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I'm currently reading about the minimal polynomial and have got this, this far:

$$ \text{Def: If $A$ is a $n \ \times n $ matrix is the minimal polynomial $q_A(x)$ the monic polynomial such that $q_A(A) = 0$}.  $$

Ex: Let $ A = \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}$ with coefficents in $ k = \mathbb{Z}_2. $ Then $ A^2 = \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix}$ and $ A^3 = I. $ The characteristic polynomial of A is $ P_A(x) = x^2 + x + 1 $, and since $ \{A , I\} $ is linearly independent it's also the minimal polynomial.

I dont get the conclusion of the bolded part.

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  • $\begingroup$ Alternatively, we may see that $x^2+x+1$ is a minimal polynomial from Hamilton-Cayley’s theorem and the fact that it is irreducible. $\endgroup$
    – fantasie
    Mar 25, 2020 at 16:00
  • $\begingroup$ Does it matter over what set of numbers? I.e $ x^2 + x + 1 $ is indeed irreducible over $ \mathbb{C} $ $\endgroup$
    – Oskar
    Mar 25, 2020 at 18:32
  • $\begingroup$ Yes, it matters. $x^2+x+1$ is irreducible over $\mathbb Z_2$, though. A minimal polynomial divides all polynomials that satisfies the matrix. But $x^2+x+1$ does not have nontrivial divisors, so the minimal polynomial is exactly itself. In $\mathbb C$, it cannot be argued like that. $\endgroup$
    – fantasie
    Mar 25, 2020 at 23:35

2 Answers 2

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The fact that $A$ and $I$ are linearly independent means that $$xA + yI \neq 0$$ for all $x, y$ where at least one is non-zero. In particular, if we take $x = 1$, then we must have, for all $y$, $$A + yI \neq 0.$$ So, for any monic polynomial $r$ of degree $1$, we must have $r(A) \neq 0$. Obviously no (constant) degree $0$ polynomial will do the trick either.

So, the minimal polynomial must be degree $2$ or more. We know that $P_A(A) = 0$, and $P_A$ is of degree $2$. From the uniqueness of the minimal polynomial (i.e. we know there can't be two polynomials of minimal degree that annihilate $A$) we can deduce that $P_A$ is the minimal polynomial too.

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  • $\begingroup$ I see, thank you. If, lets say, the minimal polynomial would be $ x^2 + x + 2 $, would it be congruent to $ x^2 + x $ since we're in $ \mathbb{Z}? $ $\endgroup$
    – Oskar
    Mar 25, 2020 at 18:36
  • $\begingroup$ Yep. In the field $\Bbb{Z}_2$, the coefficient $2$ and $0$ are the same. $\endgroup$
    – user759562
    Mar 26, 2020 at 0:54
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The characteristic polynomial of $A$ is $$\lambda^2-\lambda -1.$$ By the Cayley-Hamilton theorem, $$A^2-A-I=0.$$ Since $A$ cannot satisfy any non-trivial linear equation, the minimal poynomial of $A$ is $$X^2-X-1.$$

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  • $\begingroup$ Over $k=\Bbb Z_2$, these are the same. $\endgroup$
    – Berci
    Mar 25, 2020 at 18:45

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