Matrices and Linear Algebra

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1 Contents Quantitative methods for Economics and Business University of Ferrara Academic year

2 Contents 1 Basics

3 Contents 1 Basics

4 Contents 1 Basics

5 Contents 1 Basics

6 Contents 1 Basics

7 Part I

8 a ij := the entry on the row i and on the column j Definition A matrix is an m n array of m n scalars from R. The individual values in the matrix are called entries. Examples: A = B = The size of the array is written as m n, where Notation: m is the number of rows n is the number of columns a 11 a a 1n a 21 a a 2n A =... a m1 a m2... a mn

9 Specific cases A matrix with dimension 1 n is called row vector: x = x 1,x 2,...,x n A matrix with dimension m 1 is called column vector or simply vector: y 1 y 2 y =.. y m A unit vector is a vector of ones:

10 Special matrices If m = n, the matrix is called square. In this case we have: A matrix A is said to be diagonal if a ij = 0 for i j A diagonal matrix A may be denoted by diagd 1,d 2,...,d n where a ii = d i and a ij = 0 for i j. The diagonal matrix diag1,1,...,1 is called the identity matrix and is usually denoted by I = The diagonal matrix O = diag0,...,0 is called the zero matrix.

11 A square matrix L is said to be lower triangular if that is l ij = 0 for i < j l l 21 l 22 L =..... l n1 l n2... l nn A square matrix U is said to be upper triangular if u ij = 0 for i > j that is U = u 11 u u 1n 0 u 22 u 2n u nn

12 Definition Let A be an m n matrix. Define the transpose of A, denoted by A T, to be the n m matrix i.e. n rows and m columns with entries A T ij = a ji. In other words: a 11 a a 1n a 21 a a 2n A =... A T = a 11 a a m1 a 12 a a m2... a m1 a m2... a mn a 1n a 2n... a mn

13 Examples: We have that A = B = A T = B T = A T T = A

14 A square matrix A i.e. an n n matrix is called symmetric if a ij = a ji that is Example: The matrix A = A T = A is symmetric.

15 Equality, Addition, Scalar multiplication Two matrices A and B are equal if and only if they have the same size and a ij = b ij for all i,j. If A and B are matrices of the same size then the sum of A and B is defined by C = A+B, where c ij = a ij +b ij for all i,j. If A is any matrix and α R then the scalar multiplication B = αa is defined by b ij = αa ij for all i,j.

16 Examples π = π =

17 Matrix addition inherits many properties from R. Theorem: If A,B,C are m n matrices and α,β R, then A+B = B +A commutivity A+B +C = A+B+C associativity αa+b = αa+αb distributivity of a scalar if B = O a matrix of all zeros then A+B = A+O = A α+βa = αa+βa αβa = αβa 0A = O αo = O A+B T = A T +B T

18 Inner or scalar product Let x = x 1,...,x n and y = y 1,...,y n be two vectors. The scalar or inner product of x and y is given by x,y = x 1 y 1 +x 2 y 2 + +x n y n It is a scalar i.e. a number!!! Remark: Alternative notation for the scalar product is x,y = x y. Scalar product is defined only for vectors of the same length!!! Example: let x = 1,0,3, 1 and y = 0,2, 1,2. Then x,y = = 5

19 Matrix product Assume that A is m n and B is n p. Denote by r i A the i th row of A c j B the j th column of B The product D = AB is the m p matrix defined by that means d ij = r i A,c j B d ij = a i1 b 1j +a i2 b 2j + +a in b nj, for 1 i m, 1 j p Remark: In order to perform matrix multiplication, we need that the number of columns of A is equal to the number of rows of B!

20 Example: A = Set C = AB, then B = d 11 = r 1 A,c 1 B = = 1 d 12 = r 1 A,c 2 B = = 2 d 21 = r 2 A,c 1 B = = 11 d 22 = r 2 A,c 2 B = = 4 We obtain D = AB =

21 Properties of matrix product If AB exists, does it happen that BA exists and AB = BA? The answer is usually no. First AB and BA exist if and only if A is m n and B is n m. Even if this is so, the sizes of AB and BA are different AB is m m and BA is n n unless m = n. Example: A = 1,2 B = 1 1 AB = 1 BA = However even if m = n we may have AB BA. Example: A = B = AB = BA =

22 If AB = O, does it happen that A = O or B = O? The answer is usually no. Example: Let It happens that but and A = AB = A B B =

23 Theorem Matrix Multiplication Rules. Assume A,B, and C are matrices for which all products below make sense. Then ABC = ABC associativity AB +C = AB +AC and A+BC = AC +BC AI = A and IA = A αab = αab AO = O and OB = O AB T = B T A T

24 The simple case of 2 2 matrices Consider a 2 2 matrix a11 a A = 12 a 21 a 22 The determinant of A is defined as deta = a 11 a 22 a 12 a 21 Example: A = deta = = 2

25 The case of 3 3 matrices Consider a 3 3 matrix A = The determinant of A is defined as a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 deta = a deta 11 +a deta 12 +a deta 13 where a22 a A 11 = 23 a 32 a 33 Then a21 a A 12 = 23 a 31 a 33 a21 a A 13 = 22 a 31 a 32 deta = a 11 deta 11 a 12 deta 12 +a 13 deta 13 = a 11 a 22 a 33 a 32 a 23 a 12 a 21 a 33 a 31 a 23 +a 13 a 21 a 32 a 31 a 22

26 The same rule may be carried out by selecting any row or column of A: for each row index i i = 1,2,3, we have deta = a i1 1 i+1 deta i1 +a i2 1 i+2 deta i2 +a i3 1 i+3 deta i3 for each column index j j = 1,2,3, we have deta = a 1j 1 1+j deta 1j +a 2j 1 2+j deta 2j +a 3j 1 3+j deta 3j where A ij is the submatrix obtained from A after deleting the i th row and the j th column of A itself.

27 For instance, consider Then we have deta = 1 det It follows that A = det det 1 2 deta = = 4 Any row or column can be selected in order to evaluate the determinant: try to work on the third column and verify the result!

28 The general case of n n matrices Consider a square n n matrix a 11 a a 1n a 21 a a 2n A =.. a n1 a n2... a nn The determinant of A is defined as deta = a i1 1 i+1 deta i1 +a i2 1 i+2 deta i2 + +a in 1 i+n deta in for each row index i i = 1,2,...,n, or equivalently as deta = a 1j 1 1+j deta 1j +a 2j 1 2+j deta 2j + +a nj 1 n+j deta nj for each column index j j = 1,2,...,n, where A ij is the submatrix obtained from A after deleting the i th row and the j th column of A itself. n n

29 Some properties of the determinant Consider an n n matrix A. If two rows of A are interchanged to obtain B, then detb = deta If any row of A is multiplied by a scalar α, the resulting matrix B has determinant detb = αdeta If any two rows of A are equal, then deta = 0 If A has two rows equal up to a multiplicative constant, then deta = 0

30 If B is obtained by summing to a row or a column of A a linear combination of the other rows or columns of A, respectively, then deta T = deta detab = deta detb detαa = α n deta detb = deta If A = diagd 1,d 2,...,d n, then deta = d 1 d 2...d n. It follows that deti = = 1

31 Definition Let A be an n n square matrix. A is invertible if there exists a matrix B that satisfies the following relationship AB = BA = I Under the assumption that B exists, it is unique!!! We denote B by A 1 inverse of A. Then AA 1 = A 1 A = I It is possible to prove that: A is invertible i.e. A 1 exists deta 0 in that case, A is non singular.

32 Simple case: 2 2 matrices Consider n = 2 and a11 a A = 12 a 21 a 22 Assume that deta 0. Then A is invertible and A 1 = 1 a22 a 12 deta a 21 a 11 Example: 2 1 A = deta = A 1 = =

33 Some properties of the inverse Consider two n n matrices A, B and a scalar α R. AB 1 = B 1 A 1 αa 1 = α 1 A 1 deta 1 = 1/detA why?

34 An application: solution of linear systems Data: n n matrix A, n 1 column vector b Unknown:: n 1 column vector x Problem: Find x such that Ax = b that is equivalent to solve the following system of linear equations a 11 x 1 +a 12 x 2 + +a 1n x n = b 1 a 21 x 1 +a 22 x 2 + +a 2n x n = b 2. a n1 x 1 +a n2 x 2 + +a nn x n = b n Assume that A is non-singular i.e. deta 0, then the previous system has got a unique solution, which is obtained as thus A 1 Ax = A 1 b x = A 1 b

35 Example: The following system { 3x1 +2x 2 = 1 4x 1 +3x 2 = 2 is equivalent to with A = Ax = b b = 1 2 We have deta = 1 0, then A is invertible and A = 4 3 The solution of Ax = b is x = A b = = 7 10

36 Definition If a vector v 0 satisfies the equation Av = λv, for some scalar λ, then λ is said to be an eigenvalue of the matrix A, and v is said to be an eigenvector of A corresponding to the eigenvalue λ. Example: If 2 3 A = 3 2 and then Av = v = 1 1 = 5 1 = 5v So λ = 5 is an eigenvalue of A and v is an eigenvector corresponding to this eigenvalue.

37 The definition of eigenvector requires that v 0. The reason for this is that if v = 0 were allowed, then any number λ would be an eigenvalue since the statement A0 = λ0 holds for any λ. On the other hand, we can have λ = 0 and v 0. Example: If and then A = v = Av = = 0v So λ = 0 is an eigenvalue of A and v 0 is an eigenvector corresponding to this eigenvalue.

38 How do we find the eigenvalues and eigenvectors of a matrix? Suppose v 0 is an eigenvector of A. Then for some λ R, we have Av = λv. Then Av λv = 0 or, equivalently, This happens when A λiv = 0 deta λi = 0 Then: λ is an eigenvalue of A deta λi = 0 pλ = deta λi is a polynomial function of λ. pλ = 0 is called the characteristic equation of the matrix A

39 An example Consider Then and A = A λi = λ λ pλ = deta λi = 1 λ 2 9 = λ 2 2λ 8 = λ 4λ+2 From pλ = 0 we obtain that λ 1 = 4 and λ 2 = 2 are the eigenvalues of A. As an exercise, we want to find an eigenvector corresponding to λ 1. Thus we have to solve the linear system A 4Iv = 0, i.e v =

40 It is equivalent to Basics v1 v 2 = 0 0 This leads to the two equations 3v 1 +3v 2 = 0 and 3v 1 3v 2 = 0. Notice that the first equation is a multiple of the second one, so there is really only one equation to solve 3v 1 3v 2 = 0 The general solution to the homogeneous system is given by v 1 = v 2 = c therefore, all vectors v such that c 1 v = = c c 1 where c 0 is arbitrary are eigenvectors of A corresponding to λ 1 = 4.

41 Conclusion: In general, eigenvectors are not unique! If v is an eigenvector for A corresponding to a given eigenvalue λ, then so is cv, for any number c 0. As an exercise, find the eigenvectors of A corresponding to the other eigenvalue λ 2 = 2 in the previous example.

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