MHB Problem of the Week #114 - August 4th, 2014 Solution

[Chris L T521]

Here's this week's problem!

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Problem: Let $g\in GL_4(K)$ be the permutation matrix
\[g=\begin{pmatrix}0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0 \end{pmatrix}.\]
Compute the characteristic polynomial $P_{\text{char}}^g$. Compute the minimal polynomial $P_{\text{min}}^g$ when $K=\mathbb{F}_2$, when $K=\mathbb{F}_3$, and when $K=\mathbb{F}_5$.

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[Chris L T521]

No one answered this week's problem. You can find my solution below.

[sp]The characteristic polynomial is\[\begin{aligned}P_{\text{char}}^g &= \det(x\cdot\mathbf{1}-\phi)\\ &=\begin{vmatrix}x & -1 & 0 & 0\\ 0 & x & -1 & 0\\ 0 & 0 & x & -1\\ -1 & 0 & 0 & x\end{vmatrix}\\ &=x^4-(-1)(-1)\\ &=x^4-1.\end{aligned}\]
(One could have also determined this right away since $g$ is already in rational canonical form.) Now, in $K=\mathbb{F}_2$, $x^4-1 = (x^2+1)(x^2-1)=(x+1)^4$; in $K=\mathbb{F}_3$, $x^4-1=(x+1)(x+2)(x^2+1)$; and in $K=\mathbb{F}_5$, $x^4-1=(x+1)(x+2)(x+3)(x+4)$.
It is clear that the factors of $x^4-1$ in the cases of $K=\mathbb{F}_3$ and $K=\mathbb{F}_5$ are coprime to each other. Thus, it follows in these two cases that $P_{\min}^g = x^4-1$. In the case of $K=\mathbb{F}_2$, $P_{\min}^g$ will be the smallest polynomial that annihilates the map $(x,y,z,w)\mapsto (w,x,y,z)$ (based off of the permutation matrix given in the problem). $P_{\min}^g\neq x-1$ since in $\mathbb{F}_2$,
\[\begin{aligned} {[\phi-1]} (x,y,z,w) &= (w,x,y,z)-(x,y,z,w)\\ &= (w+x,x+y,y+z,z+w)\neq 0.\end{aligned}\]
Similarly, $P_{\min}^g\neq (x-1)^2$ since in $\mathbb{F}_2$,
\[\begin{aligned} {[(\phi-1)^2]} (x,y,z,w) &= {[\phi-1]} (w+x,x+y,y+z,z+w)\\ &= (z+x,w+y,x+z,y+w)\neq 0\end{aligned}\]
and $P_{\min}^g\neq (x-1)^3$ since in $\mathbb{F}_2$,
\[\begin{aligned} {[(\phi-1)^3]} (x,y,z,w) &= {[\phi-1]} (z+x,w+y,x+z,y+w)\\ &= (y+z+w+x,y+z+w+x,y+z+w+x,y+z+w+x)\neq0.\end{aligned}\]
However, $P_{\min}^g=(x-1)^4 = x^4-1=x^4+1$ in $\mathbb{F}_2$ since
\[\begin{aligned} {[(\phi-1)^4]} (x,y,z,w) &= {[\phi-1]} (y+z+w+x,y+z+w+x,y+z+w+x,y+z+w+x)\\ &= (0,0,0,0).\end{aligned}\]
Therefore, $P_{\min}^g=x^4-1=x^4+1$ in $\mathbb{F}_2$, $P_{\min}^g=x^4-1=x^4+2$ in $\mathbb{F}_3$ and $P_{\min}^g=x^4-1=x^4+4$ in $\mathbb{F}_5$.$\hspace{.25in}\clubsuit$[/sp]