Dominant Eigenvalues & The Power Method

Eigenvalues & Eigenvectors In linear algebra we learned that a scalar  is an eigenvalue for a square nn matrix A if there is a non-zero vector w such that Aw = w, we call the vector w an eigenvector for matrix A. The eigenvalue acts like scalar multiplication instead of matrix multiplication for the vector w. Eigenvalues are important for many applications in mathematics, physics, engineering and other disciplines. The Dominant Eigenvalue A nn matrix A will have n eigenvalues (some may be repeated). By the dominant eigenvalue we refer to the one that is biggest in terms of absolute value. This would include any eigenvalues that are complex. The matrix A to the right has as its eigenvalues the set of numbers {1,-4,2} (i.e. it is upper triangular). In absolute value this set is {|1|,|-4|,|2|}={1,4,2}. Since -4 is the largest in absolute value we say that -4 is the dominant eigenvalue. The problem we want to solve is that if we are given a matrix A can we estimate the dominant eigenvalue?

zk+1 = dominate term in Awk wk+1 = (1/zk+1) Awk The Power Method The Power Method works for matrices sort of like how the fixed point method works for functions. The iteration step is a bit different though. To explain how it works we need to introduce a bit of terminology. The dominant term of a vector v is the term that has the greatest absolute value (careful: it is the term itself not the absolute value of the term). If there are two terms that have the same absolute value you can pick either one for our purposes. dominant term=7 dominant term=-8 dominant term=-10 dominant term=6.5 The algorithm consists of the following steps. Start with an initial vector w0. Let the approximation for dominant eigenvalue be z0 the dominant term in w0. Use the iteration to the right: zk+1 = dominate term in Awk wk+1 = (1/zk+1) Awk

To get the next approximation for the dominant eigenvalue multiply the previous eigenvector by the matrix A and take that vectors dominant term. To get the next approximation for the eigenvector divide the product by it dominant term. The vector w0 given to the right is often used as the initial vector. Example: Apply the power method for 3 iterations to find z3 and w3 for the matrix A given to the right.

Power Method Convergence The Power method will not converge for a real matrix A the power method will converge to the dominate eigenvalue if the dominant eigenvalue is a real number. If the dominant eigenvalue is a complex number an initial vector with complex entries would need to be used. If the dominant eigenvalue is repeated it will find it. How this convergence can be seen is as follows. Given a nn matrix A with n eigenvalues 1,2,3,…,n with 1>2>3>…>n (i.e. 1 is the dominant eigenvalue) find a corresponding basis of eigenvectors w1,w2,w3,…,wn. Let the initial vector w0 be a linear combination of the vectors w1,w2,w3,…,wn. w0 = a1w1+a2w2+a3w3+…+anwn Aw0 = A(a1w1+a2w2+a3w3+…+anwn) =a1Aw1+a2Aw2+a3Aw3+…+anAwn (replace with eigenvalues) =a11w1+a22w2+a33w3+…+annwn Akw0 =a1(1)kw1+a2(2)kw2+…+an(n)kwn (repeat for powers of A) Akw0/(1)k-1 =a1(1)k /(1)k-1 w1+ a2(2)k /(1)k-1 w2 +…+an(n)k /(1)k-1 wn

  For large values of k (i.e. as k goes to infinity) we get the following: At each stage of the process we divide by the dominant term of the vector. If we write w1 as shown to the right and consider what happens between two consecutive estimates we get the following. Dividing by the dominant term gives something that is approximately 1.