Textbook Readings Unit Three Markov Matrices 8 3 Markov matrices are positive matrices Every entry is greater than zero Every column adds to one The largest eigenvalue is real and positive and so is its eigenvector The maximum eigenvalue controls the powers of A Example Find i ii where D is a diagonal matrix and the column of U are the eigenvectors of A Start by nding the eigenvalues of A and their corresponding eigenvectors Note Markov matrices always have an eigenvalues of one so start there and nd the eigenvector by nding the solution to A I x 0 use the trace to nd the other eigenvalue Now we have U and U and D and we know that A UD U so to get through this calculation quickly start on the far right and work your way to the right so we can plug in for when n is in nity to get the steady state Raising A to a power We can write A S S where S is the eigenvector matrix and is the diagonal eigenvalue matrix When we raise A to a power we get A S S The steady state is equal to the eigenvector corresponding to the eigenvalue of one The Consumption Matrix These matrices relate input and output Example Find a vector p such that p Ap y We can rewrite that as p I A y p I A y Now the question is whether I A is invertible We need to nd when I A is a non negative matrix If 1 then I A has negative entries If 1 then I A fails to exist If 1 then I A is nonnegative as desired Differential Equations 6 3 Differential equation Au u t u t u has the solution u t Ce If u t and v t are solutions so is Cu t Dv t We know this is true by using the geometric series I A I A A A This series converges if all eigenvalues of A have 1 We will need n constants like C and D to match the n components of u 0 Our rst step is to nd n pure exponential solutions u e x by using Ax x Solve linear constant coef cient equation by exponentials e x when Ax x is an eigenvalue of A and x is the eigenvector All components of this special solution u e x share the same e The solution grows when eigenvalues are greater than one and decays when eigenvalues are less than one Example Solve u t Au starting from u 0 Complete solution Example Solve u t Au knowing the eigenvalues of A are 1 2 and 3 Step one nd the eigenvectors associated with the given eigenvalues Step two using u 0 nd the coef cients in for each of the eigenvectors Step three Find the exponential solutions Step four Combine these exponential solutions and the coef cients to get u t Overall Sc u 0 where S is the eigenvector matrix and c are the coef cients Second Order Equations Second order equations contain second derivatives Example 0 my by ky That equation is linear with constant coef cients m b k We can substitute y e while keeping in mind that each derivative brings down a factor Example 0 e m b k This equation for has two roots for the eigenvalues This means the equation for y has two solutions Their combinations c y c y give the complete solution unless the two eigenvalues are equal In linear algebra we turn the scalar equation into a vector equation Example This equation converts to The Exponential of a Matrix Matrix exponential Its t derivative is Its eigenvalues are Diagonalize Review of the Key Ideas The equation u t Au is linear with constant coef cients starting with u 0 Its solution is usually a combination of exponential involving each and x u t The constants c c are determined by u 0 Sc The equation u t approaches zero meaning it approaches stability if every eigenvalue is negative The solution is always u t with the matrix exponential e Equations with y t reduces to u t by combining y and y into u Symmetric Matrices 6 4 For projection onto a plane in R the plane is full of eigenvectors where Px x The other eigenvectors are perpendicular to the plane where Px 0 The eigenvectors are real 1 1 0 Three eigenvectors can be chosen perpendicular to each other Symmetric Matrices The eigenvectors of P P are perpendicular unit vectors Question What is special about Ax x when A is symmetric We are looking for special properties of the eigenvalues and the eigenvectors when A A The diagonalization A S S will re ect the symmetry of A Symmetric matrices have only real eigenvalues The eigenvectors can be chosen orthonormal These n orthonormal eigenvectors go into the columns of S Every symmetric matrix can be diagonalized Its eigenvector matrix S becomes an orthogonal matrix Q Orthogonal matrices have Q Q Note When we choose orthonormal vectors S S is in its special and particular form Q Q for symmetric matrices Sum of rank one matrices A The n products x x are projection matrices so we get that A P P where P is the projection onto the eigenspace Complex Eigenvalues of Real Matrices A non symmetric matrix can easily produces eigenvalues and eigenvectors that are complex In this case Ax x and its complex conjugate Ax x are not the same Example Because A is an orthogonal matrices we know that 1 Eigenvalues versus Pivots The eigenvalues of A are very different from the pivots For eigenvalues we solve det A I 0 For pivots we use elimination Product of the pivots is equal to the product of the eigenvalues which is also equal to the determinant For symmetric matrices the pivots and the eigenvalues have the same sign A LDL where L is a lower triangular matrix and L is a upper triangular matrix All Symmetric Matrices are Diagonalizable When no eigenvalues of A are repeated the eigenvectors are sure to be independent so A can be diagonalized There are always enough eigenvectors to diagonalize A Every square matrix factor into A QTQ where T is upper triangular Review of the Key Ideas A symmetric matrix has real eigenvalues and perpendicular eigenvector Diagonalization becomes A Q Q with an orthogonal matrix Q All symmetric matrices are diagonalizable even with repeated eigenvalues The signs of the eigenvalues match the signs of the pivots when A is symmetric Every square matrix can be triangularized by A QTQ Positive De nite Matrices 6 5 Positive De nite Matrices symmetric matrices with positive eigenvalues All eigenvalues are positive All pivots are greater than zero All determinants are greater than zero A has the form B B for some B with independent columns x A x 0 except x 0 If A and B are symmetric positive de nite so is A B x A B x is simple x Ax x Bx If the columns of B are independent …
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