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I would like to know how to set up the matrix $(A - \lambda I)$ for a matrix over $\Bbb Z_2$ so I can figure out its determinant on my way to the characteristic equation. I am stuck on pretty much the first step. I think I understand, but please let me know if I have this wrong.

So I have the matrix:

$$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ \end{bmatrix} $$

Since $\Bbb Z_2$ doesn't allow for subtraction, I am guessing that I start by doing this $(A + \lambda I)$ instead.

So I would end up with

$$ \begin{bmatrix} 1+\lambda & 1 & 1 & 1 \\ 0 & 1+\lambda & 0 & 0 \\ 1 & 1 & 0+\lambda & 1 \\ 0 & 1 & 0 & 1+\lambda \\ \end{bmatrix} $$

Then from there I can just find the $\det(A+\lambda I)$ like I would for any $4\times 4$ matrix.

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  • $\begingroup$ What do you mean by "$\Bbb Z_2$ doesn't allow for subtraction"? $\endgroup$ May 23, 2014 at 4:20
  • $\begingroup$ Just that. You can't subtract. You can add ( you can multiply too and maybe divide by 1, but not by 0 of course...but these are trivial). So 1+1=0. Now imagine if 1-1=0, think about it. If you could subtract, what would happen to 0-1?? there is no such thing as -1 in Z2. Now, if you go about writing mod 2 after everything, no problem, subtract away, just remember to write down mod 2 everywhere... $\endgroup$
    – Travis
    May 23, 2014 at 4:33
  • $\begingroup$ You can definitely subtract. It just so happens that $1=-1$ in this particular ring. You can always subtract in a ring. $\endgroup$ May 23, 2014 at 4:35

2 Answers 2

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You're doing it right. $Z_2$ does allow subtraction. But the result is the same as addition i.e $$-1 \mod 2 = 1 \mod 2$$ You're doing all the math in mod 2. That is $(7+6) \mod 2= (7 \mod 2) + (6 \mod 2) = 1+0 = 1$. So take the determinant of the matrix like you would any other matrix. If the coefficient is odd, its $1 \mod 2$. If it's even, its $0 \mod 2$. So, $$(3x^4 + 2x^3 - 2x + 5) \mod 2 = (x^4 + 1) \mod 2$$

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The characteristic polynomial is $p(\lambda) = (\lambda - 1)^2({\lambda}^2 - \lambda - 1)$.

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  • $\begingroup$ I did the math on $\mathbb{R}$, sorry, I didn't read the problem. $\endgroup$
    – Alexei0709
    May 23, 2014 at 3:12

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