EA = I 3 = E = i=1, i k

Size: px

Start display at page:

Download "EA = I 3 = E = i=1, i k"

Isabel Craig
6 years ago
Views:

1 MTH5 Spring 7 HW Assignment : Sec.., # (a) and (c), 5,, 8; Sec.., #, 5; Sec.., #7 (a), 8; Sec.., # (a), 5 The due date for this assignment is //7. Sec.., # (a) and (c). Use the proof of Theorem. to obtain the inverse of each of the following elementary matrice (a) A = ; (b) B = Solution. (a) Let E be the elementary matrix corresponding to switching first and third rows of A. Then we obtain E by doing this elementary operation on the identity matrix I which is in this case just A. Thus, E = A and EA = I = Therefore, A = E = A. (b) Let E be the elementary matrix corresponding to replacing third row of B by times the first row of B plus the third row of B. Then we have EB = I so that B = E. To we obtain E by doing this elementary operation on the identity matrix I which is E = Sec.., #5. Prove that E is an elementary matrix if and only if E t is. Proof. Let E ij M n (F ), i, j n denote the n n matrices with zero entries except for the ith row, jth column entry which has a in it, in particular, Eij T = E ji. Let I n denote the n n identity matrix in M n (F ). We consider the three types of elementary matrices E obtained from an elementary row [column] operation on I n = n E ii and show that that their transposes are also elementary i= row [column] matrices. () Type elementary row [column] operation: If E switches the kth row [column] of I with the lth row [column] of I n then either k = l and E = I n = E t or k l and E = E ii + E kl + E lk = E t.. i=, i k,l [exactly the same for the column operation]. () Type elementary row [column] operation: If E multiplies the kth row [column] of I n by the nonzero scalar c then E = E ii + ce kk = E t [exactly the same for the column operation]. () i=, i k Type elementary row [column] operation: If E adds a nonzero scalar c times row [column] k to row [column] p (i.e., cr k + R p R p ) [i.e., cc k + C p C p ]

2 from I n then E = n E ii + ce kp [E = n E ii + ce pk ] and i= i= ( ) t E t = E ii + ce kp = n Eii t + cekp t = n E ii + ce pk i= i= which is just the type elementary row operation of E t adds scalar c times row p to row k (i.e., cr k + R p > R p ) from I n [E t = n i= i= E ii + ce kp ]. Therefore, since (E t ) t = E, this proves the statement. Sec.., #. Let A be an m n matrix. Prove that if B can be obtained from A by an elementary row [column] operation, then B t can be obtained from A t by the corresponding column [row] operation. Proof. Suppose that B can be obtained from A by an elementary row [column] operation then there exists and elementary matrix m m [n n] matrix E such that EA = B [AE = B] implying that A t E t = B t [E t A t = B t ]. By the proof of the previous problem, E t is the corresponding column [row] operation for the elementary row [column] operation on I n that gave us E. Thus, by Theorem., B t can be obtained from A t by the corresponding column [row] operation. This completes the proof. Sec.., #8. Prove that if a matrix Q can be obtained from a matrix P by an elementary row operation, then P can be obtained from Q by an elementary row operation of the same type. Proof. If Q = EP or Q = P E for some elementary matrix E then P = E Q or P = QE. Thus, to complete the proof we need only prove that E exists and is an elementary matrix of the same type as E. We use the results in the proof to problem #5 together with the facts: for i, i, k, l n we have the identities E t ij = E ji, E ij E kl = δ jk E il. Consider the three types of elementary matrices E obtained from an elementary row [column] operation on I n = n i= E ii and show that that their inverses are also elementary row [column] matrices of the same type. () Type elementary row [column] operation: If E switches the kth row [column] of I with the lth row [column] of I n then either k = l and E = I n = E t with E = E = I n = ( E ) t = (E t ) = E t or k l and E = E ii + E kl + E lk = E t i=, i k,l and E = E = ( E ) t = (E t ) = E t [exactly the same for the column operation]. () Type elementary row [column] operation: If E multiplies the kth row [column] of I n by the nonzero scalar c then E = E ii +ce kk = E t and E = ( E ) t = (E t ) = i=, i k i=, i k E ii + c E kk [exactly the same for the column operation]. () Type elementary row [column] operation: If E adds a nonzero scalar c times row [column] k to row [column] p (i.e., cr k + R p R p )

3 [i.e., cc k + C p C p ] from I n then E = n E ii + ce kp [E = n E ii + ce pk ] and i= E = n E ii + ( c) E kp i= which is just the type elementary row operation of E adds scalar c times row k to row p (i.e., cr k + R p > R p ) from I n [E = n E ii + ( c) E pk ]. This proves the statement. Sec.., #. Use elementary row and column operations to transform each of the following matrices into a matrix D satisfying the conditions of Theorem., and then determine the rank of each matrix. Solution. (a) A = R R (a) A = C+C C C +C C (b) B = R+R R R +R R C+Cj Cj, j=, C +C C i= i= = D. By Theorem. the rank of A is the rank of D which is. (b) B = R+R R 5 R+R R C+C C 5 C C = D. By Theorem. the rank of B is the rank of D which is. Sec.., #5. Suppose that A and B are matrices having n rows. Prove that M (A B) = (MA MB) for any m n matrix M. Proof. Let f j, j n be the standard ordered basis vectors for F n and

4 e j, j n be the standard ordered basis vectors for F n Then { M (Aej ) j n, M (A B) f j = M (Be j ) n + j n { (MA) ej j n, = (MB) e j n + j n { (MA) ej j n, (MA MB) f j = (MB) e j n + j n. This completes the proof. Sec.., #7 (a). Determine if the following linear system has a solution (a) x + x x + x = x + x + x = x + x + x + x = Solution. This is equivalent to finding whether or the linear system Ax = b has a solution, where A =, x = x x x x, b = By Theorem. of Sec.., this system has a solution if and only if rank A = rank (A b). Thus, we compute now these ranks. First, the second and third rows of A are linearly independent as they are not scalar multiples of each other whereas the third row minus the second row is the first row implying the dimension of the row space of A is implying that rank A =. Second, (A b) = R +R R R+R R R+R R. and since elementary operations do not change the rank of a matrix with the latter matrix having rank this implies rank (A b) = = rank A. Therefore the system (a) does not have a solution.

5 Sec.., #8. Let T : R R be defined by T (a, b, c) = (a + b, b c, a + c). For each vector v in R, determine whether v R (T ). Solution. The problem is asking us to compute the range of the linear transformation T = L A, i.e., left-multiplication by A, where A = We note that v R (T ) if and only if rank (A v) = rank A; the rank of a matrix does not change by applying an elementary operation on it. Thus, A = R+R R R +R R. implies and for any we have (A v) = v v v R +R R rank A = v = v v v R+R R v v v v + v v v v v implying Therefore, rank (A v) = rank A = if and only if v v + v =. R (T ) = v v v R : v v + v =. Sec.., # (a). Use Gaussian elimination to solve the following system of linear equations x + x x = (a) x + x + x = x + 5x x =. 5

6 Solution. To solve this we write this system as (A b), where A =, b =, 5 and use Gaussian elimination: (A b) = 5 R R R R R +R R R+R R R +R R 5 R+R R R+R R which implies that the solution is x x = x x = R+R R. Sec.., #5. Let the reduced row echelon form of A be B = 5. Determine A if the first, second, and fourth columns of A are a =, a =, and a =. By Theorem. of Sec.., the rank of B is r =, the first, second, and fourth columns of B, namely, b = e =, b = e =, and b = e = are the standard ordered basis vectors for F r, the third and

7 fifth columns of B are b = b 5 = 5 = b 5b + b, = b b + b, and hence if we denote a j, j =,..., 5 the ordered columns of A then the remaining columns are a = a 5a + a = 5 + = = a 5 = a a + a = = implying that + + = 7 9 A = (a a ) =

Matrix Algebra. Matrix Algebra. Chapter 8 - S&B

Matrix Algebra. Matrix Algebra. Chapter 8 - S&B Chapter 8 - S&B Algebraic operations Matrix: The size of a matrix is indicated by the number of its rows and the number of its columns. A matrix with k rows and n columns is called a k n matrix. The number