Introduction to linear algebra 24--Markov matrix, Fourier series

This paper is Gilbert Strang's introductory course note on linear algebra. Course Address: http://v.163.com/special/opencourse/daishu.html 24th class: Markov Matrix, Fourier seriesThis lecture explains the application of eigenvalue, Markov matrix theory, Fourier series (it is the ingenious application of projection matrix)
Markov matricesSatisfies two properties: 1) All elements are greater than or equal to 0; 2) The columns of all matrices are added equal to 1. (This property guarantees a eigenvalues of 1, why.) The power of Markov matrix is Markov matrix. Markov matrices are associated with probabilistic ideas, which are related to probabilities in nature. The eigenvalues and eigenvectors of Markov matrices are considered. It's steady state, what is steady state. What conditions the eigenvalues meet. In fact, the stationary problem is the eigenvector problem of that eigenvalue.
The characteristic value of the Markov matrix: 1) λ=1 is a characteristic value of it, which corresponds to the X1 of all the elements of the eigenvector is non-negative, 2) all other eigenvalues |λi|<1; so the steady state of the power is c1x1, The steady state is determined by a characteristic vector with a eigenvalue of 1 (because the steady state is determined by the eigenvector from the UK expansion formula).
Why is it that the column addition of the matrix equals 1 guarantees that there is a characteristic value of 1. Assuming there is already a eigenvalue of 1, then a-1i, the Markov matrix translates a unit matrix, and if proving a-1i is singular, then 1 is indeed its eigenvalue. The a-1i matrix adds 0 to each column, indicating that the linear combination of line vectors can get 0, which indicates the linear correlation, singular matrix. Vector (1 1 1) in its left 0 space, a eigenvector with a characteristic value of 1 is in its 0 space (because of ax1=x1, (a-i) x1=0). What is the relationship between a and a T's eigenvalues. They are the same. Because the matrix a T is equal to the determinant of a, it can be proven.
Markov matrix Application A is a Markov matrix, always unchanged, representing the population migration of two states. The total number remained the same. Time t from K to K+1, 0.9 of Californians left +0.2 of Massachusetts migrated in = Now Californians. 0.1 of Californians migrated out of +0.8 of the Massachusetts people left = Now the Massachusetts people. Consider steady state. Assuming the initial state of t=0, 0 people in California, Massachusetts 1000 people. What is the population change after K-step? To answer this question, we need to consider the eigenvalues and eigenvectors of a. Solution obtained: λ1=1,x1= (2 1); λ2=0.7,x2= (-1 1). Then the solution gets C1=1000/3,C2=2000/3.
Fourier seriesThe projection problem leads to the projection problem of the Fourier series with n standard orthogonal basis qnxn, the base vector q1,q2...qn, then any vector v in space can be obtained by this standard orthogonal base class linear combination: V=x1q1+x2q2+...+xnqn, now to know X1, or x2 is how much, Can be expressed by the expansion of the vector to expand to the standard orthogonal base, which is in the projection, because this group of base is a standard orthogonal base, so x1,x2. has a computational formula for the solution. Seek x1 When the Q1 and the formula in any one of the product will be able to get X1. Thus the coefficients of each base vector in the expansion of V can be obtained. Can also be expressed in matrix form: that is QX=V,X=Q-1V=QTV Fourier series: Known f (x) =a0+a1cosx+b1sinx+a2cos2x+b2sin2x+ ..., infinite dimension, but the key nature or orthogonal, the orthogonality of sin and Cos is still established, This makes the Fourier series meaningful, which is the Fourier series. The Fourier series compares the equation above the vector space, is the function space f (x) replaces the vector space V, the orthogonal function replaces the orthogonal vector q1,q2 ... The meaning of the orthogonal function here is that the inner product of two functions equals 0. (instead of summing with integrals) now with the infinite orthogonal base of the function space, what needs to be done now is to expand the function to the base, requiring a number of coefficients. With the same vector space, the equation of Fourier series coefficients can be obtained by multiplying the orthogonal base components on the left and right sides.