3
$\begingroup$

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.

$\endgroup$
5
  • 1
    $\begingroup$ You could do a more general form of diagonalization. See the normal forms. $\endgroup$ Jun 20, 2018 at 5:09
  • 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 $$ e^A = \sum_{n=0}^\infty \frac{A^n}{n!}. $$ $\endgroup$
    – Math1000
    Jun 20, 2018 at 5:12
  • $\begingroup$ Related: Why does the $n$-th power of a Jordan matrix involve the binomial coefficient?. $\endgroup$ Jun 20, 2018 at 7:13
  • 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. $\endgroup$ Jun 20, 2018 at 9:10
  • $\begingroup$ The generic formula for n =3 is fairly straightforward by the C-H theorem. $\endgroup$ Jun 21, 2018 at 15:54

3 Answers 3

3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ And what is the P matrix? $\endgroup$
    – Sonal_sqrt
    Jun 20, 2018 at 5:49
  • $\begingroup$ @PiyushDivyanakar: it plays the same role as the $P$ in the diagonalizable case. $\endgroup$
    – robjohn
    Jun 20, 2018 at 5:55
  • 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-H
    Jun 20, 2018 at 5:56
  • $\begingroup$ ok thanks a lot $\endgroup$
    – Sonal_sqrt
    Jun 20, 2018 at 5:57
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

  1. block-diagnoal,
  2. Jordan,
  3. permutation

or many other things which would save calculations.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.