ENGG 2013 Unit 11 RowRank Feb 2011 Last

ENGG 2013 Unit 11 Row-Rank Feb, 2011.

Last week • Caesar Cipher – Modulo arithmetic: divide and take the remainder – Encryption: y = x + 3 mod 26 – Decryption: x = y – 3 mod 26 • Hill cipher – Matrix with entries 0, 1, 2, …, or 25. – Need to compute matrix inverse mod 26 kshum ENGG 2013 2

Today • More examples on linear algebra mod n • Rank of a matrix – Row-rank – How to compute row-rank by RREF kshum ENGG 2013 3

TASTE FROM DISCRETE MATH kshum ENGG 2013 4

Continuous vs discrete vs finite 2 0. 25 Continuous – 1 0 1 e The real number line Discrete 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, …. . Finite, for example 0, 1. (binary bit) kshum ENGG 2013 5

Example of Finite mathematics A B C D E F G H I J K L M 0 1 2 3 4 5 6 7 8 9 10 11 12 N O P Q R S T U V W X Y Z 13 14 15 16 17 18 19 20 21 22 23 24 25 Fix a one-to-one correspondence between the English alphabets and the integers mod 26. Caesar’s cipher: shift a letter to the right by 3, x x+3 mod 26 kshum ENGG 2013 6

A psychological barrier • In discrete and finite math. , we need to forget the fractions and decimal nimbers. – In application like Caesar cipher, 0. 25 has no meaning. • Geometry is often missing. We need to redefine some operations like taking inverse in a pure algebraic way. kshum ENGG 2013 7

Review • What is the meaning of “square root of 5”? kshum ENGG 2013 8

Square root of a positive real number • Geometric meaning: 1 0 1 2 5 • Algebraic meaning: “Square root of 5” is a real number x which satisfies the equation x 2 = 5. kshum ENGG 2013 9

Review • What is the meaning of “ 1/3”? • What is the meaning of “(2/5)-1” kshum ENGG 2013 10

Multiplicative inverse of a real number • Geometric meaning: 1/3 is the length obtained if we divide a line segment of unit length into three equal parts. 0 1/3 1 • Algebraic meaning: “One third” is a real number x which satisfies the equation 3 x = 1. kshum ENGG 2013 11

Multiplicative inverse of a real number • Geometric meaning: What is (2/5)-1? 2/5 Area = 1 x • Algebraic meaning: “(2/5)-1” is a real number x which satisfies the equation (2/5)x = 1. (2/5)-1 = 5/2 = 2. 5 kshum ENGG 2013 12

What is a number? • We will adopt the viewpoint that A number is the root of some polynomial (with integer coefficients). • For example: – “ 2” is a number which satisfies x 2 – 2=0. – “ 3/4” is a number which satisfies 4 x – 3=0. – “ 10” is a number which satisfies x – 10=0 More precisely, the above numbers are “algebraic number”. Some numbers such as and e are not algebraic, and e are examples of “transcendental numbers”. kshum ENGG 2013 13

Square root mod 10 • Geometric meaning: N/A • Algebraic meaning: Square root of 5 mod 10 is an integer x between 0 and 9 which satisfies x 2 = 5 mod 10. Example: 5 mod 10 is equal to 5. Example: 4 mod 10 is either 2 or 8. Example: 3 mod 10 does not exist kshum ENGG 2013 14

Multiplicative inverse mod 10 • Geometric meaning: N/A • Algebraic meaning: What is the multiplicative inverse of 3 mod 10? What we are going to do is: If we can find an integer x between 0 and 9 such that 3 x=1 mod 10, then we say that this x is the multiplicative inverse of 3 mod 10, and write 3 -1 = x mod 10. kshum ENGG 2013 15

Find it by exhaustive search Simply try all 10 possibilities one by one: 3 8 = 24 = 4 mod 10 3 1 = 3 mod 10 3 9 = 27 = 7 mod 10 3 2= 6 mod 10 3 0 = 27 = 0 mod 10 3 3 = 9 mod 10 3 4 = 12 = 2 mod 10 Answer: 3 -1 = 7 mod 10 3 5 = 15 = 5 mod 10 3 6 = 18 = 8 mod 10 3 7 = 21 = 1 mod 10 Aha! kshum ENGG 2013 16

Multiplicative inverse mod 5 • Geometric meaning: N/A • Algebraic meaning: What is the multiplicative inverse of 3 mod 5? Try to find an integer x between 0 and 4 such that 3 x=1 mod 5. If succeed, we say that this x is the multiplicative inverse of 3 mod 5, and write 3 -1 = x mod 5. kshum ENGG 2013 17

Find it by exhaustive search Simply try all 5 possibilities one by one: 3 1 = 3 mod 5 3 2= 6 = 1 mod 5 3 3 = 9 = 4 mod 5 3 4 = 12 = 2 mod 5 3 0 = 0 mod 5 Answer: 3 -1 = 2 mod 5 kshum ENGG 2013 18

Multiplicative inverse mod n in general • Given an integer b and integer n. • Definition: If we can find an integer x between 0 and n-1 such that bx=1 mod n, then we say that this x is the multiplicative inverse of b mod n, and write b-1 = x mod n. Two keywords kshum ENGG 2013 19

First keyword “if”: inverse may not exist • What is the multiplicative inverse of 2 mod 10? – Try to find an integer x between 0 and 9, such that 2 x=1 mod 10. 2 1 = 2 mod 10 2 7 = 14 = 4 mod 10 2 2 = 4 mod 10 2 8 = 16 = 6 mod 10 2 3 = 6 mod 10 2 4 = 8 mod 10 2 5 = 10 = 0 mod 10 2 6 = 12 = 2 mod 10 kshum 2 9 = 18 = 8 mod 10 2 0 = 0 mod 10 None of them is equal to 1. 2 -1 mod 10 does not exist. ENGG 2013 20

Property 1 • If the greatest common divisor of b and n are larger than 1, then we cannot find any multiplicative inverse of b mod n. • Reason: – Let d be the greatest common divisor of b and n. Suppose d > 1. – For any integer x, bx is a multiple of d. But n is also a multiple of d. – The remainder of n divided by bx is also a multiple of d, which ENGG 2013 by assumption is not 1. kshum 21

Property 2, “the”: multiplicative inverse is unique • Suppose that the gcd of b and n is 1. (This assumption makes sense because of property 1). • Suppose that b has two multiplicative inverses mod n, say x 1 and x 2, such that bx 1= 1 mod n and bx 2 = 1 mod n. • Let q 1 be the quotient when we divide bx 1 by n and q 2 be the quotient when we divide bx 2 by n. We have: – bx 1 = q 1 n + 1, – bx 2 = q 2 n + 1. • Subtracting the above two equations, we get b(x 1 – x 2) = (q 1 – q 2) n • b(x 1 – x 2) is divisible by n • Because n and b has no common factor, (x 1 – x 2) is divisible by n. • Because x 1 and x 2 are both between 0 and n – 1 , we must have x 1 = x 2 • Therefore, there is only one solution to bx = 1 mod n. kshum ENGG 2013 22

Example: Multiplicative inverse mod 10 • 0 -1 mod 10 does not exist. • 5 -1 mod 10 does not exist, because gcd(5, 10)=2. • 1 -1 = 1 mod 10, because 1 1=1 mod 10. • 6 -1 mod 10 does not exist, because gcd(6, 10)=2. • 2 -1 mod 10 does not exist, because gcd(2, 10)=2. • 3 -1 = 7 mod 10, because 3 7=21=1 mod 10. • 4 -1 mod 10 does not exist, because gcd(4, 10)=2. kshum • 7 -1 = 3 mod 10, because 7 3=21= mod 10. • 8 -1 mod 10 does not exist, because gcd(8, 10)=2. • 9 -1 = 9 mod 10, because 9 1=81=1 mod 10. ENGG 2013 23

2 x 2 Hill cipher • Encryption matrix a, b, c, d are integers between 0 and 25. • Encryption is done by matrix multiplication • Need to find a decryption matrix D such that kshum ENGG 2013 24

Formula for 2 x 2 matrix inverse This is an integer between 0 and 25. The multiplicative inverse calculated as in pp. 12 -15 • Example: Suppose that the encryption matrix is given by Task: Find the matrix inverse of E mod 26. kshum ENGG 2013 25

Find the multiplicative inverse of det by exhaustive search • det(E) = 2 3 – 1 9= – 3 = 23 mod 26. • Find an integer x between 0 and 25 such that 23 x = 1 mod 26. • By property 1, we only need to search for x which has no common factor with 26. 23 1 = 23 mod 26 23 3 = 17 mod 26 23 5 = 11 mod 26 23 7 = 5 mod 26 kshum 23 9 = 25 mod 26 23 11 = 19 mod 26 We can skip 13 23 15 = 7 mod 26 23 17 = 1 mod 26 ENGG 2013 Stop. The multiplicative inverse of 23 is found. 26

The inverse mod 26 • det(E) -1 = 23 -1 = 17 mod 26, – because 23 17=391 = 15 26+1 • The inverse of E mod 26 is • Check: kshum ENGG 2013 27

Encryption by Hill cipher • The encryption and decryption matrix must be kept secret. • Encryption of “HATE”: – convert the English letters to integers between 0 and 25 HATE 7, 0, 19, 4. – group the letters into blocks of length 2, and then multiply by the matrix E. – Convert 14, 7, 22, 5 to the ciphertext “OHWF” A B C D E F G H I J K L M 0 1 2 3 4 5 6 7 8 9 10 11 12 N O P Q R S T U V W X Y Z 13 14 15 16 17 18 19 20 21 22 23 24 25 kshum ENGG 2013 28

Decryption by Hill cipher • Decryption of “OHWF”, – convert the English letters to integers between 0 and 25 OHWF 14, 7, 22, 5. – group the letters into blocks of length 2, and then multiply by the decryption matrix D. – Convert 7, 0, 19, 4 to “HATE”. A B C D E F G H I J K L M 0 1 2 3 4 5 6 7 8 9 10 11 12 N O P Q R S T U V W X Y Z 13 14 15 16 17 18 19 20 21 22 23 24 25 kshum ENGG 2013 29

RANK kshum ENGG 2013 30

Back to the “real” world • In the remainder of this lecture, all variables are real numbers. • In a system of linear equations, the notion of “rank” measures the number of essentially different equations. They are multiple of each other. We can remove one of them without altering the solutions. kshum ENGG 2013 31

A larger example • In a large system of linear equations, how to find and remove redundant equations? kshum ENGG 2013 32

The representation using augmented matrix • Recall that the augmented matrix representation of a system of linear equations. The third equation is the sum of the first and second equation The third row is equal to the sum of the first and second rows kshum ENGG 2013 33

Recall: Linearly dependence • Definition: m vectors v 1, v 2, …, vm (each of them have n components) are linearly dependent if there are m real numbers, a 1, a 2, …, am, not all zero, such that the linear combination is equal to zero a 1 v 1 + a 2 v 2 + … + a m v m = 0 kshum ENGG 2013 34

Simple fact • If one of the m vectors v 1, v 2, …, vm is the zero vector 0, then v 1, v 2, …, vm are linearly dependent. • Reason: Suppose v 1= 0. 1 v 1 + 0 v 2 + … + 0 vm = 0 Not all coefficients are zero. In fact, the first coefficient is not zero. kshum ENGG 2013 35

Theorem v 1, v 2, …, vm are linearly dependent if and only if one of them is a linear combination of the others. Proof ( ) Suppose that v 1, v 2, …, vm are linearly dependent. By definition, we can find m real numbers, a 1, a 2, …, am, not all zero such that a 1 v 1+ a 2 v 2+ … + am vm= 0. Suppose that ai 0. Then vi = (-1/ai)(a 1 v 1+ …+ai-1 vi-1+ ai+1 vi+1+ … + am vm) The i-th vector vi is a linear combination of the others. kshum ENGG 2013 36

Theorem v 1, v 2, …, vm are linearly dependent if and only if one of them is a linear combination of the others. Proof ( ) Suppose that vi is a linear combination of the other vector: vi = c 1 v 1+ …+ci-1 vi-1+ ci+1 vi+1+ … + cm vm for some constants c 1, …, ci-1, ci+1, …cm. Then 0 = c 1 v 1+ …+ci-1 vi-1– vi+ ci+1 vi+1+ … + cm vm The i-th coefficient is non-zero (it is equal to -1). The zero vector can be expressed as a linear combination of v 1, v 2, …, vm, in which not all coefficients are zero. kshum ENGG 2013 37

Another way to say the same thing • If v 1, v 2, …, vm are linearly independent then none of them can be written as a linear combination of the others, and vice versa. kshum ENGG 2013 38

Definition: Row-rank of a matrix • Given a matrix M, the row-rank of M is the maximum number of linearly independent rows in M. • Example: – Because third row is the sum of the other two rows, i. e. , is a linear combination of the other two rows. , by theorem in p. 32, the three rows are linearly dependent. – The first and second row are linearly independent. – Therefore the maximum number of linearly independent rows is 2. – the row-rank of this matrix is equal to 2. kshum ENGG 2013 39

Example • What is the row-rank of kshum ENGG 2013 ? Ans: 2 ? Ans: 4 40

Theorem Elementary row operations do not affect the row-rank Proof: Let A be an n m matrix. Let the i-th row of a matrix A be ri. 1. Exchanging two rows does not affect the maximal number linearly independent rows. 2. Pick any k rows in A. Because of (1) above, we may exchange the rows and re-label them as r 1, r 2, …, rk. We want to see the effect of multiplying the p-th row (1 p k) by a nonzero constant c. kshum ENGG 2013 41

Proof (step 2) Elementary row operations do not affect the row-rank Suppose r 1, r 2 … rp, …, rk are linearly dependent. By definition a 1 r 1+a 2 r 2 +…+ ap rp+ …+ ak rk = 0. (not all a 1, a 2, …, ak are zero). Divide an multiply simultaneously by c, a 1 r 1+a 2 r 2 +…+ (ap/c) (c rp)+ …+ak rk = 0. Not all a 1, a 2, …, ap/c, …, ak are zero. Therefore r 1, r 2 , …, crp, …, rk are linearly dependent. kshum ENGG 2013 42

Proof (step 2 cont’d) Suppose r 1, r 2 … crp, …, rk are linearly dependent. By def. , b 1 r 1+b 2 r 2 +…+ bp (crp)+ …+ bk rk = 0. (not all b 1, b 2, …, bk are zero). Just bring the constant c out of the parenthesis, b 1 r 1+b 2 r 2 +…+ (cbp) rp+ …+bk rk = 0. Not all b 1, b 2, …, cbp, …, bk are zero. r 1, r 2, …, rp, …, rk are linearly dependent. Therefore, r 1, r 2, …, rp, …, rk are linearly dependent if and only if r 1, r 2, …, crp, …, rk kshum ENGG 2013 43

Proof (step 3) 3. What is the effect of adding d times the 2 -nd row to the 1 -st row? (Because exchange of rows does not affect the row-rank, considering adding d times r 2 to r 1 is sufficient. There is no loss of generality. ) Suppose r 1, r 2 …, rk are linearly independent. By def. , e 1 r 1+e 2 r 2 +…+ ek rk = 0 is possible only when e 1= e 2= …= ek = 0. kshum ENGG 2013 44

Proof (step 3) Now we want to check whether (r 1+dr 2), r 2 , r 3 , …, rk are linearly independent. If f 1(r 1+dr 2)+f 2 r 2 +f 3 r 3 +… …+ fk rk = 0, for some constants f 1, f 2, …, fk then f 1 r 1+(f 1 d+f 2)r 2 +f 3 r 3 +… …+ fk rk = 0. But this is possible only if f 1= (f 1 d+f 2)= …= fk = 0. f 2=0 kshum All coefficients f 1, f 2 …, fk are 0 (r 1+dr 2), r 2 , r 3, rk are linearly independent. ENGG 2013 45

Proof (step 3 reverse direction) Suppose r 1+dr 2, r 2 …, rk are linearly independent. By def. , g 1(r 1+dr 2)+g 2 r 2 +… …+ gk rk = 0 is possible only when g 1= g 2= …= gk = 0. We want to check whether r 1, r 2 , r 3, …, rk are linearly independent. If h 1 r 1+h 2 r 2 +…+ hk rk = 0, then h 1(r 1+dr 2–dr 2)+h 2 r 2 +…+ hk rk = 0 h 1(r 1+dr 2)+(–h 1 d+h 2)r 2 +…+ hk rk = 0. But this is possible only when h 1= (–h 1 d+h 2) = … = hk = 0 h 2=0 kshum All coefficients h 1, h 2… hk are 0 r 1, r 2 , r 3, …, rk are linearly independent. ENGG 2013 46

An algorithm for computing row-rank • Algorithm for computing the row-rank of M. 1. Apply elementary row operations to M, and transform it to a matrix in RREF. 2. Count the number of non-zero rows in the resulting RREF. • Because kshum – Row reductions does not change the row-rank. – The row-rank of a matrix in reduced row-echelon form is easy to calculate: just count the number of nonzero rows. ENGG 2013 47

RREF for rank computation • For mxn matrix, computing the row-rank of a matrix using the definition in p. 39 requires checking 2 m-1 combinations. This is not feasible even if the number of rows, m, is moderately large. • By reduction to RREF, the running times is of the order m 3. This is polynomial time in m. • However, in Matlab, the computation of rank is done by SVD (singular value decomposition), because SVD is more stable numerically. (type “doc rank” in Matlab for more details) kshum ENGG 2013 48

Notes p For modulo arithmetic, you can find out more in p http: //en. wikipedia. org/wiki/Modular_arithmetic p http: //en. wikibooks. org/wiki/High_School_Mathematics_Extensions/Primes/Modular_Arithm etic p Or any introductory book on number theory p In some books “a=b mod n” is written as “a b mod n”, (pronounced as “a is congruent to b mod n”) p In some webpages or books, you may see the symbols They stands for the set of integers and the set of rational numbers respectively. p The letter “Z” comes from German “Zahlen”, which means “numbers”. “Q” comes from English “quotients”. http: //en. wikipedia. org/wiki/Integer kshum ENGG 2013 49