Solucionario-de-algebra-lineal-Kolman-octava-edicion.pdf

# B show that the sum and difference of two lower

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(b) Show that the sum and difference of two lower triangular matrices is lower triangular. (c) Show that if a matrix is upper and lower triangular, then it is a diagonal matrix. Solution. (a) As A above, let B = [ b ij ] be also an upper triangular matrix, S = A + B = [ s ij ] be the sum and D = A - B = [ d ij ] be the difference of these matrices. Then, for every i > j we have s ij = a ij + b ij = 0 + 0 = 0 respectively, d ij = a ij - b ij = 0 - 0 = 0. Thus the sum and difference of two upper triangular matrices is upper triangular. Example. 1 2 3 0 1 2 0 0 1 + 3 2 1 0 3 2 0 0 3 = 4 4 4 0 4 4 0 0 4 , sum of two upper triangular matrices, which is also upper triangular; 1 0 0 2 1 0 3 2 1 - 1 0 0 1 1 0 1 1 1 = 0 0 0 1 0 0 2 1 0 , difference of two lower trian- gular matrices, which is also lower triangular. (b) Similar. (c) If a matrix is upper and lower triangular, then the entries above the main diagonal and the entries below the main diagonal are zero. Hence all the entries off the main diagonal are zero and the matrix is diagonal. T.6. (a) Show that if A is an upper triangular matrix, then A T is lower triangular. (b) Show that if A is an lower triangular matrix, then A T is upper triangular. Solution. (a) By the definition of the transpose if A T = [ a T ij ], and A is an upper triangular matrix, a ij = 0 for every i > j and so a T ji = a ij = 0.

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13 Hence A T is lower triangular. (b) Similar. Page 37-38. T.4. Show that the product of two diagonal matrices is a diagonal matrix. Solution. Just verify that a 11 0 ... 0 0 a 22 ... 0 ... ... ... ... 0 0 ... a nn b 11 0 ... 0 0 b 22 ... 0 ... ... ... ... 0 0 ... b nn = a 11 b 11 0 ... 0 0 a 22 b 22 ... 0 ... ... ... ... 0 0 ... a nn b nn . T.5. Show that the product of two scalar matrices is a scalar matrix. Solution. Just verify that a 0 ... 0 0 a ... 0 ... ... ... ... 0 0 ... a b 0 ... 0 0 b ... 0 ... ... ... ... 0 0 ... b = ab 0 ... 0 0 ab ... 0 ... ... ... ... 0 0 ... ab . Short solution. Notice that any scalar matrix has the form a.I n = a 0 ... 0 0 a ... 0 ... ... ... ... 0 0 ... a . Then, obviously ( a.I n )( b.I n ) = ( ab ) .I n shows that prod- ucts of scalar matrices are scalar matrices. T.6. (a) Show that the product of two upper triangular matrices is an upper triangular matrix.
14 CHAPTER 1. MATRICES (b) Show that the product of two lower triangular matrices is a lower triangular matrix. Sketched solution. (a) A direct computation shows that the product of two upper triangular matrices a 11 a 12 ... a 1 n 0 a 22 ... a 2 n ... ... ... ... 0 0 ... a nn b 11 b 12 ... b 1 n 0 b 22 ... b 2 n ... ... ... ... 0 0 ... b nn is upper triangular too. Indeed, this is a 11 b 11 a 11 b 11 + a 12 b 22 ... a 11 b 1 n + a 12 b 2 n + ... + a 1 n b nn 0 a 22 b 22 ... a 22 b 2 n + a 23 b 3 n + ... + a 2 n b nn ... ... ... ... 0 0 ... a nn b nn , an upper triangular matrix. Complete solution. Let P = [ p ij ] = AB be the product of two upper triangular matrices. If i > j then p ij = n k =1 a ik b kj = i - 1 k =1 a ik b kj + n k = i a ik b kj = ( a i 1 b 1 j + ... + a i,i - 1 b i - 1 ,j ) + ( a ii b ij + ... + a in b nj ). Since both matrices A and B are upper triangular, in the first sum the a ’s are zero and in the second sum the b ’s are zero. Hence p ij = 0 and P is upper triangular too.

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• Fall '16
• jjaa
• Diagonal matrix, Det, Solution.

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