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I have some matrix $J$, which has a decomposition into the diagonal matrix $D$, and the nilpotent matrix $N$. It is almost exactly like a Jordan block, except with ones on the subdiagonal even when the $\lambda$ on the diagonal change - plus, it's transposed. I.e, $J,D,N$ look like this:

$$J=\begin{bmatrix}\lambda_1&0&0&0&\cdots&0\\1&\lambda_2&0&0&\cdots&0\\0&1&\lambda_3&0&\cdots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&0\\0&0&0&\cdots&\lambda_{n-1}&0\\0&0&0&\cdots&1&\lambda_n\end{bmatrix}D=\begin{bmatrix}\lambda_1&0&0&0&\cdots&0\\0&\lambda_2&0&0&\cdots&0\\0&0&\lambda_3&0&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots&0\\0&0&0&\cdots&\lambda_{n-1}&0\\0&0&0&\cdots&0&\lambda_n\end{bmatrix}N=J-D$$

The Taylor series (incidentally the function is $\exp(tJ$) I am centering around $tD$:

$$f(tJ)=\sum_{k=0}^\infty\frac{f^{(k)}(tD)}{k!}\cdot (tJ-tD)^k=\sum_{k=0}^\infty\frac{f^{(k)}(tD)}{k!}\cdot (tN)^k$$

But the order of the multiplication $f^{(k)}(tD)\times (tN)^k$ is extremely important, since $N,D$ don't commute. Which way is correct, rigorously, and why? It seems to me that the extension of the Taylor series to matrix functions requires some extra thought!

Edit - perhaps it is a requirement on Taylor series that the two summands commute? Is this the case, meaning I must split $tJ$ some other, commutative, way? I know the Jordan-Chevalley decomposition states that such a commutative split exists, but I've seen nothing about how one actually finds $J=S+N:SN=NS%$.

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  • $\begingroup$ The Jordan decomposition is more of a theoretical tool. Using it for numerical computations is a bad idea. $\endgroup$
    – greg
    Commented Aug 9, 2021 at 16:25
  • $\begingroup$ In this case specifically it would be useful to me - probably anyway, I assume a matrix as simple as mine has an acceptable form. The Taylor series question still stands though, @greg $\endgroup$
    – FShrike
    Commented Aug 9, 2021 at 16:29

1 Answer 1

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$ \def\l{\lambda}\def\p{\partial} \def\A{{\cal A}}\def\B{{\cal B}}\def\C{{\cal C}} \def\L{\left}\def\R{\right}\def\LR#1{\L(#1\R)} \def\vec#1{\operatorname{vec}\LR{#1}} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\gradLR#1#2{\LR{\frac{\p #1}{\p #2}}} \def\hess#1#2#3{\LR{\frac{\p #1}{\p #2}\bigg|_{#3}}} $Taylor series with matrix variables do not work the way you think they do.

For example, consider the second-order series of $f(X)$ centered at $X_0$ $$\eqalign{ f(X) &= \frac1{0!}f(X_0) \\ &+\; \frac1{1!}\LR{X-X_0}:\hess{f}{X}{X=X_0} \\ &+\; \frac1{2!}\LR{X-X_0}\star\LR{X-X_0}:\,:\hess{^2f}{X^2}{X=X_0} \\ }$$ In this expansion, $(:,\, :\,:)$ denote the double and quadruple contraction products, $(\star)$ denotes the dyadic (aka tensor) product, and the terms in parentheses are second and fourth order tensors.

In general, the $n^{th}$ order Taylor series of a function of $m^{th}$ order tensors will involve tensors and contraction products of order $mn$.

The Taylor series as derived in undergraduate calculus is only possible because scalars are $0^{th}$ order tensors, i.e. $\;m=0$.

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  • $\begingroup$ Right... Do you have a reference for this? I can appreciate the (very stong) intuition behind Taylor's theorem, but I would be interested in seeing a proof / link to a proof of the Taylor series for matrices. I appreciate the comment about the $n$th derivatives being $2^n$ order tensors, that has made me realise a key difference between the typical Taylor and matrix Taylor series... I wonder if your statement can show the result on Wikipedia's page, that a matrix function of a Jordan block is this $\endgroup$
    – FShrike
    Commented Aug 10, 2021 at 19:43
  • $\begingroup$ I proved it to myself using what I thought was a valid Taylor expansion, but maybe the genuinely valid Taylor expansion proves it too. I would do this myself, but I'm unfamiliar with this term "contraction product" $\endgroup$
    – FShrike
    Commented Aug 10, 2021 at 19:44
  • $\begingroup$ $\def\dss{\mathop{\bullet}\limits}\def\c#1{\color{red}{#1}}$Here's an example of a triple contraction product between a 4th & 5th order tensor $$\eqalign{ C &= A \dss^3 B \\ C_{imn} &= \sum_{\c{j}=1}^J\sum_{\c{k}=1}^K\sum_{\c{\ell}=1}^L A_{i\c{jk\ell}} B_{\c{jk\ell}mn} \\ }$$ And here's an example of a dyadic product between the same tensors $$\eqalign{ F &= A \star B \\ F_{ijk\ell pqrmn} &= A_{ijk\ell} B_{pqr mn} \\ }$$ $\endgroup$
    – greg
    Commented Aug 10, 2021 at 21:05
  • $\begingroup$ Thank you. What is the definition of the order of contraction? Can you take an $k$th order product on two tensors of order $m,n$ where $k\lt m\le n$? $\endgroup$
    – FShrike
    Commented Aug 10, 2021 at 21:08
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    $\begingroup$ That answer only applies to the case that $x$ is a vector in ${\mathbb R}^{k}\,$ and all of "ugliness" is lurking inside of the term $(h\cdot\nabla)^n\;$ $\endgroup$
    – greg
    Commented Aug 10, 2021 at 21:36

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