LeLing11: Determinants.

Transcription

1 Geometria Lingotto LeLing11: Determinants C ontents: The erminant of a matrix: basic properties Computing erminats Multilinearity of the erminat Cramer s formulas R ecommended exercises: Geoling 11 Determinants The erminant of a matrix A is an extremely useful quantity denoted by (A) or A It is related to the area or volume of a parallelogram defined by vectors Let us consider the fundamentals in calculating area or volume in dimension 2 If v, w are plane vectors, let A( v, w ) be the area of the parallelogram 0, v, w, v + w The area has two important properties The first one tells what happens when a vector is multiplied by by a constant m Clearly A(m v, w ) = ma( v, w ) Ingegneria dell Autoveicolo, LeLing11 1 Geometria

2 Geometria Lingotto The second property is a special case of the principle of Cavalieri 1 : A( v, w ) = A( v, w + m v ) The area does not change if we add to a vector a multiple of any other vector of the parallelogram The third property (self-evident yet important) says that the unit square has area 1 These three properties serve as starting point for defining the erminant of a matrix Definizione 01 The erminant 2 is a map associating a number (A) to a square matrix A = that satisfies: (i) m = m, ie multiplying a row by m has the effect of multi- plying the erminant by the same factor, (ii) + mr j se i j, ie adding a multiple of a row to another row does not alter the erminant, 1 Bonaventura Cavalieri, italian mathematician and student of Galileo Galilei 2 Also denoted A Ingegneria dell Autoveicolo, LeLing11 2 Geometria

3 Geometria Lingotto (iii) = 1, ie the identity s erminant is one Notice that if a row of A is zero the erminant vanishes, because by the second property we may multiply by 0 = m that row without changing the matrix, but the erminant should be zero This obviously states that a parallelogram has zero area when one of its sides is zero The three properties allow to compute erminants using Gauß-Jordan Esempio 02 Compute = = = = = = = Thus = Recall that Gauß-Jordan sometimes swaps the rows When computing a erminant, this corresponds to changing the overall sign, as explained in the Teorema 03 The erminant changes sign when tow rows are exchanged R j = R j Ingegneria dell Autoveicolo, LeLing11 3 Geometria

4 Geometria Lingotto Proof R j R j R j = R j + R j + R j Esempio 04 For instance : = = = = = 4 Once the Gauß step is finished we might not need to proceed with the Jordan bit If one row is zero, the erminant vanishes If not, along the diagonal we have only 1, which won t alter the erminant during the Jordan step This proves the following: Proposizione 05 The erminant of a triangular matrix is the product of the diagonal entries d 1 d d 2 = d d 2 0 1d 2 d n, = d 1d 2 d n 0 0 d n d n Proof If the product d 1 d 2 d n is non-zero the claim follows by the previous discusion If one of the d i is zero, the rank is not maximal and at the end of Gauß we shall find one null row at least Since a matrix is invertible iff its rank is maximal, we have Proposizione 06 A matrix A is invertible iff (A) 0 To find a formula 3 we note that the erminant is a linear map with respect to the rows of the matrix: 3 This will be the familiar one used in many textbooks, called Laplace s formula Ingegneria dell Autoveicolo, LeLing11 4 Geometria

5 Geometria Lingotto Proposizione 07 Let A be n n and,, its rows If the row A i is a linear combination of k rows B 1,, B k, ie A i = c 1 B 1 + c 2 B c k B k, then (A) = k c k B i i=1 Proof It is enough to prove the claim when k = 2 and c 1 = c 2 = 1, ie we shall prove it for a row that is linear combination of to rows Then assume A i = with arbitrary rows B, C We consider the two cases (A) 0, (A) = 0 separately (A) 0 : then B, D are linear combinations of A i, hence there are numbers b i, c i such that B = b 1 +, b n, C = c c n Thus j b ja j + j c ja j b i A i + c i A i The last equality follows because we can (with patience) use the erminant s property (ii) to get rid of all terms with no contribution Then using property (i) we obtain: b i A i + c i A i = (b 1 +c 1 ) A i = b i A i +c i A i = b i A i + c i A i j b ja j + j c ja j Ingegneria dell Autoveicolo, LeLing11 5 Geometria

6 Geometria Lingotto Again the last = descends from property (ii) Hence + C (A) = 0 : Suppose, A 2,, Âi,, 4 are LI It is clear hat the three erminants below are zero:,, C In this way we may assume, A 2,, Âi,, are LI If B, C span{, A 2,, Âi,, } the three erminants are still zero Thus we can suppose B / span{, A 2,, Âi,, }, so (, A 2,, B,, ) is a basis of the row space In particular Therefore: C = c 1 + c 2 A c i B + + c n B + c 1 + c 2 A c i B + + c n B + c i B where as usual the last = is a consequence of using property (ii) repeatedly 4 The symbol Âi means that A i is not present in the expression Ingegneria dell Autoveicolo, LeLing11 6 Geometria

7 Geometria Lingotto So, B + c i B = (1 + c i ) + c i = + c 1 + c 2 A c i B + + c n + C Binet s formula and consequences Now comes a crucial relation in the theory of erminants: (AB) (A)(B) known as formula of Binet 5 or product formula for the erminant The proof uses an important trick One thinks of the matrix A as a variable X, and considers the function f(x) (XB) The latter associates to a matrix X a number, the erminant of the product XB the key observation is that f satisfies properties (i) and (ii) In fact, from the point of view of rows the product equals X 1 B X 1 B X 2 B X 2 B XB =, so f(xb) clearly satisfies (i) and (ii) Since f X n B X n B does not satisfy (iii), for f(id) (B), we can divide f by (B) if (B) 0 Now the map f(x) = satisfies (i), (ii) and (iii), which implies (by definition f(x) (B) of the erminant) (X) = f(x) = f(x) (XB) The claim is proven in the (B) (B) case (B) 0 Eventually, if (B) = 0, neither B is invertible, but nor is XB, ie (B) = 0 implies (XB) = 0 5 French mathematician Ingegneria dell Autoveicolo, LeLing11 7 Geometria

8 Geometria Lingotto This trick is also useful to prove that (A) (A t ) Setting f(x) (X t ), we shall prove that f satisfies (i), (ii), (iii) hence f(x) = (X) Property (iii) holds trivially But it satisfies (i) as well, for multiplying a column by m gives m times the erminant But multiplying a column is the same as multiplying X on the right by an elementary matrix R Since (XR) (X)(R) and (R) = m, f satisfies (i) In a similar fashion one proves that f satisfies (ii) Let us summarise Teorema 08 Let A, B be square matrices: (AB) (A)(B) and consequently (A 1 ) = 1 (A) (A t ) (A) Finally we see that the erminant behaves in the same way if we look at row or at columns, thanks to the fact that (A t ) (A), in other words (i) and (ii) hold also when applied to columns This leads to the Proposizione 09 Let A be n n with columns C 1,, C n If C i is a linear combination of k columns B 1,, B k, C i = c 1 B 1 + c 2 B c k B k, then (A) = k c k ( ) C 1 B i C n i=1 The classical application are the so-called Cramer s formulas Corollary 010 (Cramer s formulas) Let (A B) be a non-homogeneous linear system where the square n n matrix A is invertible Call, A 2,, the columns of A Then the solution X = (x i ) of (A B) (unique!, for X = A 1 B) is given by: x i ( A i 1 B A i+1 ) (A) ie the coordinate x i of the solution X is the quotient of the erminant of the matrix obtained from A by replacing the column A i with B over B, Ingegneria dell Autoveicolo, LeLing11 8 Geometria

9 Geometria Lingotto Proof Since X = (x i ) is a solution iff j x ja j = B, ( A i 1 B A i+1 ) ( A1 A i 1 j x ja j A i+1 ) = ( A i 1 x i A i A i+1 ) = xi (A) Ingegneria dell Autoveicolo, LeLing11 9 Geometria