ENGG2013 Unit 11 Row-Rank Feb, 2011.

Last week Caesar Cipher Hill 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 ENGG2013

Today More examples on linear algebra mod n Rank of a matrix Row-rank How to compute row-rank by RREF kshum ENGG2013

TASTE FROM DISCRETE MATH kshum ENGG2013

Continuous vs discrete vs finite 0.25 2 e Continuous –1 1 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 ENGG2013

Example of Finite mathematics B C D E F G H I J K L M 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 ENGG2013

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 re-define some operations like taking inverse in a pure algebraic way. kshum ENGG2013

Review What is the meaning of “square root of 5”? kshum ENGG2013

Square root of a positive real number Geometric meaning: Algebraic meaning: “Square root of 5” is a real number x which satisfies the equation x2 = 5. 1 1 2 5 kshum ENGG2013

Review What is the meaning of “1/3”? What is the meaning of “(2/5)-1” kshum ENGG2013

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. Algebraic meaning: “One third” is a real number x which satisfies the equation 3x = 1. 1/3 1 kshum ENGG2013

Multiplicative inverse of a real number Geometric meaning: What is (2/5)-1? 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 Area = 1 2/5 x kshum ENGG2013

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 x2 – 2=0. “3/4” is a number which satisfies 4x – 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 ENGG2013

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 x2 = 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 ENGG2013

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 3x=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 ENGG2013

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 3-1 = 7 mod 10 Answer: 35 = 15 = 5 mod 10 36 = 18 = 8 mod 10 37 = 21 = 1 mod 10 Aha! kshum ENGG2013

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 3x=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 ENGG2013

Find it by exhaustive search Simply try all 5 possibilities one by one: 31 = 3 mod 5 32= 6 = 1 mod 5 3-1 = 2 mod 5 Answer: 33 = 9 = 4 mod 5 34 = 12 = 2 mod 5 30 = 0 mod 5 kshum ENGG2013

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 ENGG2013

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 2x=1 mod 10. 27 = 14 = 4 mod 10 21 = 2 mod 10 22 = 4 mod 10 28 = 16 = 6 mod 10 23 = 6 mod 10 29 = 18 = 8 mod 10 24 = 8 mod 10 20 = 0 mod 10 25 = 10 = 0 mod 10 None of them is equal to 1. 2-1 mod 10 does not exist. 26 = 12 = 2 mod 10 kshum ENGG2013

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 by assumption is not 1. kshum ENGG2013

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 x1 and x2, such that bx1= 1 mod n and bx2 = 1 mod n. Let q1 be the quotient when we divide bx1 by n and q2 be the quotient when we divide bx2 by n. We have: bx1 = q1 n + 1, bx2 = q2 n + 1. Subtracting the above two equations, we get b(x1 – x2) = (q1 – q2) n  b(x1 – x2) is divisible by n Because n and b has no common factor, (x1 – x2) is divisible by n. Because x1 and x2 are both between 0 and n – 1 , we must have x1 = x2 Therefore, there is only one solution to bx = 1 mod n. kshum ENGG2013

Example: Multiplicative inverse mod 10 0-1 mod 10 does not exist. 1-1 = 1 mod 10, because 11=1 mod 10. 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. 5-1 mod 10 does not exist, because gcd(5,10)=2. 6-1 mod 10 does not exist, because gcd(6,10)=2. 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. kshum ENGG2013

2x2 Hill cipher Encryption matrix Encryption is done by matrix multiplication Need to find a decryption matrix D such that a,b,c,d are integers between 0 and 25. kshum ENGG2013

Formula for 2x2 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 ENGG2013

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 23x = 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9 = 25 mod 26 233 = 17 mod 26 2311 = 19 mod 26 We can skip 13 235 = 11 mod 26 2315 = 7 mod 26 237 = 5 mod 26 2317 = 1 mod 26 Stop. The multiplicative inverse of 23 is found. kshum ENGG2013

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: \mathbf{D}=17\begin{bmatrix} 3 & -9\\ -1 & 2 \end{bmatrix}=17 \begin{bmatrix} 3 & 17 \\ 25 & 2 \end{bmatrix} = 153 & 408 \\ 391 & 17 \end{bmatrix}= 25 & 3\\ 9 & 8 \end{bmatrix} \bmod 26 kshum ENGG2013

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 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 ENGG2013

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 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 ENGG2013

RANK kshum ENGG2013

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 ENGG2013

A larger example In a large system of linear equations, how to find and remove redundant equations? kshum ENGG2013

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 \begin{align*} x+y+z&= 1 \\ 2x+y-z&= 1 \\ 3x+2y&= 2 \\ \end{align*} The third row is equal to the sum of the first and second rows kshum ENGG2013

Recall: Linearly dependence Definition: m vectors v1, v2, …, vm (each of them have n components) are linearly dependent if there are m real numbers, a1, a2, …, am, not all zero, such that the linear combination is equal to zero a1 v1+ a2 v2+ … + am vm= 0 kshum ENGG2013

Simple fact If one of the m vectors v1, v2, …, vm is the zero vector 0, then v1, v2, …, vm are linearly dependent. Reason: Suppose v1= 0. 1 v1+ 0 v2+ … + 0 vm= 0 Not all coefficients are zero. In fact, the first coefficient is not zero. kshum ENGG2013

Theorem v1, v2, …, vm are linearly dependent if and only if one of them is a linear combination of the others. Proof () Suppose that v1, v2, …, vm are linearly dependent. By definition, we can find m real numbers, a1, a2, …, am, not all zero such that a1 v1+ a2 v2+ … + am vm= 0. Suppose that ai0. Then vi = (-1/ai)(a1 v1+ …+ai-1 vi-1+ ai+1 vi+1+ … + am vm) The i-th vector vi is a linear combination of the others. kshum ENGG2013

Theorem v1, v2, …, 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 = c1 v1+ …+ci-1 vi-1+ ci+1 vi+1+ … + cm vm for some constants c1, …,ci-1, ci+1,…cm. Then 0 = c1 v1+ …+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 v1, v2, …, vm, in which not all coefficients are zero. kshum ENGG2013

Another way to say the same thing If v1, v2, …, vm are linearly independent then none of them can be written as a linear combination of the others, and vice versa. kshum ENGG2013

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 the third row is the sum of the other two rows, i.e., is a linear combination of the other two rows., by the 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 ENGG2013

Example What is the row-rank of ? What is the row-rank of ? Ans: 2 kshum ENGG2013

Theorem Proof: Let A be an nm matrix. 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. Exchanging two rows does not affect the maximal number linearly independent rows. Pick any k rows in A. Because of (1) above, we may exchange the rows and re-label them as r1, r2, …, rk. We want to see the effect of multiplying the p-th row (1  p  k) by a nonzero constant c. kshum ENGG2013

Proof (step 2) Elementary row operations do not affect the row-rank Suppose r1, r2 … rp, …, rk are linearly dependent. By definition a1r1+a2r2 +…+ ap rp+ …+ ak rk = 0. (not all a1, a2, …, ak are zero). Divide an multiply simultaneously by c, a1r1+a2r2 +…+ (ap/c) (c rp)+ …+ak rk = 0. Not all a1, a2, …, ap/c, …, ak are zero. Therefore r1, r2 ,…, crp, …, rk are linearly dependent. kshum ENGG2013

Proof (step 2 cont’d) Suppose r1, r2 … crp, …, rk are linearly dependent. By def., b1r1+b2r2 +…+ bp (crp)+ …+ bk rk = 0. (not all b1, b2, …, bk are zero). Just bring the constant c out of the parenthesis, b1r1+b2r2 +…+ (cbp) rp+ …+bk rk = 0. Not all b1, b2, …, cbp, …, bk are zero.  r1, r2,…, rp, …, rk are linearly dependent. Therefore, r1, r2,…, rp, …, rk are linearly dependent if and only if r1, r2,…, crp, …, rk kshum ENGG2013

Proof (step 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 r2 to r1 is sufficient. There is no loss of generality.) Suppose r1, r2 …, rk are linearly independent. By def., e1r1+e2r2 +…+ ek rk = 0 is possible only when e1= e2= …= ek = 0. kshum ENGG2013

Proof (step 3) Now we want to check whether (r1+dr2), r2 , r3 , …,rk are linearly independent. If f1(r1+dr2)+f2r2 +f3r3 +… …+ fk rk = 0, for some constants f1, f2, …, fk then f1r1+(f1d+f2)r2 +f3r3 +… …+ fk rk = 0. But this is possible only if f1= (f1d+f2)= …= fk = 0. All coefficients f1, f2 …, fk are 0 (r1+dr2), r2 ,r3,rk are linearly independent. f2=0 kshum ENGG2013

Proof (step 3 reverse direction) Suppose r1+dr2, r2 …, rk are linearly independent. By def., g1(r1+dr2)+g2r2 +… …+ gk rk = 0 is possible only when g1= g2= …= gk = 0. We want to check whether r1, r2 ,r3,…,rk are linearly independent. If h1r1+h2r2 +…+ hk rk = 0, then h1(r1+dr2–dr2)+h2r2 +…+ hk rk = 0 h1(r1+dr2)+(–h1d+h2)r2 +…+ hk rk = 0. But this is possible only when h1= (–h1d+h2) = … = hk = 0 h2=0 All coefficients h1, h2… hk are 0 r1, r2 ,r3,…,rk are linearly independent. kshum ENGG2013

An algorithm for computing row-rank Algorithm for computing the row-rank of M. Apply elementary row operations to M, and transform it to a matrix in RREF. Count the number of non-zero rows in the resulting RREF. Because 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. kshum ENGG2013

RREF for rank computation For mxn matrix, computing the row-rank of a matrix using the definition in p.39 requires checking 2m-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 m3. 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 ENGG2013

Notes For modulo arithmetic, you can find out more in http://en.wikipedia.org/wiki/Modular_arithmetic http://en.wikibooks.org/wiki/High_School_Mathematics_Extensions/Primes/Modular_Arithmetic Or any introductory book on number theory In some books “a=b mod n” is written as “ab mod n”, (pronounced as “a is congruent to b mod n”) In some webpages or books, you may see the symbols They stands for the set of integers and the set of rational numbers respectively. The letter “Z” comes from German “Zahlen”, which means “numbers”. “Q” comes from English “quotients”. http://en.wikipedia.org/wiki/Integer kshum ENGG2013