Linear Algebra/The Permutation Expansion/Solutions

Solutions These summarize the notation used in this book for the - and - permutations. Problem 1 Compute the determinant by using the permutation expansion. <li> <li> Answer <li> This matrix is singular. <li> This matrix is nonsingular. Problem 2 Compute these both with Gauss' method and with the permutation expansion formula. <li> <li> Answer <li> Gauss' method gives this and permutation expansion gives this. <li> Gauss' method gives this and the permutation expansion gives this. Problem 3 Use the permutation expansion formula to derive the formula for determinants. Answer Following Example 3.6 gives this. Problem 4 List all of the -permutations. Answer This is all of the permutations where the ones where the ones where and the ones where . Problem 5 A permutation, regarded as a function from the set to itself, is one-to-one and onto. Therefore, each permutation has an inverse. <li> Find the inverse of each -permutation. <li> Find the inverse of each -permutation. Answer Each of these is easy to check. <li> <TR> <TD style="border-right:2px black solid; border-bottom:2px solid black;"> permutation    <TD style="border-bottom:2px solid black;">       <TD style="border-bottom:2px solid black;">    <TR> <TD style="border-right:2px solid black;">inverse    <TD>       <TD>    <li> <TR> <TD style="border-right:2px solid black; border-bottom:2px solid black;"> permutation    <TD style="border-bottom:2px solid black;">       <TD style="border-bottom:2px solid black;">       <TD style="border-bottom:2px solid black;">       <TD style="border-bottom:2px solid black;">       <TD style="border-bottom:2px solid black;">       <TD style="border-bottom:2px solid black;">    <TR> <TD style="border-right:2px solid black;"> inverse    <TD>       <TD>       <TD>       <TD>       <TD>       <TD>    Problem 6 Prove that is multilinear if and only if for all and , this holds. Answer For the "if" half, the first condition of Definition 3.2 follows from taking and the second condition follows from taking . The "only if" half also routine. From the first condition of Definition 3.2 gives and the second condition, applied twice, gives the result. Problem 7 Find the only nonzero term in the permutation expansion of this matrix. Compute that determinant by finding the signum of the associated permutation. Answer To get a nonzero term in the permutation expansion we must use the entry and the entry. Having fixed on those two we must also use the entry and the entry. The signum of is because from the two row swaps and will produce the identity matrix. Problem 8 How would determinants change if we changed property (4) of the definition to read that ? Answer They would all double. Problem 9 Verify the second and third statements in Corollary 3.13. Answer For the second statement, given a matrix, transpose it, swap rows, and transpose back. The result is swapped columns, and the determinant changes by a factor of . The third statement is similar: given a matrix, transpose it, apply multilinearity to what are now rows, and then transpose back the resulting matrices. Problem 10 Show that if an matrix has a nonzero determinant then any column vector can be expressed as a linear combination of the columns of the matrix. Answer An matrix with a nonzero determinant has rank so its columns form a basis for . Problem 11 True or false: a matrix whose entries are only zeros or ones has a determinant equal to zero, one, or negative one. Strang 1980 Answer False. Problem 12 <li> Show that there are terms in the permutation expansion formula of a matrix. <li> How many are sure to be zero if the entry is zero? Answer <li> For the column index of the entry in the first row there are five choices. Then, for the column index of the entry in the second row there are four choices (the column index used in the first row cannot be used here). Continuing, we get . (See also the next question.) <li> Once we choose the second column in the first row, we can choose the other entries in ways. Problem 13 How many -permutations are there? Answer Problem 14skew-symmetric A matrix is skew-symmetric if , as in this matrix. Show that skew-symmetric matrices with nonzero determinants exist only for even . Answer In the exponent must be even. Problem 15 What is the smallest number of zeros, and the placement of those zeros, needed to ensure that a matrix has a determinant of zero? Answer Showing that no placement of three zeros suffices is routine. Four zeroes does suffice; put them all in the same row or column. Problem 16 If we have data points and want to find a polynomial passing through those points then we can plug in the points to get an equation/ unknown linear system. VandermondeThe matrix of coefficients for that system is called the Vandermonde matrix. Prove that the determinant of the transpose of that matrix of coefficients equals the product, over all indices with , of terms of the form . (This shows that the determinant is zero, and the linear system has no solution, if and only if the 's in the data are not distinct.) Answer The case shows what to do. The pivot operations of and give this. Then the pivot operation of gives the desired result. Problem 17block matrix A matrix can be divided into blocks, as here, which shows four blocks, the square and ones in the upper left and lower right, and the zero blocks in the upper right and lower left. Show that if a matrix can be partitioned as where and are square, and and are all zeroes, then . Answer Let be , let be , and let be . Apply the permutation expansion formula Because the upper right of is all zeroes, if a has at least one of among its first column numbers then the term arising from is (e.g., if then is ). So the above formula reduces to a sum over all permutations with two halves: first are rearranged, and after that comes a permutation of . To see this gives , distribute. Problem 18 Prove that for any matrix there are at most distinct reals such that the matrix has determinant zero (we shall use this result in Chapter Five). Answer The case shows what happens. Each term in the permutation expansion has three factors drawn from entries in the matrix (e.g., and ), and so the determinant is expressible as a polynomial in of degree . Such a polynomial has at most roots. In general, the permutation expansion shows that the determinant can be written as a sum of terms, each with factors, giving a polynomial of degree . A polynomial of degree has at most roots. ? Problem 19 The nine positive digits can be arranged into arrays in ways. Find the sum of the determinants of these arrays. Trigg 1963 Answer This is how the answer was given in the cited source. When two rows of a determinant are interchanged, the sign of the determinant is changed. When the rows of a three-by-three determinant are permuted, positive and negative determinants equal in absolute value are obtained. Hence the determinants fall into groups, each of which sums to zero. Problem 20 Show that Silverman Trigg 1963 Answer This is how the answer was given in the cited source. When the elements of any column are subtracted from the elements of each of the other two, the elements in two of the columns of the derived determinant are proportional, so the determinant vanishes. That is, ? Problem 21 Let be the sum of the integer elements of a magic square of order three and let be the value of the square considered as a determinant. Show that is an integer. Trigg Walker 1949 Answer This is how the answer was given in the cited source. Let have magic sum . Then and . Hence, adding rows and columns, ? Problem 22 Show that the determinant of the elements in the upper left corner of the Pascal triangle has the value unity. Rupp Aude 1931 Answer This is how the answer was given in the cited source. Denote by the determinant in question and by the element in the -th row and -th column. Then from the law of formation of the elements we have Subtract each row of from the row following it, beginning the process with the last pair of rows. After the subtractions the above equality shows that the element is replaced by the element , and all the elements in the first column, except , become zeroes. Now subtract each column from the one following it, beginning with the last pair. After this process the element is replaced by , as shown in the above relation. The result of the two operations is to replace by , and to reduce each element in the first row and in the first column to zero. Hence and consequently References D. L. (proposer) Silverman C. W. (solver) Trigg Quickie 237 Mathematics Magazine American Mathematical Society Jan. 1963 36 1. Gilbert Strang Linear Algebra and its Applications 2nd Hartcourt Brace Javanovich 1980 C. W. (proposer) Trigg Quickie 307 Mathematics Magazine American Mathematical Society Jan. 1963 36 1 77. R. J. (solver) Walker C. W. (proposer) Trigg Elementary Problem 813 American Mathematical Monthly American Mathematical Society Jan. 1949 56 1. H. T. R. (solver) Aude C. A. (proposer) Rupp Problem 3468 American Mathematical Monthly American Mathematical Society Jun.-July 1931 37 6 355.