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Mathematical Methods – WS 2021/22 5– Determinant – 1 / 29 Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 2 / 29 Properties of a Volume Determinant

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Mathematical Methods – WS 2021/22 5– Determinant – 1 / 29 Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 2 / 29 Properties of a Volume Determinant What is a Determinant? We want to “compute” whether n vectors in Rn are linearly dependent and measure “how far” they are from being linearly dependent, resp. Idea: Chapter 5 Two vectors in R2 span a parallelogram: Determinant vectors are linearly dependent area is zero ⇔ We use the n-dimensional volume of the created parallelepiped for our function that “measures” linear dependency. Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 1 / 29 Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 2 / 29 Properties of a Volume Determinant We define our function indirectly by the properties of this volume. The determinant is a function which maps an n n matrix × A = (a ,..., a ) into a real number det(A) with the following I Multiplication of a vector by a scalar α yields the α-fold volume. 1 n properties: I Adding some vector to another one does not change the volume. (D1) The determinant is linear in each column: I If two vectors coincide, then the volume is zero. det(..., ai + bi,...) = det(..., ai,...) + det(..., bi,...) I The volume of a unit cube is one. det(..., α ai,...) = α det(..., ai,...) (D2) The determinant is zero, if two columns coincide: det(..., ai,..., ai,...) = 0 (D3) The determinant is normalized: det(I) = 1 Notations: det(A) = A | | Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 3 / 29 Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 4 / 29 Example – Properties Determinant – Remarks (D1) I Properties (D1)–(D3) define a function uniquely. 1 2 + 10 3 1 2 3 1 10 3 (I.e., such a function does exist and two functions with these properties are identical.) 4 5 + 11 6 = 4 5 6 + 4 11 6 7 8 + 12 9 7 8 9 7 12 9 I The determinant as defined above can be negative. So it can be seen as “signed volume”. 1 3 2 3 1 2 3 · I We derive more properties of the determinant below. 4 3 5 6 = 3 4 5 6 · · 7 3 8 9 7 8 9 I Take care about the notation: · Do not mix up A with the absolute value of a number x . | | | | (D2) I The determinant is a so called normalized alternating multi-linear 1 2 1 form. 4 5 4 = 0 7 8 7 Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 5 / 29 Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 6 / 29 Further Properties Further Properties (D4) The determinant is alternating: (D5) The determinant does not change if we add some multiple of a det(..., a ,..., a ,...) = det(..., a ,..., a ,...) column to another column: i k − k i det(..., ai + α ak,..., ak,...) = det(..., ai,..., ak,...) 1 2 3 1 3 2 1 2 + 2 1 3 1 2 3 4 5 6 = 4 6 5 − · 7 8 9 7 9 8 4 5 + 2 4 6 = 4 5 6 · 7 8 + 2 7 9 7 8 9 · Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 7 / 29 Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 8 / 29 Further Properties Further Properties (D6) The determinant does not change if we transpose a matrix: (D7) det(A) = 0 columns (rows) of A are linearly independent 6 ⇔ A ist regular T ⇔ det(A ) = det(A) A ist invertible ⇔ Consequently, (D8) The determinant of the product of two matrices is equal to the all statements about columns hold analogously for rows. product of their determinants: det(A B) = det(A) det(B) 1 2 3 1 4 7 · · 4 5 6 = 2 5 8 (D9) The determinant if the inverse matrix is equal to the reciprocal of 7 8 9 3 6 9 the determinant of the matrix: 1 1 det(A− ) = det(A) Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 9 / 29 Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 10 / 29 Further Properties 2 2 Matrix × (10) The determinant of a triangular matrix is the product of its diagonal elements: a11 a12 = a11 a22 a12 a21 a21 a22 · − · a11 a12 a13 ... a1n 0 a22 a23 ... a2n 0 0 a33 ... a3n = a11 a22 a33 ... ann . · · · · . .. 1 2 . = 1 4 2 3 = 2 3 4 · − · − 0 0 0 . ann (11) The absolute value of the determinant det(a ,..., a ) is the | 1 n | volume of the parallelepiped spanned by the column vectors a1,..., an. Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 11 / 29 Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 12 / 29 3 3 Matrix: Sarrus’ Rule Source of Error × 1 2 3 1 2 Determinants of 4 4 matrices must be computed by means of 1 5 9 + 2 6 7 + 3 4 8 × 4 5 6 4 5 = · · · · · · transformation into a triangular matrix or by Laplace expansion. 7 5 3 8 6 1 9 4 2 = 0 7 8 9 7 8 − · · − · · − · · There is no such thing like Sarrus’ rule for 4 4 matrices. × Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 13 / 29 Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 14 / 29 Transform into Triangular Matrix Example – Transform into Triangular Matrix (1) Transform into upper triangular matrix similar to Gaussian 1 2 3 1 2 3 1 2 3 elimination. 4 5 6 = 0 3 6 = 0 3 6 − − − − 7 8 9 7 8 9 0 6 12 − − I Add a multiple of a row to another row. (D5) 1 2 3 I Multiply a row by some scalar α = 0 and the determinant by the = 0 3 6 = 1 ( 3) 0 = 0 1 6 − − · − · reciprocal . (D1) 0 0 0 α I Exchange two rows and switch the sign of the determinant. (D4) 0 2 4 1 2 3 1 2 3 1 2 3 = 0 2 4 = 0 2 4 (2) Compute the determinant as the product of its diagonal elements. − − 2 5 6 2 5 6 0 1 0 (Property D10) 1 2 3 1 1 = 0 2 4 = 1 2 ( 4) = 4 −2 · −2 · · · − 0 0 4 − Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 15 / 29 Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 16 / 29 Laplace Expansion Expansion along the First Row Laplace expansion along the k-th column and i-th row, resp.: 1 2 3 n n 1+1 5 6 i+k i+k 4 5 6 = 1 ( 1) det(A) = ∑ aik ( 1) Mik = ∑ aik ( 1) Mik · − · 8 9 · − · − 7 8 9 i=1 k=1 1+2 4 6 where M is the determinant of the (n 1) (n 1) submatrix + 2 ( 1) ik − × − · − · 7 9 which we obtain by deleting the i-th row and the k-th column of A. It is called a minor of A. 4 5 + 3 ( 1)1+3 · − · 7 8 We get the signs ( 1)i+k by means of a chessboard pattern: − + + = 1 ( 3) 2 ( 6) + 3 ( 3) − · − − · − · − + = 0 − − + + − Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 17 / 29 Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 18 / 29 Expansion along the Second Column Laplace and Leibzig Formula Laplace expansion allows to compute the determinant recursively: 1 2 3 The deterimant of a matrix is expanded into a sum of 1+2 4 6 k k k 4 5 6 = 2 ( 1) × · − · 7 9 determinants of (k 1) (k 1) matrices. 7 8 9 − × − For an n n matrices we can repeat this recursion step n times and 2+2 1 3 × + 5 ( 1) yield a summation of n! products of n numbers each: · − · 7 9 n 1 3 + 8 ( 1)3+2 det(A) = ∑ sgn(σ) ∏ aσ(i),i · − · 4 6 σ n i=1 ∈S where is the permutation group of order n. n = 2 ( 6) + 5 ( 12) 8 ( 6) S − · − · − − · − This formula is shown here just for completeness. = 0 Its explanation is out of the scope of this course. Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 19 / 29 Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 20 / 29 Adjugate Matrix Product A A T · ∗ In Laplace expansion the factors A = ( 1)i+k M are called the ik − ik 2 3 2 4 2 4 cofactors of a : ik 5 6 − 5 6 2 3 n det(A) = a A ∑ ik · ik 0 2 4 i=1 T 1 3 0 4 0 4 A A∗ = 1 2 3 · · − 2 6 2 6 − 1 3 2 5 6 The matrix formed by these cofactors is called the cofactor matrix A∗. 1 2 0 2 0 2 Its transpose A T is called the adjugate of A. ∗ 2 5 − 2 5 1 2 A A ... A 11 21 n1 ... 0 2 4 3 8 2 4 0 0 T A12 A22 An2 − − adj(A) = A∗ = . .. = 1 2 3 0 8 4 = 0 4 0 = A I . · − | | · 2 5 6 1 4 2 0 0 4 A1 A2n ... Ann − n Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 21 / 29 Josef Leydold – Mathematical Methods – WS 2021/22 5– Determinant – 22 / 29 Product A A T Cramer’s Rule for the Inverse Matrix · ∗ T Product of the k-th row of A∗ by the j-th column of A = (a1,..., an): We get a formula for the inverse of Matrix A: n n T i+k 1 1 T A∗ A = ∑ Aik aij = ∑ aij ( 1) Mik A− = A∗ · kj · · − A · h i i=1 i=1 | | [expansion along k-th column] = det(a1,..., aj ,..., an) k-th column This formula is not practical for inverting a matrix .
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