Trouble with Matrix Exponentials

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In summary: Hope that helps!In summary, the method used to find the matrix exponential of B = {{2, 1}, {0, 2}} by separating it into a diagonal and nilpotent matrix may not always work if the matrices do not commute. In order to check the validity of the solution, one can use the formula A = PDP^-1 and compute e^Dt to find the matrix exponential of A.
  • #1
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}}

This answer is incorrect. The correct answer is

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!
 
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  • #2
Blanchdog95 said:
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}##.
 
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  • #3
Isaac0427 said:
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.
 
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  • #4
Isaac0427 said:
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.
 
  • #5
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}##.
 

1. What are matrix exponentials?

Matrix exponentials are a mathematical operation where a square matrix is raised to a power, similar to how a number is raised to a power. The result is another matrix with the same dimensions as the original matrix.

2. What is the significance of matrix exponentials?

Matrix exponentials have numerous applications in science and engineering, particularly in fields such as physics, chemistry, and economics. They are used to model dynamic systems and solve differential equations, among other things.

3. What are some common challenges when working with matrix exponentials?

One of the main challenges is calculating the matrix exponential, which can be computationally intensive for large matrices. Another challenge is determining the properties of a matrix exponential, such as its eigenvalues and eigenvectors.

4. How can matrix exponentials be used to solve real-world problems?

Matrix exponentials can be used to model and predict the behavior of complex systems, such as population growth, chemical reactions, and economic trends. They can also be used to design control systems and optimize processes.

5. Are there any limitations to using matrix exponentials?

One limitation is that not all matrices have a well-defined exponential. Additionally, the properties of a matrix exponential can be difficult to interpret and apply in certain situations. It is important to carefully consider the assumptions and limitations when using matrix exponentials in real-world problems.

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