maximum eigenvalue of a matrix with rank one update

There is a (real symmetric) matrix A(t) updated recursively by

A(t)=(t-1)/t*A(t-1)+1/t*a(t)*a(t)',

where a(t) is a column vector.

Suppose the maximum eigenvalue of A(t-1) is known; then is there any efficient method to compute the maximum eigenvalue of A(t)?

Many thanks!

Yang

[Matt J]

Is t>1?

[Yang]

Yes. "t" actually represents time or iteration.

Answer 1

[Roger Stafford]

Yang, you appear to expect some direct relationship between the maximum eigenvector of A(t-1) and that of A(t), possibly also involving vector a(t). However, the maximum eigenvalue of A(t) is actually dependent on all the eigenvalues and all the eigenvectors of A(t-1), so the relationship would have to be very complicated. I see no better way of determining the maximum eigenvalue of A(t) than calling on the 'eig' or 'eigs' function directly, in spite of its being the result of a recursion.

If vector a were not dependent on t, the limiting case as t approaches infinity, would be just the rank one matrix a*a' itself with its single nonzero eigenvalue and corresponding eigenvector proportional to vector a. However you have presumably used the a(t) notation to indicate that a changes with changing t, so even that is untrue.

[Yang]

Roger, thank you very much for your response.

I have also realized that there is not a closed-form expression connecting the largest eigenvalue of A(t) and A(t-1), unless some special structures are assumed.