Lets start by writing

# ( ( ln(2), 1, 1, 1 ), ( 0, ln(2), 1, 1), ( 0, 0, ln(2), 1 ), ( 0, 0, 0, ln(2) ) )#

in the form #ln(2) I_4 +A#, where #I_4# is the #4\times 4# identity matrix and #A# is the upper triangular matrix

# A = ( (0, 1, 1, 1 ), ( 0, 0, 1, 1), ( 0, 0, 0, 1 ), ( 0, 0, 0, 0 ) )#

Since #I_4# commutes with all #4 times 4# matrices in general, and #A# in particular, the exponential matrix is simply

#exp(ln(2) I_4 +A) = exp(ln(2) I_4) exp(A) = e^{ln(2)}I_4 exp(A) = 2 exp(A)#

So the calculation essentially reduces to the determination of #e^A#. Note that the standard approach using diagonalization does not work here, since #A# is **not** diagonalizable. However, the characteristic equation for #A# is easily seen to be

#lambda^4 =0#

and thus the Cayley-Hamilton theorem says that

#A^4 = 0#

so that the infinite series for #e^A# truncates :

#e^A = I+A+1/{2!} A^2 + 1/{3!}A^3#

It is easy to calculate

#A^2 = ((0,0,1,2), (0,0,0,1), (0,0,0,0), (0,0,0,0)) and A^3 = ((0,0,0,1),(0,0,0,0),(0,0,0,0),(0,0,0,0))#

Thus

#e^A = ((1,1,3/2,13/6),(0,1,1,3/2),(0,0,1,1),(0,0,0,1)) #

Our solution is twice this matrix!