Most positive eigenvalue and corresponding eigenvector of a real symmetric matrix

Question

Is there any way to compute the most positive eigenvalue and eigenvector using power method?

Illustrated as follows,

for example, let

| 1 0 0 | A= | 0 -4 0 | | 0 0 3 |

whose eigenvalues are apparently 1, -4 and 3. When I apply power method to A, I end up finding eigenvalue with highest magnitude and hence I get 4 ( or -4 ) as result. But I need a way to find the most positive eigenvalue, i.e., 3 in this example.

Answer 1

Found out a way...

Say the power method returns the eigenvalue of A with highest magnitude but is negative, which shall be represented by 'b', then we try finding out eigenvalues of the matrix (A-bI). Eigenvalues of that matrix would be increased by a value of abs(b), leaving the eigenvectors unchanged. Hence the eigenvalues of the new matrix would all be essentially non-negative and thus applying power method to find the dominant eigenvalue would give us the most positive eigenvalue, but its magnitude increased by abs(b). Thus our required most positive eigenvalue of A would be obtained by subtracting abs(b) from the eigenvalue found out from new matrix.

Illustrated as below,

A - matrix for which we need to find most positive eigenvalue and corresponding eigenvector b - eigenvalue of A with highest magnitude but is negative, found using power method. Please note 'b' is negative B=A-b*I where I identity matrix b' - eigenvalue of B with highest magnitude (essentially non-negative) again found using power method. our required eigenvalue be 'req', i.e., the most positive eigenvalue. req = b'+ b

or

req = b' - abs(b)

eigenvector corresponding to 'req' would be the required eigenvector