Talk:Matrix exponential

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:

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

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

Well, the stated answer is wrong on its face, as $e^{0M}=I$. — 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:

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

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)

It looks like the two examples were copied from somewhere - it refers to examples "earlier in the article" where the exponential matrix is calculated, but this example doesn't exist. I can't tell if this is referring to a part of the wiki that was removed, or whether the example was just copied in it's entirety from another source. 130.215.95.157 (talk) 18:45, 5 March 2008 (UTC)

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)

Continuity, etc.

For any two matrices X and Y, we have

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

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

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

(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

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

or, using X and Y,

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

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

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)

Arbitrary field?

The article mentions calculations over an arbitrary field. I think this should be changed since it gives the impression that there exists an exponential map over arbitrary fields.T3kcit (talk) 19:55, 16 December 2007 (UTC)

Agreed and changed. Solian en (talk) 14:37, 23 April 2008 (UTC)

Transition matrix

The transition matrix used here does not jive with what I understand is a matrix of eigenvectors. The rows need not sum to one. Is this a different matrix? If so, I haven't found an entry for the transition matrix used as a product with the diagonal matrix of eigenvalues and the inverse transition matrix to solve for the matrix exponential. The internal link does not clearly address this usage.John (talk) —Preceding comment was added at 02:31, 26 February 2008 (UTC)

The term transition matrix here means the matrix associated with the similarity transform that puts the matrix A into Jordan form; it has little to do with the meaning explained in transition matrix. I doubt this term is used very often, so I reformulated the text to avoid this term. Hope this helps. -- Jitse Niesen (talk) 14:52, 29 February 2008 (UTC)

Computing the matrix exponential (general case, arbitrary field)

There is no reference for the X = A + N decomposition over an arbitrary field. I couldn't find one in my textbooks (except for the Jordan decomposition in C). The French version points to the Dunford decomposition, which requires the matrix's minimal polynomial to have all its roots in the field. I believe that the decomposition does not hold in general. If no one objects, I will specify the conditions under which it exists. Solian en (talk) 15:04, 2 April 2008 (UTC)

Unfortunately, we've now obscured the fact that a complex matrix always has a unique A + N decomposition. I think we should drop consideration of arbitrary fields altogether until someone wants to handle them properly (probably in another section). At any rate it isn't at all clear when the article switches from the complex case to more general fields. -- Fropuff (talk) 16:30, 23 April 2008 (UTC)

Commutativity

Does anybody know if the equation exp(A+B)=exp(A) exp(B) implies AB=BA? Franp9am (talk) 10:18, 20 June 2008 (UTC)

It does not. A counterexample (from Horn & Johnson) is given by

$A={\begin{bmatrix}0&0\\0&2\pi {\textrm {i}}\end{bmatrix}}\quad {\text{and}}\quad B={\begin{bmatrix}0&1\\0&2\pi {\textrm {i}}\end{bmatrix}}.$

A and B do not commute, but exp(A+B) = exp(A) exp(B) = the identity matrix. I added something to the article. -- Jitse Niesen (talk) 10:59, 20 June 2008 (UTC)

---

Note added; sorry if I'm not following conventions here, I haven't commented before. --Tom

It's also not even correct in general to say that exp(A+B)=exp(A)exp(B). For a simple counterexample, take nilpotent matrices A and B that sum to a nonsingular A+B, say

$A={\begin{bmatrix}0&1\\0&0\end{bmatrix}}\quad {\text{and}}\quad B={\begin{bmatrix}0&0\\-1&0\end{bmatrix}}.$

The power series for $exp(A)={\begin{bmatrix}1&1\\0&1\end{bmatrix}}$ and $exp(B)={\begin{bmatrix}1&0\\-1&1\end{bmatrix}}$ terminate after the second terms, but $exp(A+B)$ never terminates at all: $exp(A+B)={\begin{bmatrix}cos(1)&sin(1)\\-sin(1)&cos(1)\end{bmatrix}}$.

Horn and Johnson give a similar example in Topics in Matrix Analysis, p. 434-435. Automatic Tom (talk) 22:18, 9 August 2008 (UTC)

The above appears to be correct. Michael Hardy (talk) 21:46, 9 August 2008 (UTC)

---

The reference to the Horn and Johnson Topics in Matrix Analysis book quotes a theorem that says if $A$ and $B$ contain only algebraic entries then you can only say that $e^{A}e^{B}=e^{B}eA$ if and only if $A$ and $B$ commute. This is different than saying $e^{A+B}=e^{A}e^{B}$ if and only if $A$ and $B$ commute. — Preceding unsigned comment added by Robleroble (talk • contribs) 08:23, 6 March 2013 (UTC)

Matrix Exponential via Laplace transform

I don't know how to properly code the math behind this, but the page is missing a section on solving for the matrix exponential via the Laplace transform.

${\mathcal {L}}^{-1}\{e^{At}\}=(sI-A)^{-1}$

Substituting t=1 in the above equation will yield the matrix exponential of A.

99.236.42.178 (talk) 01:29, 3 November 2008 (UTC)

I am afraid this is correct but aggressively meaningless, as is section 2.5 which I propose to eliminate. That section is a circular restatement of true but jubilantly unusable facts. Unless you provided an explicit situation where a generalization of the Laplace transform for matrices provided explicit answers not available to the Taylor expansion of the exponential, or, in the truly astounding case, the Magnus expansion mentioned earlier, I don't see what the point of that section could possibly be. Cuzkatzimhut (talk) 19:55, 1 November 2013 (UTC)

I'm not sure how this is "aggressively meaningless". If the point of section 2 is to give methods for the computation of the matrix exponential, then you can do it by the Laplace transform in just the way it was written. This isn't a generalization of the Laplace transform for matrices, merely the element-wise application of the inverse Laplace to $(sI-A)^{-1}$. That being said, I don't know if this page should aim to provide every possible method for computing the matrix exponential. Zfeinst (talk) 01:54, 2 November 2013 (UTC)

OK, proceed to justify your statement by illustrating how the compact bottom line formula of the LT yields the above trivial formula of 2.4.1 for a general 2x2 matrix. It could make that section meaningful. As I indicated above, all this LT rigmarole is a circular way of rewriting the matrix exponential in terms of its power expansion, which is how it was defined in the first place, accepting application of the LT and its convergence peculiarities on the space of matrices. The connection to the differential equation is already detailed in section 1. You need a LT to handle an exponential? What would you believe the student of matrix exponentials would have lost, had she not read this subsection (2.5)? Cuzkatzimhut (talk) 11:55, 2 November 2013 (UTC)

I proceed to eliminate the LT tautology, and effectively assign its section number to the preceding subsection, 2.4.1, which evidently merits promotion to a self-standing section. Cuzkatzimhut (talk) 00:58, 5 November 2013 (UTC)

proof

It would be useful to sketch how some of the properties (eg. exp(M+N)=exp(M)exp(N) if ..) are proved. Cesiumfrog (talk) 07:59, 13 March 2010 (UTC)

Exponential of Sums Error & Lie Product Formula

I have a couple issues with this statement: "The converse is false: the equation eX + Y = eXeY does not necessarily imply that X and Y commute. However, the converse is true if X and Y contain only algebraic numbers and their size is at least 2×2 (Horn & Johnson 1991, pp. 435–437)"

• "Horn & Johnson" should be omitted in favor of a link to the actual reference at the bottom. The reference is actually to "Topics in Matrix Analysis", whereas lots of people are used to thinking of "Matrix Analysis" when they see just "Horn & Johnson".
• I read that section of Topics in Matrix Analysis this morning and what it actually says is that, if X and Y are NxN matrices, N >= 2, containing only algebraic numbers, then e^Xe^Y = e^Ye^X if and only if X and Y commute. Nothing to do with e^{X+Y}. It's also more of a side note in the H&J book; it's not proved or even stated as a theorem, lemma, or even a numbered equation.

On a related note, it may be helpful to some people (I know it would have been helpful to me!) to put a link somewhere here to the Lie product formula article, which says lim_{m\rightarrow\infty} (e^{A/m}e^{B/m})^m = e^{A+B} for ANY A and B.

I don't want to just make these changes because a) I don't really have time and b) I'm hoping someone who can take care to double check and integrate it with the rest of the article can do it.

--M0nstr42 (talk) 01:50, 30 September 2010 (UTC)

Exponential skew symmetric matrices

I added the section on skew symmetric matrices. My proofs are my own independent work - I have seen the results, however, given by a few different authors (only after I went through all the work did I bother to look anything up, derp). None of the sources prove my results. Does anyone know of anything more elegant than what I did? Lhibbeler (talk) 02:38, 4 November 2010 (UTC)

@ User:Arthur_Rubin, see Jean Gallier and Dianna Xu, "Computing Exponentials of Skew Symmetric Matrices And Logarithms of Orthogonal Matrices", International Journal of Robotics and Automation, 2000. My results are *not* original research. —Preceding unsigned comment added by Lhibbeler (talk • contribs) 04:19, 4 November 2010 (UTC)

Ahem. You said they were original research. As for the polynomial equations for the exponentials of the skew-symmetric matrices, it's not that interesting a result. It follows from some of the things which should be in [[matrix polynomial]; as a two-dimensional skew-symmetric matrix W has eigenvalues $\pm i\alpha$, and exp(W) has eigenvalues $\exp(\pm i\alpha )$. Hence, if

$P(x)=\cos \alpha +{\frac {sin(\alpha )}{\alpha }}x$,

then P(W) = exp(W). — Arthur Rubin (talk) 04:54, 4 November 2010 (UTC)

I said my proofs are original as far as I know- I have seen *the results* but not the derivations. I was trying to explain where the formulas come from rather than just throw up a formula. And who are you to say what is interesting? The results are interesting and important to mechanicians- using that formula is more than 10 times faster on my laptop than using Pade approximants. When you have a large number of skew matrices and you need to calculate their exponentials (rotation updates in crystal plasticity, for example), an order of magnitude decrease in computation time is, in fact, very interesting. It sounds like you have more beef with the matrix polynomial page than with my edit. If you still object to my revision, then get rid of the collapsible windows with the proofs. Besides, I have put two dents in the statement that opens the section "Finding reliable and accurate methods to compute the matrix exponential is difficult, and this is still a topic of considerable current research in mathematics and numerical analysis." Lhibbeler (talk) 05:19, 4 November 2010 (UTC)

You could list those formulas as examples in the "Alternative" section. Perhaps that section could be promoted, but, you really haven't provided anything else. — Arthur Rubin (talk) 05:49, 4 November 2010 (UTC)

Rechecking the 3D formula:

$P(x)=1+{\frac {\sin \alpha }{\alpha }}x+{\frac {1-\cos \alpha }{\alpha ^{2}}}x^{2}$

$P(0)=1=\exp(0)$

$P(\pm i\alpha )=\cos \alpha \pm i\sin \alpha =\exp(\pm i\alpha )$

Yep, it's the equation from the "Alternative" section. — Arthur Rubin (talk) 05:55, 4 November 2010 (UTC)

What about this

For a 2x2 matrix A one can write the exponent as follows:

$e^{A}={\sqrt {e^{trA}}}({\frac {sinh(q)}{q}}A+(cosh(q)-{\frac {tr(A)sinh(q)}{2q}})I)$

where

$q={\sqrt {-det(A-{\frac {trA}{2}}I)}}$

and I is the 2x2 identity matrix, and sinh(0)/0 is set as 1.

This doesn't seem to be in the article and might be interesting for those looking for a closed expression. —Preceding unsigned comment added by 95.117.203.24 (talk) 12:15, 22 January 2011 (UTC)

Is that correct if A has its two eigenvalues equal, but is not a multiple of the identity? If it is, it seems notable, but is sufficiently complex as to require a source. — Arthur Rubin (talk) 15:07, 22 January 2011 (UTC)

Never mind, it's accurate. Still, a better expression might be:

$e^{A}=e^{\frac {\operatorname {tr} A}{2}}\left(\left(\cosh q\right)I+{\frac {\sinh q}{q}}\left(A-{\frac {\operatorname {tr} A}{2}}I\right)\right)$

where

$q={\sqrt {-\det \left(A-{\frac {\operatorname {tr} A}{2}}I\right)}}$

Which can be seen by

$e^{\lambda }\approx e^{\frac {\operatorname {tr} A}{2}}\left(\cosh q+{\frac {\sinh q}{q}}\left(\lambda -{\frac {\operatorname {tr} A}{2}}\right)\right)$

at the eigenvalues λ of A. — Arthur Rubin (talk) 16:31, 22 January 2011 (UTC)

Hello. Yes, that case leads to q = 0. I don't have a source for this equation but the proof is not hard: one writes $A={\begin{bmatrix}a&b\\c&d\end{bmatrix}}={\begin{bmatrix}{\frac {a+d}{2}}&0\\0&{\frac {a+d}{2}}\end{bmatrix}}+{\begin{bmatrix}{\frac {a-d}{2}}&b\\c&{\frac {d-a}{2}}\end{bmatrix}}=B+C$. Since B and C commute we have e^A = (e^B)(e^C), and since B and C^2 are diagonal matrices, both e^B and e^C can be calculated directly.

Unfortunately this doesn't work so for 3x3 and bigger matrices. Maybe your reasoning will lead to a similar formula for bigger matrices, I'll have to think about it! —Preceding unsigned comment added by 95.117.203.24 (talk) 16:50, 22 January 2011 (UTC)

Included, now, as an alternative to the alternative section for 2×2 matrices. — Arthur Rubin (talk) 17:22, 22 January 2011 (UTC)

Applications to quantum computing?

I didn't see anything in this article about its applications to quantum computing. In particular, with Hermitian matrices, Pauli operators, and rotations about the Bloch Sphere. There's a bunch about this in the book I'm reading: An Introduction to Quantum Computing (Phillip Kaye, Raymond Laflamme, and Michelle Mosca - 2007). Just thought I'd point this out in case anyone feels it's important to mention.

24.212.140.98 (talk) 21:55, 22 January 2012 (UTC)

A method I was taught, but cannot find external references

Hello, I was taught a method by a Hungarian mathematician and I'm not even sure what it is called. The problem is to find $e^{At}$. If you do not care about $t$ then replace it with a 1 when done. I'm not sure if anyone else is aware of this method or if it is even useful. So I'm putting it here under the talk section of the article.

Anyways I will include an example and steps below. Given $A={\begin{pmatrix}1&1&0&0\\0&1&1&0\\0&0&1&-1/8\\0&0&1/2&1/2\end{pmatrix}}$

First, compute the eigenvalues. $\lambda _{1}=3/4$ and $\lambda _{2}=1$ each with an algebraic multiplicity of two.

Now, take the exponential of each eigenvalue multiplied by $t$: $e^{\lambda _{i}t}$. Multiply by an unknown matrix $B_{i}$. If the eigenvalues have an algebraic multiplicity greater than 1, then repeat the process, but multiply by a factor of $t$ for each repetition. If one eigenvalue had a multiplicity of three, then there would be the terms: $B_{i_{1}}e^{\lambda _{i}t},B_{i_{2}}te^{\lambda _{i}t},B_{i_{3}}t^{2}e^{\lambda _{i}t}$. Sum all terms.

In our example we get:

$e^{At}=B_{1_{1}}e^{\lambda _{1}t}+B_{1_{2}}te^{\lambda _{1}t}+B_{2_{1}}e^{\lambda _{2}t}+B_{2_{2}}te^{\lambda _{2}t}$

$e^{At}=B_{1_{1}}e^{3/4t}+B_{1_{2}}te^{3/4t}+B_{2_{1}}e^{1t}+B_{2_{2}}te^{1t}$.

So how can we get enough equations to solve for all of the unknown matrices? Differentiate with respect to $t$.

$e^{At}=B_{1_{1}}e^{3/4t}+B_{1_{2}}te^{3/4t}+B_{2_{1}}e^{1t}+B_{2_{2}}te^{1t}$

$Ae^{At}=3/4B_{1_{1}}e^{3/4t}+\left(3/4t+1\right)B_{1_{2}}e^{3/4t}+1B_{2_{1}}e^{1t}+\left(1t+1\right)B_{2_{2}}e^{1t}$

${\begin{aligned}A^{2}e^{At}=&(3/4)^{2}B_{1_{1}}e^{3/4t}+\left((3/4)^{2}t+(3/4+1\cdot 3/4)\right)B_{1_{2}}e^{3/4t}+B_{2_{1}}e^{1t}+\left(1^{2}t+(1+1\cdot 1)\right)B_{2_{2}}e^{1t}\\=&(3/4)^{2}B_{1_{1}}e^{3/4t}+\left((3/4)^{2}t+3/2\right)B_{1_{2}}e^{3/4t}+B_{2_{1}}e^{t}+\left(t+2\right)B_{2_{2}}e^{t}\end{aligned}}$

${\begin{aligned}A^{3}e^{At}=&(3/4)^{3}B_{1_{1}}e^{3/4t}+\left((3/4)^{3}t+((3/4)^{2}+(3/2)\cdot 3/4))\right)B_{1_{2}}e^{3/4t}+B_{2_{1}}e^{1t}+\left(1^{3}t+(1+2)\cdot 1\right)B_{2_{2}}e^{1t}\\=&(3/4)^{3}B_{1_{1}}e^{3/4t}+\left((3/4)^{3}t+27/16)\right)B_{1_{2}}e^{3/4t}+B_{2_{1}}e^{t}+\left(t+3\cdot 1\right)B_{2_{2}}e^{t}\end{aligned}}$

Since the these equations must be true, regardless the value of $t$, we set $t=0$. Then we can solve for the unknown matrices.

${\begin{aligned}I=&B_{1_{1}}+B_{2_{1}}\\A=&3/4B_{1_{1}}+B_{1_{2}}+B_{2_{1}}+B_{2_{2}}\\A^{2}=&(3/4)^{2}B_{1_{1}}+(3/2)B_{1_{2}}+B_{2_{1}}+2B_{2_{2}}\\A^{3}=&(3/4)^{3}B_{1_{1}}+(27/16)B_{1_{2}}+B_{2_{1}}+3B_{2_{2}}\end{aligned}}$

This can be solved using linear algebra (don't let the fact the variables are matrices confuse you). Once solved using linear algebra you have:

${\begin{aligned}B_{1_{1}}=&128A^{3}-366A^{2}+288A-80I\\B_{1_{2}}=&16A^{3}-44A^{2}+40A-12I\\B_{2_{1}}=&-128A^{3}+366A^{2}-288A+80I\\B_{2_{2}}=&16A^{3}-40A^{2}+33A-9I\end{aligned}}$

Plugging in the value for $A$ gives:

${\begin{aligned}B11=&{\begin{pmatrix}0&0&48&-16\\0&0&-8&2\\0&0&1&0\\0&0&0&1\end{pmatrix}}\\B12=&{\begin{pmatrix}0&0&4&-2\\0&0&-1&{\frac {1}{2}}\\0&0&{\frac {1}{4}}&-{\frac {1}{8}}\\0&0&{\frac {1}{2}}&-{\frac {1}{4}}\end{pmatrix}}\\B21=&{\begin{pmatrix}1&0&-48&16\\0&1&8&-2\\0&0&0&0\\0&0&0&0\end{pmatrix}}\\B22=&{\begin{pmatrix}0&1&8&-2\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}}\end{aligned}}$

So the final answer would be:

${e}^{t\,A}={\begin{pmatrix}{e}^{t}&t\,{e}^{t}&\left(8\,t-48\right)\,{e}^{t}+\left(4\,t+48\right)\,{e}^{\frac {3\,t}{4}}&\left(16-2\,t\right)\,{e}^{t}+\left(-2\,t-16\right)\,{e}^{\frac {3\,t}{4}}\\0&{e}^{t}&8\,{e}^{t}+\left(-t-8\right)\,{e}^{\frac {3\,t}{4}}&-{\frac {4\,{e}^{t}+\left(-t-4\right)\,{e}^{\frac {3\,t}{4}}}{2}}\\0&0&{\frac {\left(t+4\right)\,{e}^{\frac {3\,t}{4}}}{4}}&-{\frac {t\,{e}^{\frac {3\,t}{4}}}{8}}\\0&0&{\frac {t\,{e}^{\frac {3\,t}{4}}}{2}}&-{\frac {\left(t-4\right)\,{e}^{\frac {3\,t}{4}}}{4}}\end{pmatrix}}$

This method was taught as it is nearly the same procedure to calculate $A^{t}$. Except that instead of taking the derivative to generate more equations, you can increment $t$ (which is useful for discrete difference equations). Also, replace using $e^{(\lambda _{i}t)}$ with $\lambda _{i}^{t}$. — Preceding unsigned comment added by 150.135.222.237 (talk) 18:57, 6 December 2013 (UTC)

I confess I don't know a name for it, and I have not stumbled on it in just this form in matrix analysis books. You might register to Wikipedia and clean it up and add it as a subsection near the end. It is quite tasteful in its immediacy, as it does not require diagonalizability, as manifest in your example of a neat defective matrix. The ansatz is a most general one for the generic polynomial dictated by the Cayley-Hamilton theorem, in your case a cubic; the "Wronskian" trick for multiple roots ensures linear independence, and is equivalent to the analysis of the Matrix_exponential#Evaluation_by_Laurent_series subsection in the article. In fact, I wonder if you wished to adapt the example to a shorter section in the same notation, i.e. the ansatz exp(tA) = B_α exp (tα) + B_β exp(tβ) , yielding the same expression after solving this expression and its first derivative at t=0 for the Bs in terms A and I, arguably faster. It is a pretty method, and you'd be quite welcome to contribute it to the article (be bold, etc). Cuzkatzimhut (talk) 23:22, 15 December 2013 (UTC)

Actually, for distinct eigenvalues, it is just Sylvester's formula for the exponential, where the simplicity of the l.h.s. allows for ready indirect evaluation of the B_i s, the Frobenius covariants projecting onto the eigenspace corresponding to eigenvalues λ_i . For the general, non-diagonalizable case you are addressing, it is just about Buchheim's generalization. It' is well-worth covering in the article. Cuzkatzimhut (talk) 19:36, 16 December 2013 (UTC)

I am actually registered (I didn't sign in when I originally posted this), but following wikipedia's best practices, I cannot post "original research." Thus I cannot in good conscience post this as a subsection until someone can give this method a name or at least reference it elsewhere. Maybe this will help; this method is derived from the method of undetermined coefficients for linear differential equations. How to handle repeated roots (eigenvalues) follows from the same explanation here (https://www.khanacademy.org/math/differential-equations/second-order-differential-equations/complex-roots-characteristic-equation/v/repeated-roots-of-the-characteristic-equation). If you think this is enough support, then we (or I) can post it. Alternatively, if you can dig up the resources that discuss this being Bucheim's generalization of Sylvester's formula, then we can post it. .Mouse7mouse9 22:22, 29 December 2013 (UTC) — Preceding unsigned comment added by Mouse7mouse9 (talk • contribs)

I also wanted to mention there is a discrete variant (from solving discrete difference equations), allowing one to take symbolic matrix powers (raising a square matrix to an unknown scalar variable). I mention it here (https://en.wikipedia.org/wiki/Talk:Matrix_multiplication#Matrix_powers_and_discrete_difference_equations). Both the above method and the linked method are very similar. The symbolic matrix power is derived from solving discrete linear difference equations (again using the method of undetermined coefficients), rather than continuous differential equations, as in the case of the matrix exponential.