# Determining the minimal polynomial of $B:=A²+A$

Let $A$ $\in$ $\mathbb{C}$ be a solution of $$g:=A^3+2A-1=0$$ I want to determine the minimal polynomial in $\mathbb{Q}$ of $A$ and $B$, if $$B:=A^2+A$$ We got a hint that the theorem of Cayley-Hamilton would help here, but I don't know how to use it in this context. I guess that $g$ is already the minimal polynomial of $A$, because it's normalized, irreducible and of course $g(A)=0$. But I don't know where to go from there. I'd be glad about any kind of help. Thank you.

• Apparently $A,B$ are complex numbers, not matrices. Then it would be much more readable to use lower case letters, as from the title this really looks as if it is a question about minimal polynomials of matrices. And mentioning Cayley-Hamilton only reinforces that impression. – Marc van Leeuwen Nov 27 '16 at 6:02

HINT: Express powers of $B$ in terms of $A$, and reduce them using the fact that $A^3+2A-1=0$. Can you find a relation between the powers of $B$?

EDIT: The fact that $A^3+2A-1=0$ tells you that $A^3=1-2A$, so \begin{eqnarray*} B^2&=&(A^2+A)^2=A^4+2A^3+A^2=A^3(A+2)+A^2\\ &=&(1-2A)(A+2)+A^2=-A^2-3A+2. \end{eqnarray*} In the same way you can express $B^3$ in terms of $A$, and reduce it to get something quadratic in $A$. Then you have expressions for $B^0$, $B^1$, $B^2$ and $B^3$ in terms of $A^0$, $A^1$ and $A^2$, so these powers of $B$ are linearly dependent, meaning that $B$ is a root of a nontrivial cubic.

• :Thanks for the hint. I calculated the first few powers of $B$, but I don't see a connection between them and $A$. Could you elaborate on that a bit further, please? – Tobi92sr Nov 24 '16 at 22:29
• @Tobi92sr Edited. Let me know if/when you figured it out, I can elaborate further if needed. – Servaes Nov 24 '16 at 22:34
• I think I've got the solution. I calculated $B^3$: $$B^3=A^2+5A-4$$ Then $1B^3+4B^3+3B^2-4=0$. So $x^3+4x^2+3x-4$ is the minimal polynomial. I hope it's correct. Thanks a lot! Your hint was very helpful. – Tobi92sr Nov 25 '16 at 1:11
• You seem to have made a mistake in your calculation of $B^3$... – Servaes Nov 25 '16 at 1:42
• Oh sorry I just made a typo. $B^3=A^2+9A-4$. Now it's correct I think. – Tobi92sr Nov 25 '16 at 14:40

You are looking for the minimal polynomial over $\def\Q{\Bbb Q}\Q$ of $a+a^2$ where $a$ is the image of $X$ in the field $F=\Q[X]/(X^3+2X-1)$. The minimal polynomial of an algebraic element is the same as te minimal polynomial of the $\Q$-linear operator defined by multiplication by that element. Here that is the linear operator on $F$ whose matrix, with respect to the obvious basis $[1,a,a^2]$, is $$M=\pmatrix{0&1&1\\1&-2&-1\\1&1&-2}$$ (the columns respectively express $a+a^2$, $a^2+a^3$ and $a^3+a^4$ on the basis $[1,a,a^2]$, using the relation $a^3=1-2a$). The minimal polynomial cannot be of degree$~2$ because $F$ has no subfields of degree$~2$, so the minimal polynomial must be the characteristic polynomial of$~M$, which is $X^3+4X^2+3X-4$.

(You can also find this polynomial by trying to find a linear relation between $v,Mv,M^2v,M^3v$ for some vector$~v$, for instance for $v=[1,0,0]$, for which the other three vectors are $[0,1,1]$, $[2,-3,-1]$, and $[-4,9,1]$. The coefficients of such a relation are $-4,3,4,1$ which again gives the polynomial $-4+3X+4X^2+X^3$. This approach amounts to directly finding a relation between $1,b,b^2,b^3$.)