Talk:Matrix exponential

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1 Example (homogeneous)
2 Column method
3 Continuity, etc.
4 Thank you

[edit] Example (homogeneous)

Could someone check the result for the matrix exponential

$e^{tM}=\begin{bmatrix} 2e^t - 2te^{2t} & -2te^{2t} & 0 \\ -2e^t + 2(t+1)e^{2t} & 2(t+1)e^{2t} & 0 \\ 2te^{2t} & 2te^{2t} & 2e^t\end{bmatrix}$

The answer that Mathematica gives is quite different:

$e^{tM}=\begin{bmatrix} \frac12 e^{2t}(1+e{2t}-2t) & -te^{2t} & e^{3t}\sinh t \\ -\frac12 e^{2t}(-1+e^{2t}-2t & (t+1)e^{2t} & -e^{3t}\sinh t \\ -\frac12 e^{2t}(-1+e^{2t}+2t & te^{2t} & e^{3t}\cosh t\end{bmatrix}$

Obviously, this would give another solution to the system as well...

Well, the stated answer is wrong on its face, as

e 0 M = I

. — Arthur Rubin | (talk) 12:28, 22 August 2006 (UTC)

So, should we change it to the Mathematica-approved answer that I wrote a few lines above?

well, acutally, I get a different form:

$e^(tM) = \begin{bmatrix} e^{3t} \cosh t - t e^{2t} & -t e^{2t} & e^{3t} \sinh t \\ -e^{3t} \sinh t + t e^{2t} & (t+1) e^{2t}& -e^{3t} \sinh t \\ e^{3t} \sinh t + t e^{2t} & t e^{2t} & e^{3t} \cosh t \end{bmatrix}$

Same values, except for a couple of typos, but it looks simpler to me. — Arthur Rubin | (talk) 00:27, 31 August 2006 (UTC)

Perhaps it's better to use the same matrix that is mentioned further up in the article? -- Jitse Niesen (talk) 02:50, 31 August 2006 (UTC)

[edit] Column method

It seems to me that the column method is the same as the method based on the Jordan decomposition, but explained less clearly. Hence, I am proposing to remove that section. -- Jitse Niesen (talk) 20:11, 24 July 2005 (UTC)

I now removed the section. -- Jitse Niesen (talk) 11:21, 3 August 2005 (UTC)

I think there is indeed A Point to doing it like that, but I need to have a close look over it again as I'm a little unfamiliar on the material and need to get acquanted with it again, and I haven't had a chance to do this. On a cursory look the removal looks okay, though... Dysprosia 09:32, 4 August 2005 (UTC)

[edit] Continuity, etc.

For any two matrices X and Y, we have

$\| e^{X+Y} - e^X \| \le \|X\| e^{\|X\|} e^{\|Y\|},$

is clearly incorrect -- just look at the case X = 0. Perhaps the equation should be

$\| e^{X+Y} - e^X \| \le e^{\|X\|} e^{\|Y\|},$

(which I changed it to), but I'm not sure that's correct, either. Arthur Rubin | (talk) 21:12, 1 February 2006 (UTC)

Whoops, I'm pretty sure I put that in. According to H&J, Corollary 6.2.32, we have

$\| e^{A+E} - e^A \| \le \|E\| e^{\|E\|} e^{\|A\|},$

or, using X and Y,

$\| e^{X+Y} - e^X \| \le \|Y\| e^{\|X\|} e^{\|Y\|}.$

Apparently I made a mistake while renaming the variables. Thanks a lot! -- Jitse Niesen (talk) 21:36, 1 February 2006 (UTC)

[edit] Thank you

Hi, I just wanted to say thank you for writing this article so clearly. I needed to quickly look up how to do matrix exponentials again and I thought I'd try Wikipedia instead of MathWorld first this time. Nice simple explainations here, and written very clearly. This thanks also goes out to all deticated wikipedians who are updating the math pages.--Johnoreo 02:21, 9 February 2006 (UTC)

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Talk:Matrix exponential

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[edit] Example (homogeneous)

[edit] Column method

[edit] Continuity, etc.

[edit] Thank you

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