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In a field $\mathbb{F}_4 = \{0, 1, a, a+1\}$ where $1 + 1 = 0$ and $a^2 = a + 1$,

I am to find the characteristic polynomial of $A = \begin{pmatrix} a+1 & 0 & 0 & a \\ 0 & 1 & 0 & 0 \\ a+1 & a & a & 1 \\ a & a & a+1 & a+1 \end{pmatrix}$.

I first swap row 1 with row 2, and take the determinant of $(A-\lambda I)$ which yields

$(A-\lambda I) = -(1-\lambda)\begin{pmatrix} a+1-\lambda & 0 & a \\ a+1 & a-\lambda & 1 \\ a & a+1 & a+1-\lambda \end{pmatrix}$,

but I just get stuck in very long expressions where I can't easily determine its eigenvalues (because I will later on write the Jordan normal form of $A$).

I arrive at, after using the fact that $1 + 1 = 0$ and $a^2 = a + 1$

$-(1-\lambda) \left( ((a+1-\lambda)(a-\lambda)(a+1-\lambda)-(a+1)) + a((a+1)^2-a(a-\lambda)) \right) = 0$

Which is the determinant of the above $3\times 3$ matrix,

and the first term I can see means one eigenvalue is $\lambda=1$, but then I get an expression I cannot make sense of inside the parentheses.

What am I doing wrong?

Anton Vrdoljak
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rawestan
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    Once you swap rows 1 and 2, you get another matrix $A'$ which is not equal to $A$. Moreover, while it is true that $\det(A)=\det(A')$, it is not the case that $\det(A-\lambda I)=\det(A'-\lambda I)$, so the characteristic polynomial you calculated is not the characteristic polynomial of $A$. – Math101 Nov 11 '23 at 17:59

1 Answers1

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You can use the Addition and Multiplication in $F_4$ (I'll highlight in color where I'm using them).

Also, I think you misplaced a parenthesis in the following line:

$(a+1-\lambda)\Big((a-\lambda)(a+1-\lambda)-(a+1)\Big)=$

$\require{cancel} (a+1-\lambda)\Big(\color{blue}{a^2}+\cancel{a}-\color{red}{\bcancel{\color{black}{2\lambda a}}}-\lambda+\lambda^2-\cancel{a}+1\Big)=({a+1}-\lambda)\Big(\color{blue}{a+1}-\lambda+\lambda^2+1\Big)=\\\color{violet}{(a+1)^2}-\color{red}{\bcancel{\color{black}{2\lambda(a+1)}}}+\lambda^2(a+\cancel 1)+a+1-\cancel{\lambda^2}-\lambda^3-\lambda=\color{lightblue}{\bcancel{\color{violet}{a}}}+\lambda^2a+\color{lightblue}{\bcancel{\color{black}{a}}}+1-\lambda^3-\lambda=\\\color{lightgreen}-\lambda^3+\lambda^2a\color{lightgreen}-\lambda+1=\lambda^3+\lambda^2a\color{lightgreen}+\lambda+1$

On the other hand,

$a\Big(\color{violet}{(a+1)^2}-a(a-\lambda)\Big) = a\Big(\color{violet}a-\color{blue}{a^2}+\lambda a\Big)=a\Big(\cancel a-\color{blue}{(\color{black}{\cancel {\color{blue}{a}}}+1)}+\lambda a\Big)=-a+\lambda \color{darkblue}a=\\\color{lightgreen}-a+\lambda\color{darkblue}{(a+1)}=\lambda(a+1)\color{lightgreen}+a$

So we get

$\det(A-\lambda I)/(1-\lambda)=\big(\lambda^3+\lambda^2a+\color{red}{\bcancel{\color{black}{\lambda}}}+1\big)+\big(\lambda(a+\color{red}{\bcancel{\color{black}1}})+a\big)=\lambda^3+\lambda^2a+\lambda a+a+1$

Since there are only four elements in $\mathbb F_4$, it's easy to check that the resulting polynomial has $a$ as a single root and $a+1$ as a double root.

Darsen
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  • Thanks a lot now it is much clearer. However in solving for the polynomial roots I'm not sure if I am doing it right;

    I try the different elements as inputs and get:

    p(0) = a + 1 != 0 p(1) = 3a + 2 = a != 0 which seems right so far,

    p(a) = 2(a+1) = 0 (also seems right since it is deg 1). p(a+1) = / ... / = 2[a(a+1) + (a+1)] = 2(a) = 0 but isn't a double root? Am i doing it wrong

     – rawestan Nov 12 '23 at 15:14
  • Nvm pretty sure I got it now:

    $p(a+1) = (a+1)^3 + (a+1)^2 \cdot a + (a+1) \cdot a + a + 1 = 2a(a+1)+2(a+1) = 2(a(a+1)+(a+1)) = 2(a+1)(a+1) = 0$

     – rawestan Nov 12 '23 at 15:49
  • To check $a+1$ is a double root either 1) you divide the polynomial by $\lambda-(a+1)$, check you get a zero reminder and then check if $a+1$ is a root in the resulting polynomial; or 2) you divide the polynomial by $\Big(\lambda-(a+1)\Big)^2$ and check you get a zero reminder. – Darsen Nov 14 '23 at 11:16