Mathematical background Daniele Carnevale Dipartimento di Ing. - PowerPoint PPT Presentation

Complex numbers Matrices Mathematical background Daniele Carnevale Dipartimento di Ing. Civile ed Ing. Informatica (DICII), University of Rome “Tor Vergata” Fondamenti di Automatica e Controlli Automatici A.A. 2014-2015 1 / 20

Complex numbers Matrices The complex number A complex number z is defined as α � Re { z } , β � Im { z } , z = α + jβ ∈ C , where α and β are real numbers, j is the imaginary unit such that j 2 =-1. A different module and phase representation is given by z = ρe jθ , ρ = || z || = Abs ( z ) , θ = z. = Arg ( z ) Then ρ is the norm of the “vector” z on the 2D-plane and θ is its angle (counter-clock wise) and � arccos ( Im ( z ) / || z || ) � if Im ( z ) ≥ 0 , Re { z } 2 + Im { z } 2 , z � || z || � − arccos ( Im ( z ) / || z || ) if Im ( z ) < 0 , Figure : Different representation of complex numbers. 2 / 20

Complex numbers Matrices The complex number cont’d The sin( · ) and cos( · ) function can be rewritten using the unitary circle on the complex plane described by z = e jθ with θ ∈ [ − π, π ] such as sin( θ ) = Im ( z ) = Im ( z ) , || z || ���� || z || =1 cos( θ ) = Re ( z ) = Re ( z ) . || z || ���� || z || =1 Figure : Module and phase representation - the unitary circle. 3 / 20

Complex numbers Matrices The complex number: algebra Given two complex numbers z 1 = α 1 + jβ 1 , z 2 = α 2 + jβ 2 , then z = z 1 + z 2 = α 1 + α 2 + j ( β 1 + β 2 ) . Figure : Addition. 4 / 20

Complex numbers Matrices The complex number: subtraction Given two complex numbers z 1 = α 1 + jβ 1 , z 2 = α 2 + jβ 2 , then z = z 1 − z 2 = α 1 − α 2 + j ( β 1 − β 2 ) . Figure : Subtraction: exercise.... 5 / 20

Complex numbers Matrices The complex number: multiplication Given two complex numbers z 1 = α 1 + jβ 1 = ρ 1 e jθ 1 , z 2 = α 2 + jβ 2 = ρ 2 e jθ 2 , then z = z 1 z 2 = ρ 1 ρ 2 e j ( θ 1 + θ 2 ) ⇒ z m 1 = ρ m 1 e jmθ 1 . Figure : Multiplication ( i → j ). 6 / 20

Complex numbers Matrices The complex number: division Given two complex numbers z 1 = α 1 + jβ 1 = ρ 1 e jθ 1 , z 2 = α 2 + jβ 2 = ρ 2 e jθ 2 , then z = z 1 = ρ 1 e j ( θ 1 − θ 2 ) . z 2 ρ 2 Another way to do it, define the conjugate of z, z , such as z = ρe jθ → z = ρe − jθ , zz = || z || 2 = ρ 2 , z = Re ( z ) − jIm ( z ); then || z 2 || 2 = ρ 1 e jθ 1 ρ 2 e − jθ 2 z 1 = z 1 z 2 = z 1 z 2 = ρ 1 e j ( θ 1 − θ 2 ) ρ 2 z 2 z 2 z 2 ρ 2 2 Figure : Division: exercise...go on the other way around... . 7 / 20

Complex numbers Matrices The complex number: other examples Further examples: z = e j π 2 = j, z = e − j π 2 = − j, z = e jπ = − 1 , z = e − jπ = − 1 , √ √ 2 2 z = e jπ/ 4 = + j = cos( π/ 4) + j sin( π/ 4) , 2 2 Figure : Conjugate complex number. 8 / 20

Complex numbers Matrices The Euler formula Given that e jθ = cos( θ ) + j sin( θ ) , then e jθ + e − jθ = cos( θ ) + j sin( θ ) + cos( − θ ) + j sin( − θ ) , (1) 2 2 = cos( θ ) + j sin( θ ) + cos( θ ) − j sin( θ ) , (2) 2 = cos( θ ) . (3) e jθ − e − jθ = cos( θ ) + j sin( θ ) − (cos( − θ ) + j sin( − θ )) , (4) 2 j 2 j = cos( θ ) + j sin( θ ) − cos( θ ) + j sin( θ ) , (5) 2 j = sin( θ ) . (6) 9 / 20

Complex numbers Matrices Useful formulas (in the next) Then, for a complex number z = ρe jθ , it holds � e jθ + e − jθ � z + z = ρ = 2 ρ cos( θ ) . ���� Euler formula Given two complex numbers z 1 = ρ 1 e jθ 1 , z 2 = ρ 2 e jθ 2 , then � e j ( θ 1 + θ 2 ) + e − j ( θ 1 + θ 2 ) � z 1 z 2 + z 1 z 2 = ρ 1 ρ 2 , = 2 ρ 1 ρ 2 cos( θ 1 + θ 2 ) . (7) Which is the trajectory depicted on the plane C by z ( t ) = ρ ( t ) e jθ ( t ) , ρ ( t ) = exp ( − 2 t ) , θ ( t ) = 5 t, (8) when t ∈ [0 , + ∞ ) ? Note that if ρ ( t ) = 1 and θ ( t ) = ωt , than cos( ωt ) + j sin( ωt ) = e jωt . 10 / 20

Complex numbers Matrices Roots of complex numbers The roots of x 2 = 1 can be easily found if x ∈ C , e.g x = ± 1 . The same for x 2 = − 1 , there not exist real x . On the complex plane, the imaginary axis allows to solve also z 2 = − 1 → z = ± j . Furthermore: z 3 = 1 : � ρ = 1 , ρ 3 e 3 jθ = 1 = e j (2 kπ ) , k ∈ Z ⇐ ⇒ (9) = 2 kπ θ 3 . Figure : Roots of z 3 = 1 . 11 / 20

Complex numbers Matrices Roots of complex numbers cont’d z 3 = 1 : � ρ = 1 , ρ 3 e 3 jθ = 1 = e j (2 kπ ) , k ∈ Z ⇐ ⇒ (10) = 2 kπ θ 3 . z 5 = − 1 : � ρ = 1 , ρ 5 e 5 jθ = − 1 = e j ( ± π +2 kπ ) , k ∈ Z ⇐ ⇒ (11) = ± π +2 kπ θ . 5 z n = − 1 ? 12 / 20

Complex numbers Matrices Algebra: Matrices Consider the matrix   a 1 , 1 a 1 , 2 . . . a 1 ,n a 2 , 1 a 2 , 2 . . . a 2 ,n     A m × n =  , . .   . .  . . a m, 1 a m, 2 . . . a m,n with m rows and n columns. If n = m the matrix A is said to be “square”, if m > n it is “tall” otherwise “fat”. Based on the coefficients of the matrix A: If a i,j = 0 for all i > j , A is said to be upper triangular If a i,j = 0 for all i < j , A is said to be lower triangular If a i,j = 0 for all i � = j , A is said to be diagonal. A square matrix A is symmetric if the transpose matrix A ′ is such that A ′ = A (see also block diagonal matrices). A is antisymmetric if A ′ = − A . The transposed matrix B = A ′ is such that b i,j = a j,i . The identity matrix I n is a square diagonal matrix n × n with elements on the diagonal equal to 1. 13 / 20

Complex numbers Matrices Algebra: Matrices operations When A and B are matrices with the same number of rows and columns, then   . . . . . . . . .     A + B = B + A = . .  .   . .  . a i,j + b i,j . . . . . . . . . . If the number of columns A is the same number of rows of B , than A × B = C ( � = B × A not even possibile in some cases) , (12)   b 1 ,j b 2 ,j   � a i, 1 �    = [ A ] i [ B ] j , c i,j = a i, 2 . . . a i,n . (13)   .  . b n,j where [ A ] i is the i − th row of the matrix A and [ B ] j is the j − th column of the matrix B. I × A = A = A × I. 14 / 20

Complex numbers Matrices Matrix determinant The determinant | · | of a matrix is defined as follows: n n � � | A | = a i,j A i,j = a i,j A i,j , (14) j =1 i =1 where A i,j is the minor of the element a i,j of the matrix A given by A i,j = ( − 1) i + j | M i,j ( A ) | , (15) and M i , j ( A ) is the sub-matrix obtained eliminating from A the i − th row and the j − th column. This definition is recursive and is based on the fact that if n = 2 then � � a 1 , 1 �� a 1 , 2 � � | A | = � = a 1 , 1 a 2 , 2 − a 1 , 2 a 2 , 1 . (16) � � a 2 , 1 a 2 , 2 � 15 / 20

Complex numbers Matrices Matrix determinant cont’d Example: � �   a 1 , 1 a 1 , 2 a 1 , 3 � � � � | A | = a 2 , 1 a 2 , 2 a 2 , 3 , �   � � � a 3 , 1 a 3 , 2 a 3 , 3 � � = a 3 , 1 ( a 1 , 2 a 2 , 3 − a 2 , 2 a 1 , 3 ) − a 3 , 2 ( a 1 , 1 a 2 , 3 − a 2 , 1 a 1 , 3 ) + a 3 , 3 ( a 1 , 1 a 2 , 2 − a 1 , 2 a 2 , 1 ) . Other properties: | A | = | A ′ | B is obtained by exchanging the position of two rows (columns) of A → | B | = −| A | B is obtained by multiplying by λ ∈ R a row (column) of A → | B | = λ | A | B is obtained adding a row (column) of A with another row (column) of A multiplied by a constant → | B | = | A | C = A × B → | C | = | A || B | . 16 / 20

Complex numbers Matrices Algebra: inverse matrix A square matrix is said to be non-singular iff its determinant is different from zero. A is non-singular iff its row vectors [ A ] i ( [ A j ] ) are independents, i.e. there not exist real coefficients c i such that n � A k = c i A i , i =1 ,i � = k which means that A k , for any k = 1 ..n , can not be defined as a linear combination of the other (subset) rows of A. The same needs to hold for the columns. For square matrix the rank of A, rank ( A ) , is the number of rows (columns) linearly independent, moreover the matrix A is non-singular iff rank ( A ) = n iff | A | � = 0 . For non-square matrix A m,n it holds rank ( A ) ≤ min { m, n } . The null space of A, ker( A ) , i.e. the subspace of vectors x such that Ax = 0 , is such that dim(ker( A )) = n − rank ( A ) . 17 / 20

Complex numbers Matrices Algebra: inverse matrix cont’d For a square non-singular matrix A, the inverse matrix A − 1 is such that A × A − 1 = I = A − 1 × A. The inverse matrix can be defined as A − 1 � adj ( A ) , (17) | A | where   A 1 , 1 . . . A n, 1 . .   adj ( A ) � . .  , A adj ( A ) = adj ( A ) A = | A | . (18)  . . A 1 ,n . . . A n,n 18 / 20

Complex numbers Matrices Matrix and linear systems Consider a known vector b ∈ R n , a matrix A ∈ R n × n and unknown x ∈ R n such that Ax = b . Are there solutions? if A is non-singular, then there exists a unique solution x = A − 1 b if rank ( A ) < n there might be no solutions or even infinite of them. 19 / 20