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ENGG2013 Unit 12 Everything is related

ENGG2013 Unit 12 Everything is related. Feb, 2011. Yesterday. Row-rank of an m  n matrix. The row rank of a matrix is the maximal number of linearly independent rows. Row-rank is an integer between 0 to m. How to use row operations and calculate row-rank by RREF.

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ENGG2013 Unit 12 Everything is related

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  1. ENGG2013 Unit 12Everything is related Feb, 2011.

  2. Yesterday • Row-rank of an mn matrix. • The row rank of a matrix is the maximal number of linearly independent rows. • Row-rank is an integer between 0 to m. • How to use row operations and calculate row-rank by RREF. • Row operations do not change the row-rank • We can see the row-rank of a RREF easily, just count the number of non-zero rows. ENGG2013

  3. Calculating row-rank from definition • Let A be a 67 matrix. • Test the linear independence for • Each of the 6 rows. (6 cases) • all pairs of rows. (15 cases) • All triples of rows. (20 cases) • All combinations of four rows (15 cases) • All combinations of five rows (6 cases) • All six rows together. (1 case) ENGG2013

  4. Calculating row-rank from definition • Example: If • any combination of 5 or more rows are linearly dependent • there is a combination of 4 rows which are linearly independent, then we say that the row rank is 4. • Example: If • any combination of 2 or more rows are linearly dependent • there is a non-zero row, then we say that the row rank is 1. ENGG2013

  5. Linear independence • Definition: Given a set of k vectors v1, v2, …, vk, each of them has n components. We call them linear independent when c1v1+c2v2+ … +c2vk = 0 is possible only if all coefficients in the linear combination are equal to zero, c1=c2= … =c2=0 Otherwise, they are linear dependent. The zero vector ENGG2013

  6. Linear equations • We need to solve a system of linear equations and show that they there is only one solution, namely, the all-zero solution. • For example, if we want to test whether are linearly independent, we write down an vector equation c1v1+c2v2+ c3vk=0, and see is c1=c2=c3=0 is the only choice for the coefficients. If yes, they are linearly independent. If there is some non-zero choice for c1,c2, and c3, then they are linearly dependent. ENGG2013

  7. RREF • We can use reduced row-echelon form to see if there is any non-trivial solution(“non-trivial” here means non-zero). 1 1 1 1 If there are free variables, then there are infinitelymany non-zero solutions ENGG2013

  8. Matrix inverse • If we write the system of linear equations in matrix from Ax=b, we can solve it by multiplying both sides (from the left) by the inverse of A, provided that the inverse exists. ENGG2013

  9. A missing link from Unit 7 • Definition: Given a square matrixA, if B is an matrix such that BA = I and AB = I, (Here “I” stands for the identity matrix, then we say tht Bis the inverse ofA, and write A-1=B. • To check that a matrix B is the inverse of A, we said in Unit 7 that t is sufficient to check either • BA = I, or • AB = I. ENGG2013

  10. Example • For example, if we want to check that the inverse of is we only need to verify that Is there any justification for this laziness? ENGG2013

  11. Left and right inverse • To streamline the presentation, we distinguish left-inverse and right-inverse of a matrix. • The notions of left- and right-inverse apply to rectangular matrix in general. • Let A be an mn matrix. • Definition: • A matrix X is called a left-inverse of A if XA = In. • A matrix Y is called a right-inverse of A if AY = Im. n  n indentity matrix m  m indentity matrix ENGG2013

  12. Left- and Right-inverse forrectangular matrix are complicated • Consider the matrix • We can check that for any a and b. Therefore, A has infinitely many right-inverse. • Also, A does not have any left-inverse. ENGG2013

  13. Row- and column-rank • As an mirror image of row-rank, we define the column-rank of a matrix M by the maximum number of linearly independent columns in M. (Here, M may be rectangular matrix, not necessarily square matrix.) • Similarly to the arguments yesterday, we can see that column operations do not change the column-rank. ENGG2013

  14. The transpose operator • The transpose operator interchange rows and columns of a matrix. • It provides a useful tool for transporting any proof and calculation of rows into proof and calculation of columns, and vice versa. • For example, • The results in p.10 are translated to The matrix has no right-inverse, but has infinitely many left-inverses. ENGG2013

  15. Determinant • From Unit 9, we also know that we can solve system of linear equations by Cramer’s rule. • We have formulae for computing the inverse using cofactors and adjoint: ENGG2013

  16. What are the relationship between them • Determinant • Rank • Left-inverse and right-inverse • RREF • Solution(s) to system oflinear equations • Linear independence ENGG2013

  17. A unifying theorem • Given an nn matrix A, the followings are logically equivalent: • A has a left-inverse. (X, s.t. XA=I.) • The only solution to Ax=0 is x=0. • A has column-rank n. (Columns of A are linearly independent) • A has a right-inverse. (Y, s.t. AY=I.) • A has row-rank n. (Rows of A are linearly independent) • The RREF of A is the nn identity matrix • A is a product of elementary matrices. • The determinant of A is non-zero. ENGG2013

  18. ENGG2013

  19. 1  2 • A has a left-inverse. (X, s.t. XA=I.) • The only solution to Ax=0 is x=0. Suppose that A has a left-inverse X, such that XA=I. Suppose that x = [x1 x2… xn]T is a solution to Ax = 0. Just multiply both side by X from the left. ENGG2013

  20. 2  3 • The only solution to Ax=0 is x=0. • A has column-rank n. (Columns of A are linearly independent) Suppose that Ax= 0 implies x=0. Let the columns of A be c1, c2,…, cn. Suppose that x1c1+x2c2+…+xncn = 0. But it’s equivalent to writing Ax=0.Hence, x1 =x2=… =xn =0.  Columns of A are linearly independent. ENGG2013

  21. 4  5 • A has a right-inverse. (Y, s.t. AY=I.) • A has row-rank n. (Rows of A are linearly independent) The reason is exactly the same as 1 2. ENGG2013

  22. 5 6 • A has row-rank n. (Rows of A are linearly independent) • The RREF of A is the nn identity matrix Suppose that the n rows of A are linearly independent. From Unit 11, we know that row operations does not affect row-rank. If the RREF of A cannot contain n-1 or less non-zero rows, then the rank of the RREF is n-1 or less, implying that A has rank n-1 or less. Therefore all rows of the RREF are not zero. The only choice is the identity matrix. ENGG2013

  23. 6  7 • The RREF of A is the nn identity matrix • A is a product of elementary matrices. Definition: An elementary matrix is what you get after applying an elementary row operation to an identity matrix. Exchange row 1and row 2 Compare with the fact that every positive integer can be factorized as a product of primes. The elementary matrices are the building blocks of invertible matrices in the same sense Add 3 times row 1to row 3 ENGG2013

  24. Elementary matrices • Definition: An elementary matrix is what you get after applying an elementary row operation to an identity matrix. Exchange row 1and row 2 Elementary matrices Add 3 times row 1to row 3 ENGG2013

  25. Row operations are “undo-able” • We can undo any row operation by another row operation, without loss of information. • First kind: To undo exchanging row i and row j, we exchange row i and row j again. • Second kind: The reverse process of multiplying row i by a non-zero factor k, is to multiply the same row by 1/k. • Third kind: The reverse process of adding c times row j to row i, is to subtract c times row j from row i. ENGG2013

  26. The same ideas in terms of matrix • Given any elementary matrix, say E, we can find another elementary matrix, say E’, such that E’ E = In, the nn identity matrix. • As we have shown in Unit 7, elementary row operation is the same as multiplying a suitable elementary matrix from the left. ENGG2013

  27. From Unit 7 • If we can row reduce matrix A to the identity matrix, then we can multiply a series of elementary matrices, say E1, E2, E3, …, Ep, from the left and obtain the identify matrix. • For each E1, E2, E3, …, Ep, we can find elementary matrix Ei’ such that Ei’ E= I. (i=1,2,…,p) • Therefore A can be written as a product of elementary matrices ENGG2013

  28. 7  1 • A is a product of elementary matrices. • A has a left-inverse. (X, s.t. XA=I.) Suppose that A can be written as a product of elementary matrices A = E1 E2 E3    Ek-1 Ek . For each elementary Ei, we can find Ei’such that Ej’ Ej = I, (for j = 1,2,…,k) We can take X as Ek’ Ek-1’    E2’ E1’ ENGG2013

  29. 7  4 • A is a product of elementary matrices. • A has a right-inverse. (Y, s.t. AY = I.) Suppose that A can be written as a product of elementary matrices A = E1 E2 E3    Ek-1 Ek . For each elementary Ei, we can find Ei’’such that Ej Ej’’= I, (for j = 1,2,…,k) We can take Y as Ek’’ Ek-1’’    E2’’ E1’’ ENGG2013

  30. The Logical Road map so far • 4 5  6  7  1  2  3 ENGG2013

  31. 3  7 • A has column-rank n. (Columns of A are linearly independent) • A is a product of elementary matrices. Apply the arguments in as in 5  6  7, to the transpose of A. 5’ AT has row-rank n. (Transpose interchange rows and column) 6’ The RREF of AT is the nn identity matrix 7’ AT is a product of elementary matrices. ENGG2013

  32. 3  7 • A has column-rank n. (Columns of A are linearly independent) • A is a product of elementary matrices. • Suppose that AT is the product of some elementary matrices, say E1, E2, … , Ek,  AT = E1  E2 E3    Ek. Take the transpose of both sides, and use the fact that • the transpose of an elementary matrix is also elementary, and • (EF)T = FTET  A = EkT  Ek-1T Ek-2T   E1T. ENGG2013

  33. Logical road map so far • 4 5  6  7  1  2  3 ENGG2013

  34. 7  8 • A is a product of elementary matrices. • The determinant of A is non-zero. We need the following observation: • Determinant of elementary matrix is nonzero • Let E be an elementary matrix and B is any matrix of the same size as E, then det(EB) = det(E) det(B) ENGG2013

  35. Elementary matrix (first kind) • First kind: Row exchange. determinant of a matrix E corresponding to row exchange is -1. • Suppose we apply a row exchange to a matrix B, and the resulting matrix is say B’. Then, det(B) = det(B’).  det(EB) = det(B’)= - det(B) = det(E)det(B) ENGG2013

  36. Elementary matrix (second kind) • Second kind: Multiply a row by a non-zero constant k. The determinant of matrix E corresponding this row multiplication is k. • Suppose we apply multiply the i-th row of matrix C to obtain matrix C’. Then, det(C’) = k det(C).  det(EC) = det(C’)= k det(B) = det(E)det(B) ENGG2013

  37. Elementary matrix (third kind) • Third kind: Add c times the i-th row to the j-th row. The determinant of matrix E corresponding to this action is 1. • Suppose we add c times i-th row of matrix D to the j-th row. Let the resulting matrix be D’. Then, det(D’) = det(D).  det(ED) = det(D’)= det(D) = det(E) det(B) ENGG2013

  38. 7  8 • A is a product of elementary matrices. • The determinant of A is non-zero. If A can be factor as a product of elementary matrices, i.e., A= E1E2   Ep Then det(A) = det(E1E2   Ep) = det(E1 (E2 E3    Ep)) = det(E1)det(E2 E3    Ep) = det(E1) det(E2) det(E3   Ep) = … = det(E1) det(E2)    det(Ep) This is non-zero because each factor is non-zero. ENGG2013

  39. Logical road map so far • 4 5  6  7  1  2  3 • Need to prove 8  any one from 1 to 7 • Let’s prove the implication 8  5 8 ENGG2013

  40. 8  5 • Logically, it is equivalent to proving the negation of (5) implies the negation of (8). ~(5): A has row-rank <n. (Rows of A are linearly dependent) ~(8): The determinant of A is zero. Suppose that the rows of A are linearly dependent. From Unit 11, we know that one row is a linear combination of the other rows. ENGG2013

  41. 8  5 • Suppose that the 1-st row is a linear combination of the other rows • r1 = c2r2 + c3r3 + c4r4 + … + cnrn where c2 to cn are some constants. By row reduction ENGG2013

  42. QED • 4 5  6  7  1  2  3 • Implications include: • Left-inverse is automatically the same as the right-inverse. • Row-rank = n if and only column-rank = n • And much more… 8 ENGG2013

  43. A question from tutorial 2 • Can we say that if the determinant is zero, then the row vectors are linearly dependent? • The answer is yes. This is a corollary of the unifying theorem in p.17. ENGG2013

  44. Theorem Three vectors are co-planar if and only if the determinant obtained by writing the three vectors together is zero. Proof:  Let the three vectors be Assume that they lie on the same plane. Let this plane ax + by + cz = 0. (a, b, and c are constants, not all zero) ENGG2013

  45. Proof () a,b,c are constants not all zero •  •  The system of linear equations (a,b, and c are the variables) has non zero solution. • because condition 2 implies the condition 8 in theorem in p.17, we can conclude that ENGG2013

  46. Proof () • For the reverse direction, suppose that • From the unifying theorem, we know that condition 8 implies condition 2, i.e., if the above determinant is zero, then has a solution (a,b,c) which is not all zero. • Therefore, it is guaranteed that we can find three real numbers, a, b, c, not all zero, such that all three points lie on the plane ax+by+cz =0. QED ENGG2013

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