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For characteristic polynomial, there is a very straight forward algorithm.

Compute $\det(\lambda I - A) = 0$

Now I have looked for the method for calculating minimal polynomial everywhere but could not find an algorithm even for a 2x2 case.

Give a simple matrix, say

$A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$

How can we calculate its minimal polynomial?

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  • $\begingroup$ you can try to solve linear equations, check whether or not \begin{align*} A&\in kI\\ A^2&\in kA+kI\\ A^3&\in kA^2+kA+kI\\ &\text{etc.} \end{align*} $\endgroup$ – yoyo Mar 6 '15 at 1:42
  • $\begingroup$ yoyo how does this help I am desperate $\endgroup$ – Carlos - the Mongoose - Danger Mar 6 '15 at 1:43
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For the 2 x 2 case, here's an algorithm: Compute the characteristic polynomial. If it's not a perfect square, then it actually IS the minimal polynomial. If it IS a perfect square, say $(x - a)^2$, then your matrix is either $aI$, in which case the minimal polynomial is $(x-a)$, or it's not, in which case the minimal polynomial is $(x-a)^2$.

Why does this work? 1. The minimal polynomial always divides the characteristic polynomial.

  1. The minimal polynomial of the Jordan form is the same as the minimal polynommial of $M$. If $M$ has two distinct eigenvalues, then its jordan form is $$ \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix} $$ and it's evident that the minimal polynomial of this is $(x - \lambda_1)(x - \lambda_2)$.

If it has duplicate eigenvalues, then the Jordan form looks like either $$ \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_1 \end{bmatrix} $$ or $$ \begin{bmatrix} \lambda_1 & 1 \\ 0 & \lambda_1 \end{bmatrix} $$ In the first case, your matrix is conjugate to $\lambda_1 I$, hence must equal $\lambda_1 I$, whose minimal polynomial is $(x - \lambda_1)$.

In the second case, the matrix doesn't satisfy that polynomial, so the min poly must have two $(x - \lambda_1)$ factors, hence must be $(x - \lambda_1)^2$.

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  • $\begingroup$ Can you illustrate with an example for my pea brain for the second case? $\endgroup$ – Carlos - the Mongoose - Danger Mar 6 '15 at 1:48
  • $\begingroup$ Any reference helps $\endgroup$ – Carlos - the Mongoose - Danger Mar 6 '15 at 1:48
  • $\begingroup$ I hope my additional comments clarify things. $\endgroup$ – John Hughes Mar 6 '15 at 1:52

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