# Exponential of (Markov Chain) Transition matrix

1. Oct 6, 2009

### NewStudent200

Hi,

I have a (markov chain) transition matrix X which I understand. In particular each row of this matrix sums to 1.

I have used this transition matrix to construct it's generator, Y. I.e. Y is the continuously compounded transition matrix,

X = exp(Y)
X*X = exp(2Y), etc

both X and Y are matrices.

I am told that the sums of Y must sum to 0, but I can not see why this should be the case. Is it obvious?

Many Thanks.

2. Oct 7, 2009

### trambolin

Possible hint maybe? $e^0=1$

3. Oct 8, 2009

### NewStudent200

Thanks.

But I have trouble visualizing this for a matrix. Is there aproof somewhere, or a text that you can recommend which gives examples and talks about the applications of taking exponentials or logs of matrices?

Many thanks,

4. Oct 8, 2009

### trambolin

It is a http://en.wikipedia.org/wiki/Matrix_exponential" [Broken] right ? If you write down the power series for it you will get a pattern.

$$X = I + Y + \frac{Y^2}{2!} + \ldots$$
Now if you sum up the rows of X it is 1. On the right hand side you already get 1 from the identity matrix. So all contributions from the remaining terms must be zero right? So I will let you think if your condition is sufficient or necessary.

Last edited by a moderator: May 4, 2017
5. Oct 8, 2009

### NewStudent200

Cool. Thanks a lot!

Thinking about matrices a little further. How does one raise a matrix to a decimal power. I know that in the case of an integer power:

X^n = S.M^n.S^-1

where S is the eigen vector matrix and M is the matrix with eigen values along the diagonal. Now if n is non integer, then does this still hold? Could we also do it via:

Y = X^n
ln(Y) = n.ln(X)
Y = exp(n.ln(X))?

much appreciated.

6. Oct 8, 2009

### g_edgar

yes, yes

also:
if $$X = S M S^{-1}$$ then $$e^X = S e^M S^{-1}$$, $$\log(X) = S \log(M) S^{-1}$$, etc.