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
$$ P A P^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}$$
Since this is a very small matrix there's actually no need to explicitly find the Jordan form by going through the routine tedious procedure. You know the characteristic polynomial is
$$p(x) = (x-1)^2(x-3)$$
and you can check that the minimal polynomial is the same. This means a few things.
- You have a single Jordan block corresponding to $3$. This is just $(3)$.
- You cannot have two Jordan blocks corresponding to $1$ since that would make the matrix diagonalizable (which it is not since the minimal polynomial does not split into distinct factors). Therefore you must have a single block of the form $\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}$
These combined gives you the form (up to order of the blocks)
$$J=\begin{pmatrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0& 0 & 3\end{pmatrix}$$
In general, for small matrices like these (anything up to $6\times 6$) you can find the Jordan form through similar types of analysis using the facts
- The geometric multiplicity of an eigenvalue is the number of blocks corresponding to it.
- The algebraic multiplicity of an eigenvalue is the sum of the total sizes of the blocks.
- The exponent of the term corresponding to an eigenvalue in the minimal polynomial is the size of the largest block.
Best Answer
Transposed matrices have the same characteristic equation and minimal polynomial. Furthermore, they have the same generalised eigenspace. With all of this in mind, we can say that $A$ and $A ^ \text{T}$ have the same Jordan chains, and hence Jordan basis. So $A$ and $A ^ \text{T}$ have the same Jordan canonical form.
Finally, similar matrices have the same Jordan matrix because if $P, Q \in \text{GL}(n, \mathbb{C})$ and $J$ is the Jordan canonical form, then we can write $$P^{-1}AP = J = Q^{-1}A^\text{T}Q$$ i.e. $$QP^{-1}APQ^{-1} = A^\text{T}.$$ But $Q, P ^{-1}$ and $P, Q^{-1}$ are invertible, so $QP^{-1}$ and $PQ^{-1}$ are also invertible. So write $R = QP^{-1}$. Then we know that $R^{-1} = (QP^{-1})^{-1} = (P^{-1})^{-1}Q^{-1} = PQ^{-1}$. So
$$RAR^{-1} = A^\text{T}$$
and we can conclude that $A$ and $A^\text{T}$ are similar.