Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Could someone please expand on

Method 9. Lagrange interpolation (page 17) at

because the summation runs from 0 to (n-1) but the eigenvalues are defined from 1 to n.

Also. is it true that this is an analytic solution? Is it practical for numerical methods? Because I have to compute exp(t*A) for SEVERAL t and its expensive. I thought that if I can get the product bit of the equation (in the paper) then I only need to "plug and chug" the exp(\lambda_j t) part (which is really cheap) for various t.

Alternatively, I could do a symbolic calculation for t,.... i have to compute exp(t*A)*v, then plot the elements of v against time.

share|cite|improve this question
See this for a discussion on the interpolation approach to defining matrix functions. – J. M. Apr 26 '12 at 4:21
The sum should probably be from $j=1$ to $n$. Often eigenvalues are indexed starting with $0$, so the author might have switched up accidentally. – anon Apr 26 '12 at 4:21
"Methods 9, 10, and 11 suffer on several accounts. They are $O(n^4)$ algorithms making them prohibitively expensive except for small $n$. If the spanning matrices $A_0,\cdots,A_{n-1}$ are saved, then storage is $n^3$ which is an order of magnitude greater than the amount of storage required by any “nonpolynomial” method..." – J. M. Apr 26 '12 at 4:27
"...Furthermore, even though the formulas which define Methods 9, 10, and 11 have special form in the confluent case, we do not have a satisfactory situation. The “gray” area of near confluence poses difficult problems which are best discussed in the next section on decomposition techniques." – J. M. Apr 26 '12 at 4:28
.... well, it seems kind of like a stupid question but.... What then should I do? My matrix is such that the diagonal elements are equal to the negative sum of the other elements along a given row (so one eigenvalue is always zero). I'm not sure how best to compute this, since its the slowest part of my computer program. Would a Schur decomposition followed by exponentiation of the middle bit be best? Where A=QUQ^-1 in the schur decomposition. The problem I have with this though, is that even if I do it, I have to repeat this process for all time values and I have to check convergence! – Squirtle Apr 26 '12 at 4:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.