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Laplace Expansion of the Determinant

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Laplace Expansion of the Determinant Geometria Lingotto. LeLing12: More on determinants. Contents: ¯ • Laplace expansion of the determinant. • Cross product and generalisations. • Rank and determinant: minors. • The characteristic polynomial. Recommended exercises: Geoling 14. ¯ Laplace expansion of the determinant The expansion of Laplace allows to reduce the computation of an n × n determinant to that of n (n − 1) × (n − 1) determinants. The formula, expanded with respect to the ith row (where A = (aij)), is: i+1 i+n det(A) = (−1) ai1det(Ai1) + ··· + (−1) aindet(Ain) where Aij is the (n − 1) × (n − 1) matrix obtained by erasing the row i and the column j from A. With respect to the j th column it is: j+1 j+n det(A) = (−1) a1jdet(A1j) + ··· + (−1) anjdet(Anj) Example 0.1. We do it with respect to the first row below. 1 2 1 4 1 3 1 3 4 3 4 1 = 1 − 2 + 1 = (4 − 6) − 2(3 − 5) + (3:6 − 5:4) = 0 6 1 5 1 5 6 5 6 1 The proof of the expansion along the first row is as follows. The determinant's linearity, proved in the previous set of notes, implies 0 1 Ej n BA C X B 2C det(A) = a1j det B . C j=1 @ . A An Ingegneria dell'Autoveicolo, LeLing12 1 Geometria Geometria Lingotto. where Ej is the canonical basis of the rows, i.e. Ej is zero except at position j where there is 1. Thus we have to calculate the determinants 0 0 ··· 0 1 0 0 ··· 0 a a ··· a a a ······ a 21 22 2(j−1) 2j 2(j+1) 2n . ··· . ··· . ··· . an1 an2 ··· an(j−1) anj an(j+1) ······ ann The whole column j does not intervene in the computation 0 0 ··· 0 1 0 0 ··· 0 a a ··· a 0 a ······ a 21 22 2(j−1) 2(j+1) 2n . ··· . ··· . ··· . an1 an2 ··· an(j−1) 0 an(j+1) ······ ann Since swapping two rows changes the sign of the determinant, 0 1 Ej 1 0 0 ··· 0 0 0 ··· 0 BA C 0 a a ··· a a ······ a B 2C j−1 21 22 2(j−1) 2(j+1) 2n 1+j det B . C = (−1) . = (−1) det(A1j) @ . A . ··· . ··· . ··· . An 0 an1 an2 ··· an(j−1) an(j+1) ······ ann Altogether, we have the expansion along the first row: det(A) = a11det(Ai1) ± · · · ± a1ndet(A1n). a b Note that the determinant = ad − bc is a special case of Laplace's formula. c d Cross product a x Given a plane vector 2 3 −!v = we wish to find an orthogonal vector −!w = . R b y x y The determinant vanishes when x = a and y = a, as confirmed by the formula a b x y x b −! b −! a = xb − ya = · . Hence w = is orthogonal to v = , a b y −a −a b a b i.e. · = 0. b −a Ingegneria dell'Autoveicolo, LeLing12 2 Geometria Geometria Lingotto. This argument generalises to space, and puts us in the position of solving the problem 0 1 a1 −! −! −! −! of finding a vector w orthogonal to two given vectors v1 ; v2 . In fact, let v1 = @b1 A, c1 0 1 0 1 a2 x −! −! v2 = @b2 A and w = @yA. Consider the determinant: c2 z x y z b1 c1 a1 c1 a1 b1 a1 b1 c1 = x − y + z b2 c2 a2 c2 a2 b2 a2 b2 c2 where the equality is a consequence of Laplace expanded along the first row. We may interpret the latter equality as a dot product 0 1 b1 c1 0 1 B b2 c2 C x y z x B C b1 c1 a1 c1 a1 b1 B a1 c1 C a1 b1 c1 = x − y + z = @yA · B− C ; b2 c2 a2 c2 a2 b2 B a2 c2 C a2 b2 c2 z B C @ a1 b1 A a2 b2 0 1 b1 c1 B b2 c2 C B C −! B a1 c1 C −! −! so w = B− C is orthogonal to both v1 and v2 . B a2 c2 C B C @ a1 b1 A a2 b2 −! 1 −! −! −! −! The vector w is called the cross product of v1 and v2 , and denoted v1 × v2 . −! Similarly, using an n × n determinant and Laplace we can find a vector w 2 Rn perpendicular to n − 1 given vectors. The cross product seen geometrically −! −! −! −! The cross product v1 × v2 is orthogonal to v1 and v2 . But as the vectors orthogonal to −! −! −! −! v1 and v2 form a straight line, to determine v1 × v2 it suffices to know its length and orientation. 1Also known as vector product, or wedge product because of the symbol ^. Ingegneria dell'Autoveicolo, LeLing12 3 Geometria Geometria Lingotto. −! −! −! −! Proposition 0.2. The length of v1 × v2 is jv1 jjv2 j sin(θ), where 0 ≤ θ < π is the angle −! −! formed by v1 and v2 . 0−! −!1 v1 × v2 −! 2 Proof. Recall that det(@ v1 A) is the volume of the parallelepiped spanned by −! v2 −! −! −! −! the three vectors v1 × v2 ; v1 ; v2 : 0−! −!1 v1 × v2 −! −! −! −! −! −! −! 2 V olume(v1 × v2 ; v1 ; v2 ) = det(@ v1 A) = jv1 × v2 j : −! v2 Notice: −! −! 0 v1×v2 1 −! −! −! −! −! −! jv1−!×v2j −! −! jv1 jjv2 j sin(θ) = Area(v1 ; v2 ) = det(@ v1 A) = jv1 × v2 j −! v2 2 −! −! 0 v1×v2 1 −! −! jv1−!×v2j Since det(@ v1 A) is positive, we conclude that the −! v2 −! −! right-hand-rule is suited to tell the orientation of v1 × v2 . −! −! −! −! Hence v1 × v2 is the unique vector orthogonal to v1 ; v2 −! −! with length equal to the area Area(v1 ; v2 ) of the paral- −! −! lelogram spanned by v1 and v2 , and whose orientation is found with the right-hand-rule. The cross product −!v × −!w can be interpreted also by a 90◦ -degree rotation. Namely, assuming that −!v ? −!w , then −!v × −!w is the vector obtained by rotating by 90◦ in the counter-clockwise direction the vector −!w on the plane orthogonal to −!v . 2because the determinant stems from the need to compute areas and volumes formed by vectors. Ingegneria dell'Autoveicolo, LeLing12 4 Geometria Geometria Lingotto. Rank and determinant: minors Recaal that the determinant makes sense for square matrices exclusively. If not square, we can compute certain quantities called minors or mini-determinants. A mini-determinant of order k for the matrix A is the determinant of a k × k matrix obtained from A by selecting k rows and k columns. For instance, any number aij is a mini-determinant of order 1. 01 8 2 1 1 2 B0 6 −3C Example 0.3. The determinant = −2 is a minor of order 2 in B C, 3 4 @3 −1 4 A 1 1 1 corresponding to choosing the first and third rows and columns. Also the determinant 1 8 2 6 −3 = 21 is a mini-determinant for A. Here's a 3rd order minor: 0 6 −3 = −1 4 3 −1 4 −87 0 1 c1 Bc C B 2C A column C = B . C is non-zero iff at least one minor is not zero. In fact, the minors @ . A cn are the numbers c1; ··· ; cn . This is a general fact, it holds for arbitrary matrices n × m; A is non-zero iff at least one mini-determinant is non-zero. Furthermore, a matrix has rank zero iff all mini-determinants are zero. This generalises, too, and is expressed by Kronecker's theorem 3. Theorem 0.4. An n × m matrix A has rank ρ(A) = k if and only if there exists a non-zero minor of order k and all minors of order > k are zero. Proof . Let Mk be a mini-determinant of order k. If there were a non-trivial linear combination among the k rows (columns) involved in building Mk , such combination would also be a non-trivial linear combination of k rows in the mini-determinant Mk . That would imply Mk = 0, so Mk = 0 if k > ρ(A). To prove that if k = ρ(A) then there is a non-zero minor of order k, we may assume A has k columns, i.e. we merely pay attention k LI columns. Since the rank is the dimension of the column space, there are k LI rows as well. Hence the k × k minor thus built cannot be trivial, for the k rows a re LI. 2 3Leopold Kronecker (1823 - 1891), german mathematician. Ingegneria dell'Autoveicolo, LeLing12 5 Geometria Geometria Lingotto. The choice of free parameters There are linear systems where some variables are free. In the Gauß-Jordan elimination method explained at the beginning of the course the free variables were the last ones: if x1; x2; ··· ; xn were the unknowns, then ··· ; xn−1; xn are those we tried to leave free, so to speak. What if we want to leave the first ones x1; x2 free? More generally, if we want that a certain subset xi1 ; xi2 ; ··· ; xil is free. In practice the easiest way is simply to change the order, i.e. we write the columns of xi1 ; xi2 ; ··· ; xil at the end of the matrix, and proceed as before. x1 + 2x2 + 7x3 + 5x4 = 1 Example 0.5. To solve ; using x1; x2 as parameters, 3x1 + 4x2 + 8x3 + 3x4 = 0 we use the matrix 7 5 1 2 1 8 3 3 4 0 The first column corresponds to x3 , the second to x4 , the third to x1 and the last one to x2 . Now, Gauß-Jordan: 5 1 2 1 5 1 2 1 7 5 1 2 1 1 7 7 7 7 1 7 7 7 7 ! ! −19 13 12 −8 ! 8 3 3 4 0 8 3 3 4 0 0 7 7 7 7 5 1 2 1 12 14 −3 1 7 7 7 7 1 0 19 19 19 −13 −12 8 ! −13 −12 8 0 1 19 19 19 0 1 19 19 19 12 14 −3 x4 + 19 x1 + 19 x2 = 19 The system is equivalent to: −13 −12 8 The solution, with free param- x3 + 19 x1 + 19 x2 = 19 eters x ; x , is: 1 2 0 1 0 1 0 1 0 1 x1 0 1 0 Bx2C B 0 C B 0 C B 1 C B C = B 8 C + x1 B 13 C + x2 B 12 C @x3A @ 19 A @ 19 A @ 19 A −3 −12 −14 x4 19 19 19 In the next example, instead, x1; x2 cannot be taken as fre parameters .
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