Let the matrix

\begin{equation} A=\begin{bmatrix} 2 & 1 & 2 \\ -2 & -1 & -4 \\\ 1 & 1 & 3 \end{bmatrix}. \end{equation}

So far I found the characteristic polynomial $C_A(x)=(2-x)(x-1)^2$, the minimal polynomial $m_A(x)=(2-x)(x-1)$ and the Jordan matrix form is

\begin{equation} J=\begin{bmatrix} 2 & * & *\\ * & 1 & * \\\ * & * & 1 \end{bmatrix}. \end{equation}

Does someone could tell me from what I did how it possible to find the rational canonical form of $A$?


There are two "rational canonical forms", one obtained using elementary divisors and another obtained by invariant factors.

The invariant factors form a list where each term divides the next, their product is the characteristic polynomial, and the minimal polynomial is the largest one.

Therefore in this problem, $(x-2)(x-1)$ is the largest invariant factor. The other invariant factor is $(x-1)$ simply because there is nothing left over.

The rational canonical form is obtained by putting the companion matrices corresponding to the invariant factors on the main diagonal.

So the rational canonical form of $A$ given by the invariant factors is: $$\begin{pmatrix} 0 & -2 & 0\\ 1 & 3 & 0\\ 0 & 0 & 1 \end{pmatrix}_.$$

The matrix $$\begin{pmatrix} 0 & -2 \\ 1 & 3 \end{pmatrix}_.$$

Is the companion matrix of $(x-2)(x-1)$.


  • $\begingroup$ It is easy to obtain the invariant factors from the elementary divisors, and there is a great algorithm for computing the elementary divisors of a matrix in the book "Linear Algebra" by Friedberg, Insel, and Spence. $\endgroup$ – Ken Duna Apr 29 '16 at 14:38
  • $\begingroup$ Is it necessary to use this algorithm to obtain the elementary divisor? $\endgroup$ – user332681 Apr 29 '16 at 14:45
  • $\begingroup$ No. This is just a convenient way, and the way I like to do it. There are methods to find the invariant factors, such as the Smith-Normal Form. However, I find that very cumbersome. math.stackexchange.com/questions/340352/… Here is a link to a question about computing rational canonical forms that has an answer (by the guy who wrote the book the questioner was studying from) on how to computing the invariant factors. $\endgroup$ – Ken Duna Apr 29 '16 at 14:48
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    $\begingroup$ So far, no one has said anything mathematically incorrect. We just have a disagreement on terminology. If you do not want to refer to the companion matrix of elementary divisors decomposition as a rational canonical form, that is fine and I don't think I will be able to change your mind. Let's just move on and be friends :) $\endgroup$ – Ken Duna Apr 29 '16 at 21:59
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    $\begingroup$ Fair enough, I had never seen such a definition before. Thanks for the reference! $\endgroup$ – André 3000 Apr 30 '16 at 2:06

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