## HW FOUR

1. 4.2

(a) This matrix is not positive definite, see (b) below

(b) Eigenvalues are and , so the matrix is not positive definite

(c) Power method will converge to the dominant eigenvalue, which is .

(d) Since dominant eigenvalue is , and we know that the other eigenvalue is such that , we can make a shift by . Indeed, performing power iteration to the will converge to the nondominant eigenvalue.

2. 4.4

(a) This matrix is not positive definite

(b) Using Rayleigh quotient we find eigenvalues , and .

(c) Power method will converge to the dominant eigenvalue and the vector, which will be a linear combination of two different eigenvectors corresponding to the dominant eigenvalue.

More precisely, where and are three different eigenvectors.

Cross multiplying by corresponding eigenvectors, we obtain   Performing power iteration on we will obtain the vector 3. 4.5

(a) The power will fail since it will not be able to “decide” whether to converge to eigenvector corresponding to eigenvalue or . The resulting vector will therefore converge to the linear combination of the two eigenvectors corresponding to eigenvalue and , and will alternate its direction at each of the steps.

(b) Consider the matrix , and use power method and inverse power method to compute eigenvalues and .

(c) Using will converge to the maximum positive eigenvalue of matrix . The convergence will be very slow, the error will be multiplied by When you make a shift by , the error will be multiplied by at each iteration.

4. (4.8)

(a) Eigenvalues are and . Power method will presumptuously converge to linear combination of eigenvectors corresponding to these eigenvalues and will alternate its direction.

(b) The equation (4.13) will reduce to This formula does not have desired assymptotic behavior at (c) When we choose , we are shifting eigenvalues from and to and . The eigenvalues of will be different for nonzero . Choose to find eigenvalue and corresponding eigenvector for the eigenvalue of . Choose to find eigenvalue and corresponding eigenvector to .

5. 4.12

It is easier to work with or matrix first to gain intuition. For example, type in matlab

In what follow, we assume that is even.

To find eigenvalues, write . To calculate this determinant, expand it in the first raw. The second determinant should be expanded in the last raw. The result is The roots of this are , and .

To calculate eigenvectors, write in components.

The eigenvectors corresponding to are    The eigenvector corresponding to is Similarly, the eigenvector corresponding to is 6. 4.15

(a)   So there are four eigenvalues close by absolute value to .

(b) Power method will fail since four eigenvalues are very close to each other, and , which is very close to 1

(c) If you do a shift, and , then the dominant eigenvalue is (d) The next biggest eigenvalue in the shifted matrix will be , so the error will be reduced each time by Now solve which gives .

(e) To converge to one can choose . Then the eigenvalue will be the dominant one, so power method will converge to it.