5
$\begingroup$

recently I've come across a few cases where I had to evaluate $e^{At}$ (differential equations) where A is in Frobenius canonical form. The algorithm looks like: $$e^{At} = Pe^{Jt}P^{-1}$$ so I need to do Jordan decomposition. From Frobenius canonical form I can easily write the characteristic equation => compute the roots which are eigenvalues and write the Jordan canonical form.

This was the easy part. Now the other part: compute matrices $P$ and $P^{-1}$.

Is there any standard form of those matrices if matrix A is in a Frobenius canonical form?

$\endgroup$
2
  • $\begingroup$ You might be able to use the Frobenius form directly. If the Frobenius form is A=[ 0,1; 1,1 ] with char poly xx-x-1, then xx = x+1 and exp(At) = 1 + At + (At)^2/2 + (At)^3/6 + ... = 1 + At + (A+1)/2 * t^2 + (2A+1)/6 * t^2 + ... = (1 + 1/2 t^2 + 1/6 t^2 + 2/24 t^3 + ...) + A*(t+t^2/2 + t^3/3 + t^4/8 + ... ). The coefficients in the terms are just Fibonacci numbers over factorials. So you get one matrix times a power series in t plus another matrix times a power series in t. If your matrix is n×n, then you get at most n matrices times n power series. $\endgroup$ Feb 21, 2011 at 13:31
  • $\begingroup$ This is a neat problem. I think your idea to use P directly turns out to be better than my simple recurrence. P is a very natural and nice matrix. It is probably as stable as one could expect given that one is working with JCF at all. $\endgroup$ Feb 22, 2011 at 0:27

1 Answer 1

8
$\begingroup$

The matrix P is the matrix of coefficients of certain divisors of the characteristic polynomial of the Frobenius canonical form.

In fact this works for any companion matrix, not just the companion matrix of a power of an irreducible polynomial.

Suppose a matrix A has 1s above the diagonal, and the last row is [ v0, v1, v2, v3, ..., vn ], so $$ A = \begin{bmatrix} 0 & 1 & 0 & 0 & \ldots & 0 & 0 \\ 0 & 0 & 1 & 0 & \ldots & 0 & 0 \\ 0 & 0 & 0 & 1 & \ldots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \ldots & 1 & 0 \\ 0 & 0 & 0 & 0 & \ldots & 0 & 1 \\ v_0 & v_1 & v_2 & v_3 & \ldots & v_{n-1} & v_n \\ \end{bmatrix}.$$ Then $$ A^{(n+1)} = v_0 A^0 + v_1 A^1 + v_2 A^2 + \ldots + v_n A^n$$

Suppose $x-\alpha$ is a factor of $$ f(x) = x^{(n+1)} - v_n x^n - \ldots - v_1 x^1 - v_0 x^0 $$ and write $$ f(x)/(x-\alpha) = w_n x^n + w_{n-1} x^{(n-1)} + \ldots + w_1 x^1 + w_0 x^0 $$ where of course $w_n = 1$. Then $w = [ w_0, w_1, w_2, \ldots, w_n ]$ is a left eigenvector of A, since $wA = \alpha w$.

In more detail, $$wA = [ v_0, w_0 + v_1, w_1 + v_2, \ldots, w_{n-1} + v_n ]$$ and $$f(x) = (x-\alpha)( w_n x^n + w_{n-1} x^{(n-1)} + \ldots + w_1 x^1 + w_0 x^0) = x^{n+1} - (-\alpha+w_{n-1}) x^{n} \ldots$$ so by equating coefficients of $x^n$, one gets $v_n = \alpha - w_{n-1}$ and $v_n + w_{n-1} = \alpha w_n$. Similarly, $v_{n-1} = \alpha w_{n-1} - w_{n-2}$, and in general $v_i = \alpha w_i - w_{i-1}$ so $v_i + w_{i-1} = \alpha w_i$ and $wA = \alpha w$.

Something similar works with repeated roots even, to give you Jordan blocks.

So to find your matrix P, find each root α of f(x) with multiplicity k, and then the rows of P are the coefficients (in increasing order of power of x) of f(x)/(x−α)i, from i=1 to k, with corresponding diagonal entry of the matrix just being α, and the run from 1 to k forming the Jordan block.


Example 1

Find the JCF of $$A = \begin{bmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\-1&4&-6&4\end{bmatrix}$$

The characteristic polynomial is just $f(x) = x^4-4x^3+6x^2-4x+1$ since it is a companion matrix of $f$. Of course, $f(x) = (x-1)^4$ is easy to factor.

We calculate $$f(x)/(x-1) = 1x^3 - 3x^2 + 3x - 1,$$ $$f(x)/(x-1)^2 = 0x^3+1x^2-2x+1,$$ $$f(x)/(x-1)^3 = 0x^3+0x^2+x-1,\text{ and}$$ $$f(x)/(x-1)^4 = 0x^3+0x^2+0x+1.$$

We then write down the coefficients (in reverse order, I guess): $$P = \begin{bmatrix} -1 & 3 & -3 & 1 \\ 1 & -2 & 1 & 0 \\ -1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}$$ so that $$ P A P^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}$$

$\endgroup$
1
  • 1
    $\begingroup$ Wow. Great answer! Thank you. $\endgroup$
    – kubal5003
    Feb 22, 2011 at 7:56

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.