Paul Schrimpf. September 10, 2013

Transcription

1 Systems of UBC Economics 526 September 10, 2013

2 More cardinality Let A and B be sets, the Cartesion product of A and B is A B := {(a, b) : a A, b B} Question If A and B are countable is A B countable?

3 More cardinality Theorem If A and B are countable, then so is A B Proof A and B are countable, so we can write A = {a 1, a 2, } and B = {b 1, b 2, } Consider (a 1, b 1 ) (a 1, b 2 ) (a 1, b 3 ) (a 2, b 1 ) (a 2, b 2 ) (a2, b 3 ) (a 3, b 1 ) (a 3, b 2 ) (a 3, b 3 ) Count along the diagonals

4 More cardinality Question If A 1, A 2, A 3, are each countable is n=1 A n countable?

5 More cardinality Theorem If {A n } are countable, then so is n=1 A n Proof A n is countable, so we can write A n = {a 1n, a 2n, } Consider a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 Count along the diagonals

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7 Systems of Example: In general: 5x 1 7x 2 =9 8x 1 + x 2 =0 a 11 x 1 + a 12 x a 1n x n =b 1 a 21 x 1 + a 22 x a 2n x n =b 2 a m1 x 1 + a m2 x a mn x n =b m,

8 Coefficient matrix Coefficient matrix A a 11 a 1n x 1 = a m1 a mn x n Ax =b b 1 b m Augmented coefficient matrix  a 11 a 1n b 1  = = ( Ab ) a m1 a mn b m

9 Let s t = Example: Markov model of employment { 1 if employed at time t 0 if unemployed at time t Random process described by P(s t s t 1, s t 2, ) Markovian: P(s t s t 1, s t 2, ) = P(s t s t 1 ) Probability of being employed tomorrow only depends on whether you re employed today and not the more distant past Stationary distribution: q stationary if when P(s t = 1) = q(1) and P(s t = 0) = q(0) today, then also have P(s t+1 = 1) = q(1) and P(s t+1 = 0) = q(0) tomorrow q(s) = P(s s 0 )q(s 0 ) s 0 {0,1}

10 Example: Markov model of employment Stationary distribution satisfies: q(1) =P(1 1)q(1) + P(1 0)q(0) q(0) =P(0 1)q(1) + P(0 0)q(0) 1 =q(1) + q(0) system of for unknowns q(1) and q(2) Coefficient matrix and augmented coefficient matrix =?

11 Questions to be answered: Given a system of : 1 Does any solution exist? 2 How many? 3 How can a solution be computed?

12 Equation/row operations Familiar with solving by: Substitution Elimination Use equation (row) operations that preserve set of 1 Multiply an equation by a non-zero constant, 2 Add a multiple of one equation to another, and 3 Interchange two

13 Row echelon form Row echelon form: each row begins with more zeros than the row above it or the row is all zeros Examples: ( ) a11 a 12 0 a 22 ( ) a11 a 12 a a 23 Once a system is in row echelon form it is easy to solve by back substitution

14 Systematic way to transform system of to row echelon form 1 Identify the first column to contain any non-zero elements, call this column c 2 Interchange rows so that a nonzero entry appears at the top of column c 3 Add a multiple of the first row to each of the rows below so that the entries in column c below the first row are zero 4 Repeat 1-2 on the submatrix consisting of the lower right part of the original matrix below the first row and to the right of column c Stop if this submatrix has no columns or has no rows

15 : example x 1 + x 2 + 3x 3 = 0 2x 1 + 3x 2 + 7x 3 = 9 3x 1 + 5x x 3 = 1

16 Transform the following system into row echelon form: x + 2y z =2 4y + z =5 2x 4y + 2z =1

17 rank of A is the number of nonzero rows in its row echelon form Is rank well-defined? Lemma The rank of a matrix A is always less than or equal to the number of columns of A and less than or equal to the number of rows of A Lemma Let A be a coefficient matrix and  be an augmented coefficient matrix Then ranka rankâ

18 Theorem ( ) A system of with coefficient matrix A and augmented coefficient matrix  has a solution (perhaps more than one) if and only if ranka = rankâ Proof See notes

19 Consider the system: 4y + z =5 x + 2y z =2 8y 2z = 10 What if b 3 10?

20 Multiple means infinite Lemma Suppose x 1 and x 2 are two distinct to the system of Ax = b Then the system of has (uncountably) infinitely many

21 Solution existence for any b Theorem (Solution existence) A system of with coefficient matrix A will have a solution for any choice of b 1,, b m if and only if ranka is equal to the number of rows of A Corollary For any system of with more than variables, (ie an overdetermined system) there exists a choice of b such that no exist

22 Uniqueness Theorem (Solution uniqueness) Any system of with coefficient matrix A has at most one solution for any b 1,, b m if and only if ranka equals the number of columns of A Corollary If ranka is less than the number of columns of A then either no exists or multiple exists Corollary A is nonsingular (always has a unique solution) if and only if A has an equal number of columns and rows (A is square) and has rank equal to its number of columns (or rows)

23 Summary ranka = rankâ ranka < rankâ ranka = columns(a) ranka < n 0 X 1 X X

24 y x z y x z y x 5 10 Systems of Sets of 2x + y = 1 x y z = 0 2x + y + 3z = 1 2x y + z = 2

25 Definition The set S R n is called a subspace if it is closed under (i) scalar multiplication and (ii) addition in other words, if (i) for every (x 1,, x n ) S and a R, we have (ax 1, ax 2,, ax n ) S, and (ii) for every (x 1,, x n ) S and (y 1,, y n ) S, we have (x 1 + y 1,, x n + y n ) S

26 Definition A set of vectors in R n, {x j = (x j 1,, xj n)} J j=1, is ly independent if the only solution to is c 1 = c 2 = = c J = 0 J c j x j = 0 j=1 Definition The dimension of a subspace S R n is the cardinality of the largest set of ly independent elements in S

27 Theorem (Rouché-Capelli) A system of with n variables has a solution if and only if the rank of its coefficient matrix, A, is equal to the rank of its augmented matrix, Â If a solution exists and ranka is equal to its number of columns, the solution is unique If a solution exists and ranka is less than its number of columns, there are infinite In this case the set of is of the form 1 {s + x R n : s S and Ax = b} where S is the subspace of dimension n ranka defined by S = {s R n : As = 0} and x is any single solution to Ax = b 1 A set of this form is called an affine subspace It is a subspace that has been shifted so that it no longer necessarily contains the origin

28 Example: Markov model of employment Employment s t {u, e} Markovian: P(s t s t 1, s t 2, ) = P(s t s t 1 ) Stationary distribution: (π e, π u ) such that if s t = e with probability π e and u with probability π u, then s t+1 has same distribution, ie P(e e)π e + P(e u)π u =π e P(u e)π e + P(u u)π u =π u

29 Example: Markov model of employment Stationary distribution is a solution to: Questions: (p ee 1)π e + p eu π u =0 p ue π e + (p uu 1)π e =0 π e + π u =1 1 Does any solution exist? 2 How many exist? 3 How can a solution be computed?