# Minimal polynomial for a matrix

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.

• 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. Mar 25, 2020 at 16:00
• Does it matter over what set of numbers? I.e $x^2 + x + 1$ is indeed irreducible over $\mathbb{C}$ Mar 25, 2020 at 18:32
• 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. Mar 25, 2020 at 23:35

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.
• 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 $\mathbb{Z}?$ Mar 25, 2020 at 18:36
• Yep. In the field $\Bbb{Z}_2$, the coefficient $2$ and $0$ are the same. Mar 26, 2020 at 0:54
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 $$X^2-X-1.$$
• Over $k=\Bbb Z_2$, these are the same. Mar 25, 2020 at 18:45