1. ENGG2013 Unit 11 Row-Rank
Feb, 2011.

2. 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

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

4. TASTE FROM DISCRETE MATH
kshum
ENGG2013

5. 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, (binary bit)
kshum
ENGG2013

6. Example of Finite mathematicsB 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

7. 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

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

9. 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

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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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

20. 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 mod 10 does not exist.
26 = 12 = 2 mod 10
kshum
ENGG2013

21. 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

22. 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

23. 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

24. 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

25. 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

26. 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

27. 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

28. 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

29. 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

30. RANK
kshum
ENGG2013

31. 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

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

33. 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

34. 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

35. 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

36. 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

37. 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

38. 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

39. 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

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

41. 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

42. 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

43. 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

44. 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

45. 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

46. 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

47. 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

48. 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

49. Notes For modulo arithmetic, you can find out more in
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”.
kshum
ENGG2013