# How to calculate expm of a badly scaled matrix?

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alex corno on 4 Jun 2013
I'm using expm for a matrix with big and small coefficients and I get NaN in the result. Is there a method to get the good matrix exponential even if the matrix is badly scaled? Normally there is a exponential for every matrix.
ALEX
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alex corno on 4 Jun 2013
I think I have a different problem. My matrix isn't even invertible, there are two zero lines, moreover there are some coefficients that are of order of 1e6. In my opinion at some point Matlab is doing some inversion that's why expm doesn't work properly with badly scaled matrices. In my problem I changed the units of physical parameters (distance -> µm Stresses -> kPa) and now it works. But I'm looking for some existent function (transformation) which would enable to calculate expm for any matrix (apart for huge numbers that shouldn't appear in a physical problem there's no chance that e^100 will be a coefficient in a transfer matrix)
Alexandru

Richard Brown on 5 Jun 2013
In particular, have you tried the alternative versions expmdemo1 and expmdemo3 ?

Matt J on 4 Jun 2013
Edited: Matt J on 4 Jun 2013
I think I have a different problem. My matrix isn't even invertible, there are two zero lines, moreover there are some coefficients that are of order of 1e6. In my opinion at some point Matlab is doing some inversion that's why expm doesn't work properly with badly scaled matrices.
No, it's not an invertibility issue. The reason for the NaNs is the big 1e6 order numbers. Below is an example of a non-invertible matrix on which EXPM works fine.
>> expm(diag([1,0]))
ans =
2.7183 0
0 1.0000
Matt J on 4 Jun 2013
Edited: Matt J on 4 Jun 2013
Even if the matrix is well-scaled, you could have the same problem, for example:
>> A=diag([1e6,1e6]); %a well-scaled matrix
>> expm(A)
ans =
NaN NaN
NaN NaN
The bottom line is, you must avoid matrices with large eigenvalues, and then things should be fine. Otherwise, they will not be fine.
There are any number of transformations you could consider to reduce the magnitude of the eigenvalues. Dividing the matrix by a large number is one way, but I can't know if that suits your application.

alex corno on 5 Jun 2013
Thank you for your help. I think the function balance() is what I was looking for.
Alexandru

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