Talk:Matrix exponential

message: 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)
replies: text: 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)
thread_title: Column method

message: 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)
thread_title: Thank you

message: 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)
replies: text: Agreed and changed. Solian en (talk) 14:37, 23 April 2008 (UTC)
thread_title:  Arbitrary field? 

message: 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)
replies: text: 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)
thread_title:  [[Transition matrix]] 

message: 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)
replies: text: 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)
thread_title:  Computing the matrix exponential (general case, arbitrary field) 

message: 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)
replies: text: 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)
replies: text:  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)
replies: text: 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)
thread_title:  Matrix Exponential via Laplace transform 

message: 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)
thread_title:  proof 

message: 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)
thread_title:  Exponential of Sums Error & Lie Product Formula 

message: 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)
thread_title:  Applications to quantum computing? 

message: 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^{A t} . 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} t e^{\lambda_i t}, B_{i_3} t^2 e^{\lambda_i t} . Sum all terms.
In our example we get:
e^{A t} = B_{1_1} e^{\lambda_1 t} + B_{1_2} t e^{\lambda_1 t} + B_{2_1} e^{\lambda_2 t} + B_{2_2} t e^{\lambda_2 t}
e^{A t} = B_{1_1} e^{3/4 t} + B_{1_2} t e^{3/4 t} + B_{2_1} e^{1 t} + B_{2_2} t e^{1 t}
.
So how can we get enough equations to solve for all of the unknown matrices? Differentiate with respect to t.
e^{A t} = B_{1_1} e^{3/4 t} + B_{1_2} t e^{3/4 t} + B_{2_1} e^{1 t} + B_{2_2} t e^{1 t}
A e^{A t} = 3/4 B_{1_1} e^{3/4 t} + \left( 3/4 t + 1 \right) B_{1_2} e^{3/4 t} + 1 B_{2_1} e^{1 t} + \left(1 t + 1 \right) B_{2_2} e^{1 t}
\begin{align}
A^2 e^{A t} =& (3/4)^2 B_{1_1} e^{3/4 t} + \left( (3/4)^2 t + ( 3/4 + 1 \cdot 3/4) \right) B_{1_2} e^{3/4 t} + B_{2_1} e^{1 t} + \left(1^2 t + (1 + 1 \cdot 1 )\right) B_{2_2} e^{1 t} \\  =& (3/4)^2 B_{1_1} e^{3/4 t} + \left( (3/4)^2 t + 3/2 \right) B_{1_2} e^{3/4 t} + B_{2_1} e^{t} + \left(t + 2\right) B_{2_2} e^{t}
\end{align}
\begin{align}
A^3 e^{A t} =& (3/4)^3 B_{1_1} e^{3/4 t} + \left( (3/4)^3 t + ( (3/4)^2 + (3/2) \cdot 3/4) ) \right) B_{1_2} e^{3/4 t} + B_{2_1} e^{1 t} + \left(1^3 t + (1 + 2) \cdot 1 \right) B_{2_2} e^{1 t} \\ =& (3/4)^3 B_{1_1} e^{3/4 t} + \left( (3/4)^3 t + 27/16 ) \right) B_{1_2} e^{3/4 t} + B_{2_1} e^{t} + \left(t + 3 \cdot 1 \right) B_{2_2} e^{t}
\end{align}
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{align}
I =& B_{1_1} + B_{2_1} \\
A =& 3/4 B_{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} + 2 B_{2_2} \\
A^3 =& (3/4)^3 B_{1_1} + (27/16) B_{1_2} + B_{2_1} + 3 B_{2_2}
\end{align}
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{align}
B_{1_1} =& 128 A^3 - 366 A^2 + 288 A - 80 I \\
B_{1_2} =& 16 A^3 - 44 A^2 + 40 A - 12 I \\
B_{2_1} =&-128 A^3 + 366 A^2 - 288 A + 80 I\\
B_{2_2} =& 16 A^3 - 40 A^2 + 33 A - 9 I
\end{align}
Plugging in the value for A gives:
\begin{align}
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{align}
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)
replies: text: 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) = Bα exp (tα) + Bβ 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)
replies: text: 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 Bi s, the Frobenius covariants projecting onto the eigenspace corresponding to eigenvalues λi . 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)
replies: text: 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)

text: 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.  — Preceding unsigned comment added by Mouse7mouse9 (talk • contribs) 23:26, 29 December 2013 (UTC)
replies: text: OK, I put it in the page as stated. Lots of room for improvement, especially by the father thereof, User:Mouse7mouse9. A big thanks. Cuzkatzimhut (talk) 15:34, 30 December 2013 (UTC)
thread_title:  A method I was taught, but cannot find external references 

message: We have a problem here.  Neither of the recent versions satisfies .  I don't have time to solve the equation at the moment.  — Arthur Rubin  (talk) 20:44, 23 October 2014 (UTC)
replies: text: At least for the equation given right now (where the disputed tag is placed) it does satisfy e^{0A} = I.  That doesn't necessarily mean this is the correct equation (I don't have the time to solve the equation at the moment either) but this quick complaint does not seem accurate to me. Zfeinst (talk) 21:38, 23 October 2014 (UTC)
replies: text:  I have now checked it, using the method of section 2.7: exp (tA) = B exp2t + Ct exp2t +D exp4t;  it follows that B= A−A2/4,  D = 1−A +A2/4,  and C= −4 +3A −  A2/2. It all checks in the disputed formula, so I believe you should take down the unwarranted tag. Cuzkatzimhut (talk) 13:59, 24 October 2014 (UTC)
thread_title:  Linear differential equation examples 

message: Why need the base be normal?  As noted in matrix logarithm, any non-singular complex matrix has a logarithm. 
— Although the paper states in this way, I actually agree with you that for the base matrix, just being non-singular suffice.Xuancong (talk) 03:17, 5 September 2015 (UTC)
thread_title:  Matrix-matrix exponential 

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thread_title:  External links modified (January 2018) 

message: Please include explanation/definition/link for \oplus and anything else not obvious to undergraduate mathematics students.  —DIV (120.17.110.164 (talk) 12:48, 27 March 2018 (UTC)
replies: text: Direct sum not obvious to math undergrads? Rather alarming.  Cuzkatzimhut (talk) 21:08, 27 March 2018 (UTC)
thread_title:  Nomenclature 

message: For iterative solvers (e.g. GMRES) one does not need to compute explicitly the matrix exponential e^M but rather the action of the exponential on a vector: e^M \vec{b}. One can use Krylov subspace methods to approximate the action which is for large sparse systems more efficient.
Linking to existing packages such as http://www.maths.uq.edu.au/expokit/ would be helpful.  HerrHartmuth (talk) 09:04, 6 December 2019 (UTC)
thread_title:  Matrix-vector product