How Do You Find the Inverse of an Exponential Matrix?

[chester20080]

We have a matrix with dimension NxN.For some m belongs to N,m0 we have A^m0=0.We consider the exponential matrix e^A=I+A+A^2/(2!)+A^2/(3!)+A^m/(m!).Find the inverse matrix of e^A.
I tried to write the e^A=e^A(m0)+A^m/(m!) or (e^A)^(-1)=( I+A+A^2/(2!)+A^2/(3!)+A^m/(m!))^(-1)=I+A^(-1)(I+2/3A(-1)+1/(m!)(A^(-1))^(m-1)) or (e^A)^(-1)=(e^(-1))^A=(1/e)^A=1/(e^A)=1/( I+A+A^2/(2!)+A^2/(3!)+A^m/(m!)),but I don't know how to proceed any further.How am I supposed to end up with m0 and the result of (e^A)^(-1) will equal to something(A),(A^(-1)?Please help.

 

[Mandelbroth]

We have a matrix with dimension NxN.For some m belongs to N,m0 we have A^m0=0.We consider the exponential matrix e^A=I+A+A^2/(2!)+A^2/(3!)+A^m/(m!).Find the inverse matrix of e^A.
I tried to write the e^A=e^A(m0)+A^m/(m!) or (e^A)^(-1)=( I+A+A^2/(2!)+A^2/(3!)+A^m/(m!))^(-1)=I+A^(-1)(I+2/3A(-1)+1/(m!)(A^(-1))^(m-1)) or (e^A)^(-1)=(e^(-1))^A=(1/e)^A=1/(e^A)=1/( I+A+A^2/(2!)+A^2/(3!)+A^m/(m!)),but I don't know how to proceed any further.How am I supposed to end up with m0 and the result of (e^A)^(-1) will equal to something(A),(A^(-1)?Please help.

I cannot read a word of that. Please put a little bit of formatting into that.

 

[chester20080]

Sorry for that.For convenience I have a picture of the above.

 

[D H]

Hint: What is the general form for the inverse matrix of the matrix exponential?

 

[chester20080]

I don't know.We haven't learned anything for exponential matrices.Should I consider a function or something like that?But we haven't been taught how to combine matrices and functions as well.

 

[D H]

For real and complex numbers, the exponential function is defined by $\exp(z) = \sum_{r=0}^{\infty} z^r/r!$ The matrix exponential is defined analogously: $\exp(A) = \sum_{r=0}^{\infty} A^r/r!$ Note that this assumes that $A^r$ makes sense. It doesn't make sense for an NxM matrix with N≠M. The matrix exponential is only defined for square matrices. It makes sense for square matrices because even though matrix multiplication is not commutative in general, it is for powers of a matrix: $AA^n=A^nA$, and thus $A^{n+1}$ is well-defined.

A useful identity for any two complex numbers u and v is that $\exp(u+v) = \exp(u)\exp(v)$. This identity is not necessarily the case for square matrices: $\exp(A+B)$ is not necessarily equal to $\exp(A)\exp(B)$ for any two NxN matrices A and B. It does hold if A and B commute.

You can use this identity for commuting matrices to find the inverse of $\exp(A)$.

Where did you come across this problem?

 

[chester20080]

Our professor in the university has this in a project we have to give in after the holidays for linear algebra...So the inverse of exp(A) will be found as a function of A,inverse A...what?But,yet,how will I proceed from exp(A)*exp(-A)=I (1),in order to find exp(-A),which is the inverse matrix to find?What will happen to the term A^m/(m!)?And first of all how will I show that there is the inverse matrix of exp(A) so as to begin from (1)?

 

[D H]

This is a *lot* simpler than you are making it out to be. You apparently already know that $(\exp(A))^{-1} = \exp(-A)$. So compute $\exp(-A)$. This isn't that hard! What's the general case? Next use the fact that your matrix A is nilpotent.

 

[chester20080]

Something like this:exp(-A)= I-A+A^2/2-A^3/6+...+-A^m1/m1!,where m1 is the value right before the value for which A^m=0?And now what?That's it?

 

[D H]

That's it.

 

[chester20080]

So the inverse of exp(A) equals with something that has the term m1?Why I don't feel it right?Are you sure?

 

[Dick]

So the inverse of exp(A) equals with something that has the term m1?Why I don't feel it right?Are you sure?

What doesn't feel right? Suppose $A^2=0$. Then exp(A)=I+A. exp(-A)=I-A. (I+A)(I-A)=I, can you see why?

 

[chester20080]

Yes,but the answer took me only a couple of lines to write.I mean it can't be that simple!And we have to take cases for m1 in order to determine the sign before A^m1,right?

 

[Dick]

Yes,but the answer took me only a couple of lines to write.I mean it can't be that simple!And we have to take cases for m1 in order to determine the sign before A^m1,right?

The m1 term is (-A)^m1/m1!. If you want to write that as a power of A, then you could write (-1)^m1*A^m1/m1!. No need for cases. If you want to make the problem a LOT harder you could try to prove by expanding terms that exp(A)*exp(-A)=I. But I don't think you have to do that.

 

[chester20080]

Ok,thank you!

 

[D H]

If you want to make the problem a LOT harder you could try to prove by expanding terms that exp(A)*exp(-A)=I. But I don't think you have to do that.

It's not that hard if someone has already gone to the trouble of showing that if matrices A and B commute then $\exp(A+B)=\exp(A)\exp(B)$. Set B=-A. This obviously commutes with A, so $\exp(A+B)=\exp(A)\exp(B)$ applies. With $B=-A$, the right hand side becomes $\exp(A)\exp(B) = \exp(A)\exp(-A)$ while the left hand side becomes $\exp(A+B)=\exp(A+(-A))=\exp(0)=I$. Thus $\exp(A)\exp(-A)=I$, or the inverse of $\exp(A)$ is $\exp(-A)$ -- valid for any square matrix A.

Showing that $\exp(A+B)=\exp(A)\exp(B)$ for any two matrices A and B that commute -- now that takes some work.