I wasn't able to find a straightforward answer to this online. It is straightforward when you can diagonalize the matrix but how is the non-diagonalizeable case handled? The $3 \times 3$ case is the most relevant to me, and I will have to do this using pen paper so I am looking for solutions that are easy to do manually.
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1$\begingroup$ You could do a more general form of diagonalization. See the normal forms. $\endgroup$– Cameron WilliamsJun 20, 2018 at 5:09
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4$\begingroup$ There is the Jordan-Chevally decomposition: $$A = B + N $$ where $B$ is diagonalizable, $N$ is nilpotent (i.e. $N^k=0$ for some nonnegative integer $k$), and $BN=NB$. Then $e^A = e^{B+N} = e^Be^N$ where $$ e^A = \sum_{n=0}^\infty \frac{A^n}{n!}. $$ $\endgroup$– Math1000Jun 20, 2018 at 5:12
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$\begingroup$ Related: Why does the $n$-th power of a Jordan matrix involve the binomial coefficient?. $\endgroup$– Rodrigo de AzevedoJun 20, 2018 at 7:13
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1$\begingroup$ You can try and find some other canonical transformation $$A = TCT^{-1}$$ where $C$ is sparse matrix but not diagonal. The sparser the $C$, the sparser $C^k$ will (usually) be. $C$ could be block-diagnoal, Jordan, permutation or many other things which would save calculations. $\endgroup$– mathreadlerJun 20, 2018 at 9:10
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$\begingroup$ The generic formula for n =3 is fairly straightforward by the C-H theorem. $\endgroup$– Cosmas ZachosJun 21, 2018 at 15:54
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3 Answers
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For $3\times 3$ matrix, if it can't be diagonalized, it will have Jordan forms $A=PJP^{-1}$ for following two cases $$J=\begin{pmatrix}\lambda&1&0\\ 0&\lambda&0\\ 0&0&\mu\end{pmatrix}, \ \ \ J=\begin{pmatrix}\lambda&1&0\\0&\lambda&1\\0&0&\lambda\end{pmatrix}$$ So all you need to do is figure out what will happen to $J^n$, i.e. conclude a formula for upper trangular entries.
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$\begingroup$ @PiyushDivyanakar: it plays the same role as the $P$ in the diagonalizable case. $\endgroup$– robjohn ♦Jun 20, 2018 at 5:55
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1$\begingroup$ Usually, when matrix is diagonalizable, $P$ is composed of three eigenvectors, but when it is not (geometrical multiplicity is smaller than algebraic multiplicity or in your case, the repeated eigenvalue only has one corresponding eigenvector ), $P$ is composed of generalized eigenvectors, try to spend some time figuring out how to find Jordan form, you will get it. $\endgroup$– H-HJun 20, 2018 at 5:56
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A consequence of the Cayley-Hamilton theorem is that any analytic function $f$ of a $3\times3$ matrix $A$ can be expressed in the form $a_0I+a_1A+a_2A^2$, where the coefficients are possibly constant scalar functions. Once you know the eigenvalues of $A$, finding these coefficients is a matter of solving a small system of linear equations, specifically the equations $a_0+\lambda_i a_1+\lambda_i^2 a_2 = f(\lambda_i)$. If $A$ has repeated eigenvalues, this system is underdetermined, but you can generate additional independent equations by differentiating with respect to the repeated eigenvalue. This method is often a lot less work than computing a Jordan decomposition and reassembling the result.
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You can try and find some other canonical transformation $$A = TCT^{-1}$$ where $C$ is sparse matrix but not diagonal. The sparser the $C$, the sparser $C^k$ will (usually) be. $C$ could be
- block-diagnoal,
- Jordan,
- permutation
or many other things which would save calculations.
answered Jun 20, 2018 at 9:16