Determinant Properties 3/19/2002 Math 21b, O. KnillHomework for Thursday, March 21, 6.2 Nr. 3,4,6,16*,36,40*REMINDER. The determinant of a square matrix A = aijwas defined as the sum over all possible products(−1)πa1π(1)· · · anπ(n), where (−1)πis the sign of the permutation (pattern).TWO CASES.The determinant of a diagonal or tridiagonal matrixis the product of its diagonal elements.Example: det(1 0 0 04 5 0 02 3 4 01 1 2 1) = 20.The determinant of a partitioned matrixA 00 Bisthe product det(A)det(B).Example det(3 4 0 01 2 0 00 0 4 −20 0 2 2) = 2 ∗ 12 = 24.LINEARITY OF THE DETERMINANT. If the columns of A and B are the same except for the i’th column,thendet([v1, ..., v, ...vn]] + det([v1, ..., w, ...vn]] = det([v1, ..., v + w, ...vn]] .In general:det([v1, ..., kv, ...vn]] = k det([v1, ..., v, ...vn]] .The same holds for rows.PROPERTIES OF DETERMINANTS.det(AB) = det(A)det(B) .det(A−1) = det(A)−1det(SAS−1) = det(A) .det(AT) = det(A)det(λA) = λndet(A)det(−A) = (−1)ndet(A) .If B is obtained from A by switching two rows, then det(B) = −det(A). If B is obtained by adding an otherrow to a given row, then this does not change the value of the determinant.PROOF OF det(AB) = det(A)det(B), one brings the n × n matrix [A|AB] into row reduced echelon form.Similar than the augmented matrix [A|b] was brought into the form [1|A−1b], we end up with [1|A−1AB] = [1|B].By looking at the n×n matrix to the right during the Gauss elmination, the determinant has changed by det(A).We end up with a matrix B which has determinant det(B). Therefore, det(AB) = det(A)det(B).PROBLEM. A =0 0 0 21 2 4 50 7 2 90 0 6 4. Three transpositions of rows give B =1 2 4 50 7 2 90 0 6 40 0 0 2a matrix whichhas determinant 84. Therefore det(A) = (−1)3det(B) = −84.PROBLEM. Determine det(A100), where A is the matrix1 23 16.SOLUTION. det(A) = 10, det(A100) = (det(A))100= 10100= 1 · gogool. This name as well as the gogoolplex =1010100are official. They are huge numbers: the mass of the universe for example is 1052kg and 1/101051is thechance to find yourself on Mars by quantum fluctuations. (R.E. Crandall, Scient. Amer., Feb. 1997).ROW REDUCED ECHELON FORM. Determining rref(A) also determines det(A).If A is a matrix and αiare the factors which are used to scale different rows and s is the number of times, tworows are switched, then det(A) = (−1)sα1· · · αndet(rref(A)).INVERTIBILITY. A n × n matrix A is invertible if and only if det(A) 6= 0.THE LAPLACE EXPANSION.In order to calculate by hand the determinant of n × n matrices A = aijfor n > 3, the following expansionis useful. Choose a column i. For each entry ajiin that column, take the (n − 1) × (n − 1) matrix Aijcalledminor which does not contain the i’th column and j’th row. Thendet(A) = (−1)i+1ai1det(Ai1) + · · · + (−1)i+naindet(Ain) =nXj=1(−1)i+jaijdet(Aij) .VAN DER MONDE DETERMINANT.If we define for a scalar a the vector ~a =1, a, a2, . . . , an . For n + 1 scalars a0, . . . , anwe consider the(n + 1) × (n + 1)-matrix with rows ~ai.CLAIM: det(A) =Qi>j(ai− aj). To prove this, we make a Laplace expansion with respect to the lastcolumn. To do so, we call an= x and see that the determinant is a polynomial f(x) of degree n in x.It satisfies f(a0) = f(a1) = ... = f(an−1) = 0 because in those cases, the determinant is zero. Thereforef(x) = k(x − a1) . . . , (x − an−1) for some constant k. The number k is the coefficient in front of xn: it is thevan der Monde determinant in the case n − 1. The recursion VMn=Qn>j(an− aj)VMn−1proves the claim.BRAIN TEASERS.1) What is the determinant of1 1 0 0 0 0π 1 0 0 0 00 0 π 0 0 00 0 1 π 0 00 0 0 0 π 10 0 0 0 1 1.2) Assume A = −AT. What is det(A)?3) Assume a A has integer entries and A−1has also integer entries, then det(A) = 1 or det(A) = −1.4) What values can the determinant of an orthogonal matrix have? (Remember that it satisfies AAT= 1.)5) What is the determinant of0 0 0 15 0 6 00 1 0 04 0 5 1. (Hint: Use the Laplace expansion).6) (True or False) If a matrix has integer entries, the determinant is an integer.(True or False) If the determinant is an integer, the matrix has integer entries.(True or False) A matrix with positive entries has positive determinant.QR DECOMPOSITION. Also the QR decomposition allows to compute the determinant. The determinantof Q is ±1, the determinant of R is the product of the diagonal elements of that tridiagonal matrix.HOW FAST CAN ONE COMPUTE THE DETERMINANT?. Doing the Gauss elimination needsabout n3steps. The cost to compute the determinant is therefore also of the order n3.The graph to the left shows the time Mathematica needs to cal-culate the determinant in dependence on the size of the n × nmatrix. The matrix size goes from n=1 to n=300. The best cubicfit of these data has been obtained by the least square methodfrom chapter IV .DETERMINANTS IN PHYSICS. Physicists are excited about determinants because summing over all possible”paths” is used as a quantization method. The Feynmann path integral is a ”summation” over a suitable setof paths and leads to quantum mechanics. What does it have to do with determinants? Each summand of thedeterminant can be interpreted like a contribution of a path in a finite graph with n nodes.The article of Hawking deals with adeterminant functional in physics.This Ray-Singer determinant is anumber attached to a geometry rsp.an infinite dimensional matrix (theLaplacian) attached to the geome-try. Physicists trying to glue quan-tum mechanics with general relativ-ity hope to make sense of expres-sions likeRgeidet(A(g)), where A isan operator attached to some ge-ometry g and where the integral”sums” over all possible
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