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Proof Writing Problem Set I Linear Algebra MATH 228-02 Fall 2015 September 26, 2015 SOLUTIONS 1 Proof Problem Set I September 26, 2015 MATH 228-02 1.] Let A = [aij ] be an m × n matrix. Let c, d ∈ R be scalars. Use the definition of matrix addition and scalar multiplication to show that the following distributive law holds for scalar multiplication: (c + d)A = cA + dA. Proof. Let A = [aij ] be an m × n matrix, such that aij ∈ R is the element of the matrix A in the ith row and j th column where i = 1, 2, . . . , m and j = 1, 2, . . . , n. Also, let c, d ∈ R be scalars, then c + d ∈ R is also a scalar since addition is closed in the real number system. By the definition of scalar multiplication, it follows (c + d)A = (c + d)[aij ] = (c + d)aij . From here, we know (c+d)aij = caij +daij since real numbers obey the distributive property. Hence, from the definition of matrix addition and scalar multiplication, we have (c + d)A = [caij + daij ] = [caij ] + [daij ] = c[aij ] + d[aij ] = cA + dA. We have shown (c + d)A = cA + dA. 2.] Let A = [aij ], B = [bij ], and C = [cij ] be m × n matrices. Use the definition of matrix addition to show that matrix addition is associative: (A + B) + C = A + (B + C). Proof. Let A = [aij ], B = [bij ], and C = [cij ] be m × n matrices, such that aij , bij , cij ∈ R are the elements of the matrices A, B, and C, respectively, in the ith row and j th column where i = 1, 2, . . . , m and j = 1, 2, . . . , n. By the definition of matrix multiplication, it follows that (A + B) + C = [aij ] + [bij ] + [cij ] = [aij + bij ] + [cij ] = [(aij + bij ) + cij ]. Now, we have a single matrix whose ith , j th element is (aij + bij ) + cij . Since aij , bij , and cij are real numbers and are associative, it follows that (aij + bij ) + cij = aij + bij + cij = aij + (bij + cij ). Therefore, by the definition of matrix addition, we obtain (A + B) + C = [aij + (bij + cij )] = [aij ] + [bij + cij ] = [aij ] + [bij ] + [cij ] = A + (B + C). We have shown that (A + B) + C = A + (B + C). 2 Proof Problem Set I September 26, 2015 MATH 228-02 3.] Let A = [aij ] be an m × n matrix and B = [bjk ] be an n × p matrix. Let c ∈ R be a scalar. Using the definition of matrix and scalar multiplication, prove the following equation: c(AB) = A(cB). Proof. Let A = [aij ] be an m × n matrix and B = [bjk ] be an n × p matrix, such that aij , bjk ∈ R are the elements in the ith row and j th column of A and the j th row and k th column of B, respectively, where i = 1, 2, . . . , m, j = 1, 2, . . . , n, and k = 1, 2, . . . , p. Also, let c ∈ R be a scalar. Then by the definition of matrix multiplication, we obtain n X c(AB) = c[aij ][bjk ] = c aij bjk . j=1 By the definition of scalar multiplication, we have n X c(AB) = c aij bjk . j=1 From here, we can bring the constant c into the summation since c, aij and bjk are real numbers that obey the distributive property: n X c(AB) = caij bjk . j=1 Further, since real number multiplication is associative, it follows that caij bjk = aij (cbjk ). We have n X aij (cbjk ) . c(AB) = j=1 Finally, by applying the definitions of matrix and scalar multiplication, we arrive at the desired result: n X c(AB) = aij (cbjk ) j=1 = [aij ][cbjk ] = [aij ](c[bjk ]) = A(cB). 4.] Let A = [aij ] and B = [bij ] be m × n matrices, and C = [cjk ] be an n × p matrix. Use the definition of matrix addition and multiplication to prove the following distributive law for matrices: (A + B)C = AC + BC. Proof. Let A = [aij ] and B = [bij ] be m × n matrices, and C = [cjk ] be an n × p matrix. Let aij , bij ∈ R be the elements in the ith row and j th column of the matrices A and B, respectively, where i = 1, 2, . . . , m and j = 1, 2, . . . , n. Also, let cjk ∈ R be the element in the j th row and k th column 3 Proof Problem Set I September 26, 2015 MATH 228-02 of the matrix C for j = 1, 2, . . . , n and k = 1, 2, . . . , p. Then by the definition of matrix addition, we have (A + B)C = [aij ] + [bij ] [cjk ] = (aij + bij ) [cjk ]. By the definition of matrix multiplication, we obtain n X (A + B)C = (aij + bij )cjk . j=1 Since aij , bij , and cjk are real numbers, they obey the distributive law so that (aij + bij )cjk = aij cjk +bij cjk . Furthermore, addition of the real numbers is associative, which means the summation can be written as follows: n n n n X X X X (aij + bij )cjk = aij cjk + bij cjk = aij cjk + bij cjk . j=1 j=1 j=1 j=1 From this, we have n X (A + B)C = (aij + bij )cjk j=1 n n X X = aij cjk + bij cjk j=1 j=1 n n X X = aij cjk + bij cjk j=1 j=1 = [aij ][cjk ] + [bij ][cjk ] = AC + BC, which follows from the definition of matrix addition and multiplication. We have successfully shown that (A + B)C = AC + BC. 5.] An n × n matrix is said to be upper triangular if all the entries above the diagonal are zero. More precisely, if A = [aij ], then A is upper triangular precisely when aij = 0 for all i > j. Using the definition of matrix multiplication, show that if A and B are two n × n upper triangular matrices, then the product C = AB is also upper triangular. Proof. Let A = [aij ] and B = [bjk ] be a n × n matrices, such that aij , bjk ∈ R are the elements in the ith row and j th column of A and the j th row and k th column of B, respectively, where i, j, k = 1, 2, . . . , n. Suppose A and B are upper triangular so that if i > j, then aij = 0 and if j > k, then bjk = 0. Then by the definition of matrix multiplication, the matrix C = AB is an n × n matrix denoted by C = [cik ] for i, k = 1, 2, . . . , n. We wish to show cik = 0 if i > k. Consider the following matrix product: n X aij bjk . [cik ] = [aij ][bjk ] = j=1 We note that for i ≤ k, our summation can be written as i−1 k n X X X [cik ] = aij bjk + aij bjk + aij bjk . j=1 j=i 4 j=k+1 Proof Problem Set I September 26, 2015 MATH 228-02 The first summation from j = 1 to j = i − 1 is zero as aij = 0 since i > j. Also, the summation on the right from j = k + 1 to j = n is zero as bjk = 0 since j > k. Hence, we are left with k X [cik ] = aij bjk . j=i We note that this sum has terms in it if i ≤ k. As i approaches k, the number of terms decreases, which means if i > k, no more terms are within the sum. Therefore, for i > k we have cik = 0, as desired. This shows that the matrix C = AB is upper triangular. 5