What is the limit of exp(-kM/T)?

[CarlB]

What is the limit of exp(-kM/T)?

Let M be a matrix:

\left(\begin{array}{cc}a & b \\c & d\end{array}\right)

where a,b,c and d are complex constants. Let k and T be positive real numbers. Is there a closed formula for this limit?

\begin{array}{c}\lim \\ T-&gt;0^{+}\end{array} \;\;\;<br /> e^{-kM/T}

Of course the answer will not depend on k, or T.

Carl

[edit]Hint:
http://mathworld.wolfram.com/JordanMatrixDecomposition.html

 

[Jelfish]

Is this what you meant?

\lim\limit_{T\to 0^+}e^{-kM/T}

 

[thepatientmental]

The limit of exp(-kM/T) as T approaches 0 from the positive side does not have a closed formula, as it depends on the specific values of a, b, c, and d in the matrix M. However, we can use the Jordan matrix decomposition to find the limit in terms of the eigenvalues of M. The Jordan matrix decomposition states that any square matrix can be written as the sum of a diagonal matrix and a nilpotent matrix. This means that M can be written as:

M = D + N

where D is a diagonal matrix and N is a nilpotent matrix. Since the limit does not depend on k or T, we can ignore them for now and just focus on the matrix M. We can rewrite the limit as:

\begin{array}{c}\lim \\ T->0^{+}\end{array} \;\;\;
e^{-M/T} = \begin{array}{c}\lim \\ T->0^{+}\end{array} \;\;\;
e^{-(D+N)/T}

Using the properties of exponents, we can rewrite this as:

\begin{array}{c}\lim \\ T->0^{+}\end{array} \;\;\;
e^{-M/T} = \begin{array}{c}\lim \\ T->0^{+}\end{array} \;\;\;
e^{-D/T} \cdot e^{-N/T}

As T approaches 0 from the positive side, the first term e^{-D/T} will approach 1 since D is a diagonal matrix with real eigenvalues. The second term e^{-N/T} will approach 1 since N is a nilpotent matrix, meaning that N^k = 0 for some positive integer k. Therefore, the limit of exp(-M/T) as T approaches 0 from the positive side is 1. In other words, the limit of exp(-kM/T) as T approaches 0 from the positive side is 1 for any complex constants a, b, c, and d and any positive real numbers k and T.