±1.

Now it is clear that λ = −1 must be one of the eigenvalues of A.


1


2 Characteristic polynomial

Let a, b and c be elements of a field F , and let A be the following 3 × 3 matrixover F :

A =

0 0 c

1 0 b

0 1 a

. (2.1)

Show that the characteristic polynomial for A is

x3 − ax2 − bx− c, (2.2)

and that this is also the minimal polynomial for A.

Proof. It is easy to know that

det(λI −A) =

λ 0 −c

−1 λ −b

0 −1 λ− a

= (λ− a)λ2 + b(−λ)− c,

= λ3 − aλ2 − bλ− c.

It follows that the characteristic polynomial for A is Eq.(2.2).Besides, it is easy to know that

A2 =

0 c ac

0 b c+ ab

1 a b+ a2

. (2.3)

Suppose there exist a polynomial of degree 2

p(x) = dx2 + ex+ f (2.4)

such that p(A) = 0.Assume the 3× 3 matrix B = (bij) = p(A), then

b11 = 0 ⇒ f = 0,

b31 = 0 ⇒ d+ f = 0 ⇒ d = 0,

b21 = 0 ⇒ e+ f = 0 ⇒ e = 0.

2


It follows that it is impossible to find a polynomial of degree 2 to be the annihi-lating polynomial of matrix A.

Similarly, it is obvious that any polynomial of degree 1 can not be the anni-hilate polynomial of matrix A.

In conclusion, (2.2) is the minimal polynomial for A.

3 Characteristic polynomial

Let A be the following 4× 4 real matrix:

A =

1 1 0 0−1 −1 0 0−2 −2 2 11 1 −1 0

. (3.1)

Show that the characteristic polynomial for A is

x2(x− 1)2, (3.2)

and that this is also the minimal polynomial for A.

Proof. It is easy to know that

det(λI − A) =

λ− 1 −1 0 01 λ+ 1 0 02 2 λ− 2 −1−1 −1 1 λ

= λ2(λ− 2)2.

It follows that the characteristic polynomial for A is Eq.(3.2).Furthermore, according to Cayley-Hamilton Theorem, the possible minimal

polynomials of A are (besides (3.2))

x, (3.3)

x− 1, (3.4)

x(x− 1), (3.5)

x2(x− 1), (3.6)

x(x− 1)2. (3.7)

3


It is obvious that the polynomials with degree 1, namely (3.3) and (3.4), areimpossible.

Now we need to compute A− I, A2 and (A− I)2:

A− I =

0 1 0 0−1 −2 0 0−2 −2 1 11 1 −1 −1

,

A2 =

0 0 0 00 0 0 0−3 −3 3 22 2 −2 −1

,

(A− I)2 =

−1 −2 0 02 3 0 01 1 0 00 0 0 0

.

It is easy to check that none of A(A− I), A2(A− I) and A(A− I)2 is zero.In conclusion, (3.2) is the minimal polynomial for A.

4 Diagonalizable matrix

In class, we have this theorem:

Theorem 4.1. A square matrix A is diagonalizable iff the minimal polynomial

of A can be written as

m(x) =

n∏

i=1

(x− ai), (4.1)

where ai are distinct to each other.

Since in last section, the minimal polynomial for A has multiple roots, thenA is not diagonalizable.

4


5 Nilpotent matrix

Theorem 5.1. Let V be an n-dim vector space and let T be a linear operator

on V . Suppose that there exists some positive integer k so that

T k = 0. (5.1)

Prove that

T n = 0. (5.2)

Proof. First, claim that the only eigenvalue of T is 0.By Eq.(5.1), we know that the polynomial

p(x) = xk

is one of the annihilating polynomials of T .Suppose m(x) is the minimal polynomial of T , then

m(x) | p(x),

⇒ m(x) = xm,

where m ≤ k.It implies that the only root of the minimal polynomial is 0. The roots of the

minimal polynomial are the eigenvalues, so the only eigenvalue of T is 0.Second, claim that the characteristic polynomial of T is

χT (x) = xn. (5.3)

The characteristic polynomial is monic of degree n with its roots being theeigenvalues, so if the only eigenvalue of T is 0 then we can get (5.3).

Finally, by Cayley-Hamilton Theorem, (5.3) is the annihilating polynomialof T , which implies Eq.(5.2).

5

MA 554: Homework 6

Yingwei Wang ∗


1 Jordan canonical form

Theorem 1.1. Let V be a finite-dimensional vector space over the field F and let

T be a linear operator on V . Then T is triangulable iff the minimal polynomial

for T is a product of linear polynomials over F .

Let

A =

0 1 02 −2 22 −3 2

. (1.1)

Is A similar over the field of real numbers to triangular matrix? If so, find sucha triangular matrix.

Solution:

It is easy to know thatdet(λI − A) = λ3. (1.2)

Suppose the minimal polynomial of A is p(x), then the possibilities for p(x)are

p(x) = x, or x2 or x3.

Since A 6= 0, A2 6= 0 but A3 = 0, we know that

p(x) = x3. (1.3)


1


It follows that A can be triangulable according to Theorem 1.1.Furthermore, we can find a nonsingular matrix V and upper triangular matrix

J ,

V =

2 0 10 2 0−2 2 0

, J =

0 1 00 0 10 0 0

,

such thatV −1AV = J.

2 Idempotent matrix

Proposition 2.1. Every matrix A such that A2 = A is similar to a diagonal

matrix.

Proof. First, claim that A can be triagulable. Since det(λI − A) = λ(λ − 1),then minimal polynomial p(x) of A should satisfy

p(x) | x(x− 1). (2.1)

By Theorem 1.1, A can be triagulable.Second, claim that A can be diagonalizable. We can find a nonsingular matrix

P such thatP−1AP = J,

where J is the Jordan canonical form of A.Then

P−1A2P = (P−1AP )(P−1AP ) = J2.

But we know that A2 = A, so

P−1A2P = P−1AP,

⇒ J2 = J,

⇒ J is a diagonal matrix.

Now we know that A is similar to a diagonal matrix.

Remark 2.1. Actually, just by Eq.(2.1) and Theorem 3.1, we can also conclude

that A is diagonalizable.

2


3 Diagonalizable matrix

Theorem 3.1. Let V be a finite-dimensional vector space over the field F and let

T be a linear operator on V . Then T is triangulable iff the minimal polynomial

for T has the form

p(x) = (x− c1) · · · (x− ck), (3.1)

where c1, · · · , ck are distinct elements of F .

Proposition 3.1. Let V be a finite-dimensional vector space over the field C.

Prove that T is diagonalizable iff T is annihilated by some polynomial over C

which has distinct roots.

Proof. On one hand, if T is diagonalizable, then by Theorem 3.1, the minimalpolynomial of T should be in the form Eq.(3.1). It is obvious that p(x) hasdistinct roots.

On the other hand, T is annihilated by a polynomial f(x) over C,

f(x) = (x− c1) · · · (x− cn),

where c1, · · · , cn ∈ C are distinct to each other. Then the minimal polynomialf(x) of T should satisfy

p(x) | f(x). (3.2)

It follows that p(x) must have the form as Eq.(3.1). Then by Theorem 3.1, T isdiagonalizable.

3


4 3 dimensional matrix

Proposition 4.1. Let A be a 3 × 3 matrix with real entries. Prove that if A is

not similar over R to a triangular matrix, then A is similar over C to a diagonal

matrix.

Proof. Let consider the characteristic polynomial of A,

χA(λ) = det(λI − A) = λ3 + a2λ2 + a1λ+ a0, (4.1)

where ai ∈ R, i = 0, 1, 2.We know that for polynomial with real coefficients, the complex roots must

appear in pairs. So the polynomial (4.1) at least has one real root, say λ0 ∈ R.Then χA(λ) can be written as

χA(λ) = (λ− λ0)(λ2 + b1λ+ b0), (4.2)

where bi ∈ R, i = 0, 1.Let us focus on this polynomial

f(λ) = λ2 + b1λ+ b0. (4.3)

1. If f(λ) has two real roots, then by Theorem 1.1, we know that A can betriagulable, which contradicts with the assumption on A.

2. If f(λ) has two complex roots, then χA(λ) has three distinct roots over C.By Proposition 3.1, A is diagonalizable over C.

4

MA 554: Homework 7

Yingwei Wang ∗


1 Powers of triangular matrices

Theorem 1.1. Let A = (aij)n×n be an n × n lower triangular matrix. Then any

power of A, denoted by Am = (a(m)ij ), m ∈ N, is also an n×n lower triangular matrix.

Besides, the diagonal elements of Am satisfies

a(m)ii = amii , ∀ 1 ≤ i ≤ n. (1.1)

Proof. Just by induction.Let m = 2. It is easy to check that

A2 =

a11 0 0 · · · 0a21 a22 0 · · · 0a31 a32 a33 · · · 0

· · · · · ·

an1 an2 an3 · · · ann.

a11 0 0 · · · 0a21 a22 0 · · · 0a31 a32 a33 · · · 0

· · · · · ·


=

a211 0 0 · · · 0∑2

j=1 a2jaj1 a222 0 · · · 0∑3

j=1 a3jaj1∑3

j=1 a3jaj2 a233 · · · 0

· · · · · ·∑n

j=1 anjaj1∑n

j=1 anjaj2∑n

j=1 anjaj3 · · · a2nn.

(1.2)

It implies that A2 is also an n×n lower triangular matrix with diagonal elements a2ii.


1


Suppose Am−1 = (a(m−1)ij )n×n with diagonal elements am−1

ii . Then for Am = (a(m)ij ),

we have

Am = Am−1A

=

am−111 0 0 · · · 0

a(m−1)21 am−1

22 0 · · · 0

a(m−1)31 a

(m−1)32 am−1

33 · · · 0· · · · · ·

a(m−1)n1 a

(m−1)n2 a

(m−1)n3 · · · am−1

nn .

a11 0 0 · · · 0a21 a22 0 · · · 0a31 a32 a33 · · · 0

· · · · · ·


=

am11 0 0 · · · 0∑2

j=1 a(m−1)2j aj1 am22 0 · · · 0

∑3j=1 a

(m−1)3j aj1

∑3j=1 a

(m−1)3j aj2 am33 · · · 0

· · · · · ·∑n

j=1 a(m−1)nj aj1

∑n

j=1 a(m−1)nj aj2

∑n

j=1 a(m−1)nj aj3 · · · amnn.

(1.3)

It implies that Am is also an n × n lower triangular matrix with diagonal elementsamii .

Now we are done.

2

MA 554: Homework 7

Yingwei Wang ∗


1 Projective resolution of polynomial module

In k[x, y], let I = (f(x, y), g(x, y)) where f and g are homogeneous and relative prime.Find a projective resolution of I.

Solution: Define the sequence as

0 −→ k[x, y]σ

−→ k[x, y]⊕ k[x, y]ε

−→ I −→ 0, (1.1)

where the definition of ε : k[x, y]⊕ k[x, y] → I = (f, g) is

ε(1, 0) = f, ε(0, 1) = g, (1.2)

and the definition of σ : k[x, y] → k[x, y]⊕ k[x, y] is

σ(1) = (g,−f). (1.3)

Now we can make the claims that

1. It is obvious that ε is surjective.

2. ℑ(σ) = ker(ε).

On one hand,

ker(ε) = {(h1, h2) ∈ k[x, y]⊕ k[x, y] : h1f − h2g = 0} ,

= {h(−g, f) ∈ k[x, y]⊕ k[x, y], ∀h ∈ k[x, y]} . (1.4)


1


One the other hand,

ℑ(σ) = {h(−g, f) ∈ k[x, y]⊕ k[x, y], ∀h ∈ k[x, y]} . (1.5)

3. It is obvious that both k[x, y] and k[x, y]⊕ k[x, y] are projective.

Now we know that (1.1) is a projective resolution of I.

2 Idempotent

Proposition 2.1. Let e be idempotent, i.e. e2 = e. Then eR is a projective module.

Proof. Just need to show that

eR ⊕ (1− e)R ≃ R. (2.1)

1. ∀r ∈ R, r = er + (1− e)r

2. eR ∩ (1− e)R = {0}.

Since ∀x ∈ eR ∩ (1− e)R, we have

ex = (1− e)x,

⇒ e2x = e(1− e)x = (e− e2)x = 0,

⇒ x = 0.

Now we are done.

3 Z-module

Proposition 3.1. The Q is not a projective Z-module.

Proof. Let us take the first step in the free resolution of Q:

f0 : M ։ Q, (3.1)

where M is a free Z-module. Besides, let

i : Q → Q. (3.2)

be the identity map. Then we have the following diagram:

2


Now we just need to show that there is no map

φ : Q → M, (3.3)

(as indicated by the dashed arrow in the following diagram) such that the diagramcommutes.

Let us prove this by contradiction. Suppose φ : Q → M were such a map. Then

φ(1) = (a1, a2, · · · , an, 0, · · · ),

for some a1, a2, · · · , an ∈ Z. Let

a = max{a1, a2, · · · , an}.

Then

(a1, a2, · · · , an, 0, · · · ) = φ(1) = φ

(

a+ 1

a+ 1

)

= (a+ 1)φ

(

1

a + 1

)

.

It indicates that a + 1 must divide each of a1, a2, · · · , an. However, since a + 1 > aifor each i = 1, 2, · · · , n, this is impossible.

Therefore, there is no such φ. Hence, Q is not a projective Z module.

3

MA 554: Homework 9

Yingwei Wang ∗


1 Baer sum

Definition 1.1. Given two extensions

0 −→ Dσ1−→ E1

µ1

−→ M −→ 0, (1.1)

0 −→ Dσ2−→ E2

µ2

−→ M −→ 0, (1.2)

(1.3)

we can construct the Baer sum as following.

Let

A = {(a1, a2) : µ1(a1) = µ2(a2)} ⊂ E1 ⊕ E2, (1.4)

B = {(d1, d2) : ∃d ∈ D such that σ1(d) = d1, σ2(−d) = d2} , (1.5)

C = σ1D ⊕ σ2D. (1.6)

Then

C/B ≃ D, A/C ≃M. (1.7)

And the extension

0 −→ C/B −→ A/B −→ A/C −→ 0 (1.8)

is called the Baer sum of the extensions E1 and E2.

Proposition 1.1. Consider the extension

0 −→ Z2−→ Z

π−→ Z/2Z −→ 0. (1.9)


1


Let E1 = E2 = Z, then the Baer sum of extensions E1 and E2 is

0 −→ Z −→ Z⊕ Z/2Z −→ Z/2Z −→ 0. (1.10)

Proof. By Definition 1.1, we can define

A = {(even, even)} ∪ {(odd, odd)} ⊂ Z⊕ Z, (1.11)

B = {(2n,−2n) : n ∈ Z} , (1.12)

C = 2Z⊕ 2Z. (1.13)

1. Claim that C/B ≃ Z.

Define the map ϕ1 : C −→ Z as

ϕ1(2n, 2m) = 2n− 2m. (1.14)

It is easy to know that ϕ1 is a surjective homomorphism and kerϕ1 = B. Itfollows that

C/B ≃ Z. (1.15)

2. Claim that A/C ≃ Z/2Z.

Define the map ϕ2 : A −→ Z/2Z as

ϕ2(n,m) = n, mod 2. (1.16)

It is easy to know that ϕ2 is a surjective homomorphism and kerϕ2 = C. Itfollows that

A/C ≃ Z/2Z. (1.17)

3. Claim that A/B ≃ Z⊕ Z/2Z.

Define the map ϕ3 : A −→ Z⊕ Z/2Z as

ϕ3(a1, a2) =

(

a1 + a22

, a1 mod 2

)

. (1.18)

It is easy to know that ϕ3 is a surjective homomorphism and kerϕ3 = B. Itfollows that

A/B ≃ Z/2Z. (1.19)

Now it is clear that the Baer sum of extensions E1 and E2 is

0 −→ Z −→ Z⊕ Z/2Z −→ Z/2Z −→ 0. (1.20)

2


2 Cohomology group

Definition 2.1. LetM and D be R-modules. For any projective resolution ofM , define

ExtnR(M,D) = ker d∗n+1/ Im dn, (2.1)

which is called the nth cohomology group.

Theorem 2.1. The groups ExtnR(M,D) depend only on M and D, i.e. they are inde-

pendent of the choice of projective resolution of M .

Proof.

· · · −→ Pn −→ · · · −→ P0 −→ M −→ 0

· · · −→ P ′

n −→ · · · −→ P ′

0 −→ M ′ −→ 0

· · · −→ Pn −→ · · · −→ P0 −→ M −→ 0.

Define the two homomorphisms on cohomology groups

ϕn : ExtnR(M′, D) −→ ExtnR(M,D), (2.2)

ψn : ExtnR(M,D) −→ ExtnR(M′, D) (2.3)

The mapsgn ◦ fn : Pn −→ Pn (2.4)

are now a lift of the identity map g◦f , and they are seen to induce the homomorphismsϕn ◦ ψn on the cohomology groups.

However, since the first and third rows are identical, taking the identity map fromPn to itself for all n is a particular lift of g ◦ f , and this choice clearly induces theidentity map on cohomology groups. It implies that ϕn ◦ ψn is also the identity onExtnR(M,D).

By a symmetric argument ψn ◦ ϕn is the identity on ExtnR(M′, D). Thus, the maps

ϕn and ψn are isomorphisms.

3

MA 554: Homework 10

Yingwei Wang ∗


1 x-adic norm

Definition 1.1 (Formal power series ring). Define the formal power series ring over

the field k as follows

k[[x]] = {∑

j≥0

ajxj , aj ∈ k}. (1.1)

Definition 1.2 (x-adic norm). Define the x-adic norm over the ring k[[x]] as follows

∀f(x) ∈ k[[x]], |f(x)| = 2−ordxf , (1.2)

where ordxf is the maximum integer such that xe divides the polynomial f(x).

Remark 1.1. In the definition (1.2), the constant 2 is irrelevant, and could be replaced

by any real number greater than 1.

Definition 1.3 (x-adic distance). Define the x-adic distance between two elements

f, g ∈ k[[x]] as follows

d(f, g) = |f − g| = 2−ordx(f−g). (1.3)

Remark 1.2. It is easy to know that

d(0, f) = |f |, ∀f(x) ∈ k[[x]]. (1.4)


1


Proposition 1.1. The disk centered at 0 with radius s in k[[x]] is

D(0, s) = {f ∈ k[[x]] : d(0, f) < s}. (1.5)

And the boundary of D(0, s) is

∂D(0, s) = {g ∈ k[[x]] : d(0, g) = s}. (1.6)

Prove that for any f ∈ D(0, s) and any g ∈ ∂D(0, s), the distance between f and g is

always the same, i.e.

d(f, g) = s, ∀f ∈ D(0, s), ∀g ∈ ∂D(0, s). (1.7)

In other word, any point in the disk D(0, s) is its center .

Proof. By the definition, we can get

∀g ∈ ∂D(0, s), d(0, g) = |g| = s = 2−ordxg,

⇒ ordxg = − log2 s,

⇒ g(x) = xsg(x), (1.8)

where s = − log2 s and (x, g(x)) = 1.Similarly, ∀f ∈ D(0, s), suppose d(0, f) = |f | = r, then r < s. Besides,

f(x) = xrf(x), (1.9)

where r = − log2 r and (x, f(x)) = 1.Since r < s, we can get r > s. By Eqs.(1.8)-(1.9), we can get

f − g = xs(xr−sf − g), where (xs, xr−sf − g) = 1,

⇒ ordx(f − g) = s,

⇒ d(f, g) = 2−ordx(f−g) = s. (1.10)

Now we are done.

Remark 1.3. Actually, this argument can be generalized to the ring k[[x1, x2, · · · , xn]].

2

MA 554: Homework 11

Yingwei Wang ∗


1 Basis for perpendicular space

Consider R4 with the standard inner product. Let W be the subspace of R4 consisting allof the vectors which are orthogonal to both α = (1, 0,−1, 1) and β = (2, 3,−1, 2). Finda basis for W .

Solution: Let x = (x1, x2, x3, x4) ∈ W , then

(

1 0 −1 12 3 −1 2

)

x1

x2

x3

x4

=

0000

. (1.1)

It is easy to know that two linearly independent solutions to the equation (1.1) isx = (3,−1, 3, 0) and y = (−1, 0, 0, 1).

Now we find a basis for W , which is {x, y}.

W = span {(3,−1, 3, 0), (−1, 0, 0, 1)}.

2 Identity

Proposition 2.1. Let V be an inner product space, and let α, β be vectors in V . Show

that α = β if and only if 〈α|γ〉 = 〈β|γ〉, for every γ ∈ V .


1


Proof. 1. ⇒): Suppose α = β, then

α = β,

⇒ α− β = 0,

⇒ 〈α− β|γ〉 = 0, ∀γ ∈ V,

⇒ 〈α|γ〉 − 〈β|γ〉 = 0, ∀γ ∈ V,

⇒ 〈α|γ〉 = 〈β|γ〉, ∀γ ∈ V.

2. Suppose 〈α|γ〉 = 〈β|γ〉, for every γ ∈ V , then

〈α|γ〉 = 〈β|γ〉, ∀γ ∈ V,

⇒ 〈α|γ〉 − 〈β|γ〉 = 0, ∀γ ∈ V,

⇒ 〈α− β|γ〉 = 0, ∀γ ∈ V,

⇒ α− β = 0,

⇒ α = β.

3 New inner product in R2

Let V be the inner product space consisting of R2 and the inner product whose quadraticform is defined by

‖(x1, x2)‖2 = (x1 − x2)2 + 3x2

2. (3.1)

The inner product consisting of this norm is

〈x|y〉 = (x1 − x2)(y1 − y2) + 3x2y2, (3.2)

where x = (x1, x2), y = (y1, y2) ∈ V .Let E be the orthogonal projection of V onto the subspace W spanned by the vector

(3, 4). Now do the following things.

1. Find a formula for E(x1, x2).

By Eq.(3.1), it is easy to know that ‖(3, 4)‖ = 7. Let e1 =(3,4)

‖(3,4)‖ =(

37, 47

)

, which isa normalized basis for W .

2


Now let us do the projection.

E(x1, x2) = 〈x|e1〉e1,

=

(

(x1 − x2)

(

3

7− 4

7

)

+ 3x24

7

)(

3

7,4

7

)

,

=

(

−1

7x1 +

13

7x2

)(

3

7,4

7

)

,

=

(

− 3

49x1 +

39

49x2,−

4

49x1 +

52

49x2

)

. (3.3)

2. Find the matrix of E in the standard ordered basis.

By Eq.(3.3), we know that

[

− 349

3949

− 449

5249

](

x1

x2

)

=

(

− 349x1 +

3949x2

− 449x1 +

5249x2

)

. (3.4)

It follows that the matrix form of E is[

− 349

3949

− 449

5249

]

. (3.5)

3. Find W⊥.

We just need to find e2 ∈ V such that 〈e1|e2〉 = 0 and ‖e2‖ = 1.

Let e2 = (z1, z2), then 〈e1|e2〉 = 0 implies that −(z1 − z2)17+ 3z2

47= 0, which leads

to z1 = 13z2. After normalization, we can get e2 =1√147

(13, 1).

Now we know that W⊥ = span {e2}, where e2 =1√147

(13, 1).

4. Find an orthonormal basis in which E is represented by the matrix

[

1 00 0

]

. (3.6)

Just choose the set {e1, e2} as an orthonormal basis of V . Then E is equivalent tothe matrix (3.6).

3


4 Polynomial space

Let V be the subspace of R[x] of polynomials of degree at most 3. Equip V with theinner product

〈f |g〉 =∫ 1

0

f(t)g(t)dt. (4.1)

1. Find the orthogonal complement of the subspace of scalar polynomial.

Let W = span {1} be the scalar polynomial space and p(x) ∈ W⊥, where p(x) =a3x

3 + a2x2 + a1x+ a0. Then

〈p|1〉 =∫ 1

0

p(x)dx = 0,

⇒ 1

4a3 +

1

3a2 +

1

2a1 + a0 = 0,

⇒ a0 = −1

4a3 −

1

3a2 −

1

2a1.

It follows that the basis for W⊥ can be chosen as

p1(x) = x− 1

2, p2(x) = x2 − 1

3, p3(x) = x3 − 1

4.

That is to say, W⊥ = {x− 12, x2 − 1

3, x3 − 1

4}.

2. Apply the Gram-Schmidt process to the basis {1, x, x2, x3}.

(a) Let v1 = 1, then w1 =v1

‖v1‖ = 1.

(b) Let v2 = x, then

v2 − 〈v2|w1〉w1 = x− 1

2,

⇒ w2 =x− 1

2

‖x− 12‖ = 2

√3

(

x− 1

2

)

.

(c) Let v3 = x2, then

v3 − 〈v3|w1〉w1 − 〈v3|w2〉w2 = x2 − x+1

6,

⇒ w3 =x2 − x+ 1

6

‖x2 − x+ 16‖ = 6

√5

(

x2 − x+1

6

)

.

4


(d) Let v4 = x3, then

v4 − 〈v4|w1〉w1 − 〈v4|w2〉w2 − 〈v4|w3〉w3 = x3 − 3

2x2 +

3

5x− 1

20,

⇒ w4 =x3 − 3

2x2 + 3

5x− 1

20

‖x3 − 32x2 + 3

5x− 1

20‖ = 20

√7

(

x3 − 3

2x2 +

3

5x− 1

20

)

.

In conclusion, we get an orthonormal basis, V = span {w1, w2, w3, w4}.

5 Perpendicular space

Proposition 5.1. Let S be a subset of an inner product space V . Show that (S⊥)⊥

contains the subspace spanned by S. When V is finite-dimensional, show that (S⊥)⊥ is

the subspace spanned by S.

Proof. Let S = span{S}.

1. Claim that (S⊥)⊥ ⊃ S.

For ∀α ∈ S, ∀β ∈ S⊥,

β ∈ S⊥,

⇒ 〈β|s〉 = 0, ∀s ∈ S,

⇒ 〈β|α〉 = 0, ∀α ∈ S,

⇒ α ∈ (S⊥)⊥.

It follows that (S⊥)⊥ ⊃ S.

2. Claim that if V is finite-dimensional, then (S⊥)⊥ = S.

Suppose dim(V ) = n and dim(S) = k, where k ≤ n. By theorem 5 in the textbook,we know that

dim(S⊥) = n− k,

⇒ dim((S⊥)⊥) = n− (n− k) = k.

It follows that dim((S⊥)⊥) = dim(S). Besides, since we know that (S⊥)⊥ ⊃ S, thenwe can conclude that (S⊥)⊥ = S.

5


6 Odd and even functions

Let V be the real value inner product space consisting of the space of real-valued contin-uous functions on the interval −1 ≤ t ≤ 1, with the inner product

〈f |g〉 =∫ 1

−1

f(t)g(t)dt. (6.1)

Let W be the subspace of odd functions, i.e. functions satisfying f(−t) = f(t). Find theorthogonal complement of W .

Solution:

Choose ∀g ∈ W⊥ and define the two functions

g1(t) =g(t) + g(−t)

2,

g2(t) =g(t)− g(−t)

2.

It is easy to know that g1(t) is an even function and g2(t) is an odd function. Besides,g(t) can be written as the sum of two parts, i.e.

g(t) = g1(t) + g2(t). (6.2)

Let f ∈ W , then

〈f |g〉 = 0,

⇒∫ 1

−1

f(t)g1(t)dt+

∫ 1

−1

f(t)g2(t)dt = 0,

⇒∫ 1

−1

f(t)g2(t)dt = 0, (since f(t)g1(t) is odd),

⇒ f(t)g2(t) = 0, (since f(t)g2(t) is even),

⇒ g2(t) = 0, (since f(t) is arbitrary).

Now it is clear that ∀g ∈ W⊥, g must be an even function. In other word, W⊥ is thesubspace of odd functions.

6

MA 554: Homework 12

Yingwei Wang ∗


1 Adjoint operator

Proposition 1.1. Let V be an inner product space and β, γ fixed vectors in V . Show that

Tα = 〈α|β〉γ defines a linear operator on V . Show that T has an adjoint, and describe

T ∗ explicitly.

Now suppose V is Cn with the standard inner product, β = (y1, · · · , yn) be row vectors,

and γ = (z1, · · · , zn). What is the j, k entry of the matrix of T in the standard ordered

basis? What is the rank of the matrix?

Proof. On one hand, by the definition of adjoint operator, we can get

〈Tα|w〉 = 〈α|T ∗w〉, (1.1)

for any α,w ∈ V .On the other hand, by the definition of T , we can get

〈Tα|w〉 = 〈α|β〉〈γ|w〉,

= 〈α|(

〈γ|w〉)

β〉,

= 〈α| (〈w|γ〉) β〉. (1.2)

By Eqs.(1.1)-(1.2), we can get

T ∗w = 〈w|γ〉β, ∀w ∈ V. (1.3)

Furthermore, let {e1, e2, · · · , en} be the standard ordered basis. Then

Tej = 〈ej|β〉γ = yjγ,

⇒ the matrix of T = βTγ.

Besides, the rank of this matrix is 1.


1


2 Self-adjoint operators

Proposition 2.1. Show that the product of two self-adjoint operators is self-adjoint if

and only if the two operators commute.

Proof. Suppose T1 and T2 are two self-adjoint operators, i.e.

〈T1α|β〉 = 〈α|T1β〉, (2.1)

〈T2α|β〉 = 〈α|T2β〉, (2.2)

(2.3)

for any α, β ∈ V .

1. Suppose T1T2 is also a self-adjoint operator, then

〈T1T2α|β〉 = 〈α|T1T2β〉,

⇒ 〈T2α|T1β〉 = 〈α|T1T2β〉, (since T1 is self-adjoint)

⇒ 〈α|T2T1β〉 = 〈α|T1T2β〉, (since T2 is self-adjoint)

⇒ T1T2 = T2T1, (since α, β are arbitrary).

It follows that T1 and T2 commute.

2. Suppose T1T2 = T2T1, then

〈α|T1T2β〉 = 〈α|T2T1β〉, α, β ∈ V,

⇒ 〈T1α|T2β〉 = 〈α|T2T1β〉, α, β ∈ V,

⇒ 〈T2T1α|β〉 = 〈α|T2T1β〉, α, β ∈ V,

which indicates that T2T1 is also a self-adjoint operator and so is T1T2.

2


3 Differentiation operator

Let V be the vector space of the polynomial over R of degree less than or equal to 3, withthe inner product

〈f |g〉 =

∫

1

0

f(t)g(t)dt. (3.1)

Let D be the differentiation operator on V . Find the adjoint D∗.

Solution:

First, choose a basis for V ,

V = span {e0 = 1, e1 = x, e2 = x2, e3 = x3}. (3.2)

By the definition of adjoint operator, we can get

〈Dei|ej〉 = 〈ei|D∗ej〉. (3.3)

Let D∗ej =∑

3

k=0akjek, then

〈Dei|ej〉 =3

∑

k=0

akj〈ei|ek〉. (3.4)

Let C = (〈Dei|ej〉), A = (akj), B = (〈ei|ek〉), then the system of equations (3.4) can bewritten as the matrix form

C = AB, (3.5)

where A is the matrix form of operator D∗, and

C =

0 1 1 3

2

0 1

2

2

31

0 1

3

1

2

3

4

0 1

4

2

5

3

5

, B =

1 1

2

1

3

1

41

2

1

3

1

4

1

51

3

1

4

1

5

1

61

4

1

5

1

6

1

7

.

Now it is easy to know that A = CB−1, i.e.

A =

−90 1020 −2520 1680−40 480 −1230 840−25 310 −810 560−18 228 −603 420

.

3


4 Self-adjoint unitary operator: a particular case

Let V be a finite-dimensional inner product space, and let W be a subspace of V . ThenV = W ⊕W⊥, that is, each α in V is uniquely expressible in the form α = β + γ, withβ ∈ W and γ ∈ W⊥.

4.1 Self-adjoint unitary operator

Proposition 4.1. Define a linear operator U by Uα = β−γ. Then U is both self-adjoint

and unitary.

Proof. Choose ∀x, y ∈ V , then ∃x1, y1 ∈ W and x2, y2 ∈ W⊥ such that

x = x1 + x2,

y = y1 + y2.

Note that here

〈x1|x2〉 = 0, 〈x1|y2〉 = 0,

〈y1|x2〉 = 0, 〈y1|y2〉 = 0.

1. U is self-adjoint.

On one hand,

〈U(x)|y〉 = 〈x1 − x2|y1 + y2〉,

= 〈x1|y1〉 − 〈x2|y2〉. (4.1)

On the other hand,

〈x|U(y)〉 = 〈x1 + x2|y1 − y2〉,

= 〈x1|y1〉 − 〈x2|y2〉. (4.2)

By Eqs.(4.1)-(4.2), we know that

〈U(x)|y〉 = 〈x|U(y)〉, ∀x, y ∈ V,

which means U is self-adjoint.

4


2. U is unitary. On one hand,

〈x|y〉 = 〈x1 + x2|y1 + y2〉,

= 〈x1|y1〉+ 〈x2|y2〉. (4.3)

On the other hand,

〈U(x)|U(y)〉 = 〈x1 − x2|y1 − y2〉,

= 〈x1|y1〉+ 〈x2|y2〉. (4.4)

By Eqs.(4.3)-(4.4), we know that

〈x|y〉 = 〈U(x)|U(y)〉, ∀x, y ∈ V,

which means U is unitary.

4.2 R3

If V = R3 with the standard inner product and W is the subspace spanned by (1, 0, 1),

find the matrix of U in the standard ordered basis.

Solution: It is easy to know that

W = span {(1, 0, 1)}, (4.5)

W⊥ = span {(0, 1, 0), (−1, 0, 1)}. (4.6)

Let {e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1)} be the orthonormal basis of V . Thenfor each ej, we can do the projection onto W and W⊥:

e1 =

(

1

2, 0,

1

2

)

+

(

1

2, 0,−

1

2

)

, (4.7)

e2 = 0 + (0, 1, 0), (4.8)

e3 =

(

1

2, 0,

1

2

)

+

(

−1

2, 0,

1

2

)

. (4.9)

By the definition of U , we can get

U(e1) =

(

1

2, 0,

1

2

)

−

(

1

2, 0,−

1

2

)

= (0, 0, 1), (4.10)

U(e2) = 0− (0, 1, 0) = (0,−1, 0), (4.11)

U(e3) =

(

1

2, 0,

1

2

)

−

(

−1

2, 0,

1

2

)

= (1, 0, 0). (4.12)

5


It follows that the matrix form of U is

U =

0 0 10 −1 01 0 0

. (4.13)

It is easy to check that U = UT and UUT = UTU = I.

5 Self-adjoint unitary operator: generous case

Proposition 5.1. Let V be a finite dimensional inner product space. In previous section,

we showed how to construct some linear operators on V which are both self-adjoint and

unitary. Now prove that there are no others, i.e. that every self-adjoint and unitary

operator arises from some subspace W as we described in Proposition 4.1.

Proof. Let the dimension of V is n.Suppose U is both self-adjoint and unitary operator. Consider the eigenvalues of U .

That U is self-adjoint means that all of the eigenvalues of U are real. Besides, that U isunitary means that each of the eigenvalue of U is 1 or −1.

Suppose the eigenvectors of U associated with eigenvalue 1 are

α1, α2, · · · , αk. (5.1)

Besides, suppose the eigenvectors of U associated with eigenvalue −1 are

αk+1, αk+2, · · · , αn. (5.2)

Note that 〈αi|αj〉 = 0, ∀i, j = 1, · · · , n. Furthermore, after normalization, the set{α1, α2, · · · , αn} forms an orthonormal basis of V .

For a fixed self-adjoint unitary operator U , we can choose the space W and W⊥ asfollows

W = {α1, α2, · · · , αk}, (5.3)

W⊥ = {αk+1, αk+2, · · · , αn}. (5.4)

Then for any x ∈ V , we have

x = (x1α1 + x2α2 + · · ·+ xkαk) + (xk+1αk+1 + xk+2αk+2 + · · ·+ xnαn),

= β + γ, (5.5)

U(x) = (x1U(α1) + x2U(α2) + · · ·+ xkU(αk)) + (xk+1U(αk+1) + xk+2U(αk+2) + · · ·+ xnU(αn)),

= (x1α1 + x2α2 + · · ·+ xkαk)− (xk+1αk+1 + xk+2αk+2 + · · ·+ xnαn),

= β − γ, (5.6)

6


where β ∈ W and γ ∈ W⊥.In conclusion, every self-adjoint and unitary operator arises from some subspace W

as we described in Proposition 4.1.

6 Rigid motion

If V is an inner product space, a rigid motion is any function T from V into V (notnecessary linear) such that

‖Tα− Tβ‖ = ‖α− β‖, ∀α, β ∈ V. (6.1)

One example of a rigid motion is a linear unitary operator. Another example istranslation by a fixed vector γ:

Tγ(α) = α + γ. (6.2)

6.1 Linear and unitary

Proposition 6.1. Let V be R2 with the standard inner product. Suppose T is a rigid

motion of V and that T (0) = 0. Prove that T is linear and a unitary operator.

Proof. 1. Claim that T preserve the inner product.

In Eq.(6.1), choosing β = 0 and thanks to the fact that T (0) = 0, we can get

‖Tα‖ = ‖α‖, ∀α ∈ V. (6.3)

Recall the formula

‖x− y‖2 = ‖x‖2 + ‖y‖2 − 2〈x|y〉, ∀x, y ∈ V. (6.4)

Now we can get

2〈Tx|Ty〉 = ‖Tx‖2 + ‖Ty‖2 − ‖Tx− Ty‖2,

= ‖x‖2 + ‖y‖2 − ‖x− y‖2,

= 2〈x|y〉, ∀x, y ∈ V.

7


2. T is linear.

Let us choose the standard basis in R2,

e1 = (1, 0), e2 = (0, 1).

Then, for ∀x, y ∈ V , we have

〈T (x+ y)|Tei〉 = 〈x+ y|ei〉,

= 〈x|ei〉+ 〈y|ei〉,

= 〈Tx|Tei〉+ 〈Ty|Tei〉,

= 〈Tx+ Ty|Tei〉, i = 1, 2.

It follows thatT (x+ y) = Tx+ Ty, ∀x, y ∈ V. (6.5)

Similarly, for any constant c, we have

〈T (cx)|Tei〉 = 〈cx|ei〉,

= c〈x|ei〉,

= c〈Tx|Tei〉,

which in turn yields that

T (cx) = cTx ∀ constant c and ∀x ∈ V. (6.6)

In conclusion, T is linear and a unitary operator.

6.2 Generalization

Proposition 6.2. Use the result we just proved to show that every rigid motion of R2 is

composed of a translation, followed by a unitary operator.

Proof. Let f be a rigid motion of R2, then

f(x) = f(0) + (f(x)− f(0)) . (6.7)

By Proposition 6.1, we know that g(x) = f(x)− f(0) is a unitary operator. Besides,h(x) = x+ f(0) can be view as a translation. Now by Eq.(6.7), it is clear that

f = h ◦ g, (6.8)

where h is a translation and g is a unitary operator.

8


6.3 Further generalization

Proposition 6.3. Now show that a rigid motion of R2 is either a translation followed by

a rotation, or a translation followed by a reflection, followed by a rotation.

Proof. Let V = R2 and e1 = (1, 0)T , e1 = (0, 1)T . Suppose U is a unitary operator on V ,

then

‖U(e1)‖ = 1

⇒ U(e1) =

(

cos θsin θ

)

,

and

‖U(e2)‖ = 1 & U(e2) ⊥ U(e1)

⇒ U(e2) =

(

− sin θcos θ

)

, or

(

sin θ− cos θ

)

.

Now we know that if U is a unitary operator on V , then the matrix of U in thestandard ordered basis is either

[

cos θ − sin θsin θ cos θ

]

(6.9)

or[

cos θ sin θsin θ − cos θ

]

. (6.10)

It follows that in Eq.(6.8), g is either a rotation or a reflection followed by a rotation.Furthermore, by Proposition 6.2, a rigid motion of R2 is either a translation followed

by a rotation, or a translation followed by a reflection, followed by a rotation.

9

MA 554: Homework 13

Yingwei Wang ∗


1 Diagonal matrix

For

A =

1 2 32 3 43 4 5

,

there is a real orthogonal matrix P such that P tAP = D is diagonal. Find such a diagonalmatrix D.

Solution: Since A is hermitian, the elements of D should be the eigenvalues of A.The characteristic polynomial of A is

χA(λ) = det(λI − A),

= λ3 − 9λ2 − 6λ,

= λ

(

λ− 9 +√105

2

)(

λ− 9−√105

2

)

. (1.1)

Hence, the matrix D is

D =

0 0 0

0 9+√105

20

0 0 9−√105

2

.


1


2 Normal operator

Proposition 2.1. Prove T is normal if and only if T = T1 + iT2 where T1 and T2 are

self-adjoint operators which commutes.

Proof. 1. Suppose T = T1 + iT2 is normal, where

T1 =1

2(T + T ∗),

T2 =1

2i(T − T ∗).

Then we have

TT ∗ = T ∗T,

⇒ (T1 + iT2)(Tt1 − iT t

2) = (T t1 − iT t

2)(T1 + iT2),

⇒ (T1Tt1 + T2T

t2) + i(T2T

t1 − T1T

t2) = (T t

1T1 + T t2T2) + i(T t

1T2 − T t2T1),

⇒{

T1Tt1 + T2T

t2 = T t

1T1 + T t2T2

T2Tt1 − T1T

t2 = T t

1T2 − T t2T1,

⇒ T1Tt1 = T t

1T1 = I, T2Tt2 = T t

2T2 = I, T1T2 = T2T1.

It follows that T1 and T2 are self-adjoint operators which commutes.

2. Suppose T = T1 + iT2, where T1 and T2 are self-adjoint operators and commuteswith each other. Then

T1Tt1 = T t

1T1 = I, T2Tt2 = T t

2T2 = I, T1T2 = T2T1. (2.1)

On one hand,

TT ∗ = (T1 + iT2)(Tt1 − iT t

2),

= (T1Tt1 + T2T

t2) + i(T2T

t1 − T1T

t2). (2.2)

On the other hand,

T ∗T = (T t1 − iT t

2)(T1 + iT2),

= (T t1T1 + T t

2T2) + i(T t1T2 − T t

2T1). (2.3)

By Eqs.(2.1)-(2.3), we can getTT ∗ = T ∗T. (2.4)

It follows that T is normal.

2


3 Normal operator

Proposition 3.1. Let T be a normal operator on a finite-dimensional complex inner

product space. Prove that there is a polynomial f , with complex coefficients, such that

T ∗ = f(T ).

Proof. 1. Assume T is an n× n diagonal matrix, i.e.

T =

z1. . .

zn

, T ∗ =

z1. . .

zn

.

It is obvious that TT ∗ = T ∗T .

We need find a polynomial f(z) of degree n− 1 such that T ∗ = f(T ).

We can use the so called “Lagrange interpolating polynomial”. Let us define

fk(z) =∏

j 6=k

zkzj

z − zj,

then it is easy to know that

fk(zj) = δkj, ∀k, j = 1, 2, · · ·n.

Let

f(z) =

n∑

k=1

zkfk(z). (3.1)

Then f(z) is a polynomial of degree n− 1 which satisfies

f(zk) = zk, ∀1 < k < n.

Now we know that

f(T ) =

f(z1). . .

f(zn)

=

z1. . .

zn

= T ∗.

3


2. Assume T is not a diagonal matrix. By theorem 22 on page 317 of the textbook,we can find a unitary matrix P such that

P−1TP = Λ = diag(λ1, λ2, · · · , λn),

P−1T ∗P = Λ∗ = diag(λ1, λ2, · · · , λn).

By Part 1, we know that there exists a polynomial of degree n− 1,

f(z) = an−1zn−1 + · · ·a1z + a0,

such that

Λ∗ = f(Λ) = an−1Λn−1 + · · · a1Λ + a0,

⇒ P−1T ∗P = an−1P−1Λn−1P + · · ·a1P−1ΛP + a0,

⇒ T ∗ = an−1Tn−1 + · · · a1T + a0 = f(T ).

Now we are done.

4 Quadratic form

Let f be the form on R2 given by

f((x1, x2), (y1, y2)) = x1y1 + 4x2y2 + 2x1y2 + 2x2y1. (4.1)

Find an ordered basis in which f is represented by a diagonal matrix.

Solution: Let

A =

[

1 22 4

]

,

thenf((x1, x2), (y1, y2)) = (x1, x2)A(y1, y2)

t.

The characteristic polynomial of A is

χA(λ) = det(λI −A),

= λ2 − 5λ,

= λ(λ− 5). (4.2)

4


It is easy to know that the eigenvalues and associated eigenvectors are

λ1 = 0, v1 =1√5(−2, 1)t,

λ2 = 5, v2 =1√5(1, 2)t.

Let

P =1√5

[

−2 11 2

]

, Λ =

[

0 00 5

]

,

then

P t = P−1,

A = PΛP t.

Let x = xP, y = yP , thenf(x, y) = xΛyt. (4.3)

Then f is represented by a diagonal matrix for the basis {v1, v2}.

5 Non-degenerate form

Definition 5.1. Call the form f non-degenerate if 0 is the only vector α such that

f(α, β) = 0, for all β.

Proposition 5.1. Let f be a form on an inner product space V , i.e.

f(α, β) = 〈Tf |β〉, ∀β ∈ V. (5.1)

Prove that f is non-degenerate if and only if the associate linear operator Tf is nonsingular

Proof. 1. Suppose Tf be nonsingular.

0 = f(α, β) = 〈Tf |β〉, ∀β ∈ V,

⇒ Tfα = 0,

⇒ α = 0, since Tf is nonsingular.

By Definition 5.1, f is a non-degenerate form.

5


2. Let f be a non-degenerate form.

Assume Tf is singular, then we have

∃α 6= 0, such that Tfα = 0,

⇒ 〈Tf |β〉, ∀β ∈ V,

⇒ f(α, β) = 〈Tf |β〉, ∀β ∈ V.

This contradicts the fact that f is a non-degenerate form.

Hence, Tf should be nonsingular.

6 Non-degenerate form

Proposition 6.1. Let f be a non-degenerate form on a finite-dimensional space V . Let

L be a linear function on V . Show that there exists one and only one vector β ∈ V such

that

L(α) = f(α, β), ∀α ∈ V. (6.1)

Proof. Let L be a linear function on V . By Riesz representation theorem, there existsone and only one vector γ ∈ V such that

L(α) = 〈α|γ〉, α ∈ V. (6.2)

Besides, f is a non-degenerate form, so ∃Tf such that

f(α, β) = 〈Tfα|β〉 = 〈α|T ∗f β〉, α ∈ V. (6.3)

By Proposition 5.1, Tf is nonsingular. So T ∗f is also nonsingular. It follows that the

equation T ∗f β = γ has a unique solution. We just choose this β.

6

MA 554: Homework 14

Yingwei Wang ∗


1 Local minimal point

We have a function

f(x1, x2, x3) = x2

1+ 2x2

2+ 3x2

3− 2x1x2 − 2x1x2 − 4x1x3. (1.1)

Determine whether the origin (0, 0, 0) is the local minimal or not.

Solution: Here, I have two methods to show that the origin is not the local minimal.Method ILet

A =

1 −1 −1−1 2 −2−1 −2 3

.

Then the function f can be written as

f(x) = xTAx, (1.2)

where

x =

x1

x2

x3

.

On one hand,f(0) = 0.

On the other hand, since det(A) = −7, A is not positive. We can find a sequence{xn = (1/n, 1/n, 1/n)T}∞n=1

which goes to the origin as n → ∞, such that

f(xn) = −2/n2 < f(0), ∀n.∗E-mail address : [email protected]; Tel : 765 237 7149

1


It follows that the origin can not be the minimal point of f .Method IILet us use the second derivative test.The second derivative matrix of f is

D =

2 −2 −2−2 4 −4−2 −4 6

.

It is easy to know that det(D) = −56, which means D is not positive. Hence, originis not the local minimal point of f .

2 Hilbert matrix

Let

A =

[

1 1

21

2

1

3

]

.

2.1 Positive

Let us compute the principal minors of A.

det(A1) = 1,

det(A2) =1

12.

It follows that A is positive.

2.2 Gram-Schmidt process

Let V be the space of 2× 1 real matrices with the inner product

〈X|Y 〉 = Y tAX.

Find an orthogonal basis for V , by applying the Gram-Schmidt process to the basis

X1 =

(

10

)

, X2 =

(

01

)

.

Solution:

2


e1 = X1 =

(

10

)

,

v2 = X2 − 〈X2|e1〉e1,

=

(

01

)

− 1

2

(

10

)

,

=

(

−1

2

1

)

,

e2 =v2

‖v2‖,

=

(

−√3

2√3

)

.

Here, {e1, e2} is an orthonormal basis for V .

2.3 Cholesky factorization

Let

Q = [e1, e2] =

[

1 −√3

0 2√3

]

.

It is easy to know that QTAQ = I.Let

P = Q−1 =

[

1 1

2

0 1

2√3

]

.

Then we haveA = P tP.

3 Positive matrix

Theorem 3.1. Prove that every entry on the main diagonal of a positive matrix is posi-tive.

Proof. Let A = (aij)n×n be a positive matrix. Suppose ∃ a positive integer k such that

akk ≤ 0.

3


Then consider the vector ek = (0, · · · , 0, 1, 0, · · · )T , where 1 appears in the k-th place.Then it is clear that

etkAek = akk ≤ 0.

It contradicts with the fact that A is positive.Hence, akk > 0, for ∀k.

4 Non-degenerate

Definition 4.1. A bilinear form on a finite dimensional vector space V is called non-degenerate (or non-singular) if it satisfies any one of the following conditions:

1. rank(f) = n.

2. for each non-zero α ∈ V , there is a β ∈ V such that f(α, β) 6= 0.

3. for each non-zero β ∈ V , there is an α ∈ V such that f(α, β) 6= 0.

Proposition 4.1. Let q(x1, x2) = ax2

1+ bx1x2 + cx2

2be the quadratic form associated

with a symmetric bilinear form f on R2. Show that f is non-degenerate if and only if

b2 − 4ac 6= 0.

Proof. Let

A =

[

a b2

b2

a

]

, x =

[

x1

x2

]

,

thenf(x) = xTAx.

By the definition, f is non-degenerate if and only if A is non-singular.

A is non-singular,

⇔ det(A) 6= 0,

⇔ ac− b2

46= 0,

⇔ b2 − 4ac 6= 0.

4


5 Diagonalizationn

Let q be the quadratic form on R2 given by

q(x1, x2) = ax2

1+ 2bx1x2 + cx2

2, a 6= 0.

Find an invertible linear operator U on R2 such that

(U †q)(x1, x2) = ax2

1+

(

c− b2

a

)

x2

2.

Solution: It is easy to know that

q(x1, x2) = ax2

1+ 2bx1x2 + cx2

2,

= a

(

x1 +b

ax2

)2

+

(

c− b2

a

)

x2

2. (5.1)

Let

U−1 =

[

1 ba

0 1

]

, U =

[

1 − ba

0 1

]

, x =

[

x1

x2

]

, y =

[

y1y2

]

. (5.2)

Then it easy to know thaty = U−1x, x = Uy. (5.3)

Besides, let us denote

A =

[

a bb a

]

, B =

[

a 0

0 c− b2

a

]

.

Then it is easy to know thatB = UTAU. (5.4)

Thanks to Eqs.(5.1)-(5.4), we can get

f(x) = xTAx = (Uy)TA(Uy) = yT (UTAU)y = yTBy. (5.5)

It follows that

(U †q)(x1, x2) = q(Ux) = ax2

1+

(

c− b2

a

)

x2

2, (5.6)

where U is defined in (5.2).

5

MA 554: Homework 1

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