# Trouble with Matrix Exponentials

Blanchdog95
Homework Statement:
Given the matrix A = {{1, 0}, {1, 0}}, find the matrix exponential e^(At)
Relevant Equations:
e^(Mt) = I + Mt + (Mt)^2/2! + ... + (Mt)^k/k!
e^t = 1 + t + t^2/2! + t^3/3! + ... + t^k/k!
I've attempted to solve this by separating A into a diagonal matrix D and nilpotent matrix N:
D = {{1, 0}, {0, 0}}
N = {{0, 0}, {1, 0}}

e^(At) = e^((D + N)t) = e^(Dt) * e^(Nt)

When N is raised to the second power, it becomes the zero matrix. Therefore,
e^(Nt) = I + Nt = {{1, 0}, {t, 1}}

Note that D^2 = D
e^(Dt) = I + Dt + Dt^2/2! + ... + Dt^k/k! = {{1, 0}, {0, 1}} + {{1, 0}, {0, 0}}t + {{1, 0}, {0, 0}}t^2/2! + ... + {{1, 0}, {0, 0}}t^k/k!
e^(Dt) = {{1 + t + t^2/2! + ...+ t^k/k!, 0}, {0, 1}} = {{e^t, 0}, {0, 1}}

e^(Dt) * e^(Nt) = {{e^t, 0}, {0, 1}} * {{1, 0}, {t, 1}} = {{e^t, 0}, {t, 1}}

e^(At) = {{e^t, 0}, {t, 1}}

e^(At) = {{e^t, 0}, {e^t-1, 1}}, which can be obtained without separating A into D and N by noting that A^2 = A and substituting the series representation of e^t. I can see how this is done, but what I can't see is why the method I used failed, since the textbook itself used this method to find the matrix exponential of B = {{2, 1}, {0, 2}} by separating it into a diagonal and nilpotent matrix. If someone could teach me why this method does not work for A, or if it does, where I made a mistake, I would appreciate it a lot. Thanks in advance!

e^((D + N)t) = e^(Dt) * e^(Nt)
This is your problem. The statement ##e^{X+Y}= e^{X}e^{Y}## holds if XY=YX. Otherwise, it is not necessarily true. As you have seen, it is not true in your case. Notice that in the case where
$$B=\begin{pmatrix} 2 & 1\\ 0 & 2 \end{pmatrix}=N+D= \begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}= \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}$$
N and D commute (as N=2I, and I commutes with anything), and thus you can use ##e^{X+Y}= e^{X}e^{Y}##. In your current case, however, N and D do not commute, and thus you cannot use ##e^{X+Y}= e^{X}e^{Y}##.

• Blanchdog95
This is your problem. The statement ##e^{X+Y}= e^{X}e^{Y}## holds if XY=YX. Otherwise, it is not necessarily true. As you have seen, it is not true in your case. Notice that in the case where
$$B=\begin{pmatrix} 2 & 1\\ 0 & 2 \end{pmatrix}=N+D= \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}+ \begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}$$
N and D commute (as D=2I, and I commutes with anything), and thus you can use ##e^{X+Y}= e^{X}e^{Y}##. In your current case, however, N and D do not commute, and thus you cannot use ##e^{X+Y}= e^{X}e^{Y}##.
It won't let me edit the post as I made a few mistakes, so see the above quote for my revised post.

• Blanchdog95
Blanchdog95
Much needed help.
THANK YOU SO MUCH! Now that I think about it the textbook did mention something about commutation; this probably saved me half a dozen points on my test next week.

Well, good luck on your test!

Just a quick thought-- you can easily check your work by noting that ##A=PDP^{-1}## where
$$D= \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$$
and
$$P= \begin{pmatrix} 1 & 0\\ 1 & 1 \end{pmatrix}.$$
It is easy to compute ##e^{Dt}##, and then you can just use ##e^{At}=Pe^{Dt}P^{-1}##.