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Problem: Show that $$ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $$ Are not similar over GL$_n ( \mathbb{F}_2)$

Note: We write $ \mathbb{F}_2= \text{GF}(2)$ the Galois Field, just in case people use a different notation.

I assume that this problem is ridiculously easy, but I couldn't find a satisfying way to show the statement as above. I have read a couple of similar problems on this website and found this great comment by @J.M. on How do I tell if matrices are similar?

"The simplest test you can make is to see whether their characteristic polynomials are the same. This is necessary, but not sufficient for similarity (it is related to having the same eigenvalues)."

So I did this for the above and got two characteristic polynomials that differ in their sign, hence they would not be similar. My problem now is, that with this approach I did not even make use of the statement that my invertible Matrix, lets call it $P$ has to be in $\mathbb{F}_2$.

So my most naive approach was to define a general Matrix $P$ as follows: $$P := \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ and use $A=P^{-1}BP \implies PA=BP$ where $A$ is the first Matrix on the left as stated in the problem and $B$ the latter. My next naive step was to perform the calculations $$PA=BP \iff \begin{pmatrix} b & a \\ d & c \end{pmatrix} = \begin{pmatrix} a & b \\ -c & -d \end{pmatrix} $$ Well and my last thought was that two matrices are equal if and only if their entries are equal. This gave me $b=a, d=-c$ and these entries have to be in $\mathbb{F}_2$, so I don't see how this could be a contradiction.

I would have said that now $$P = \begin{pmatrix} a & a \\ -d & d \end{pmatrix} $$ and next looked at the special case that all the entries are $1$, which would work considering the Mathematica output for P.A == H.P

Community
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Spaced
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  • I agree @seaturtles, we just started studying on this topic so I first did have a look at some of the other related posts. Some use the Jordan approach to solve this, which we didn't cover yet so I thought there must be a 'elementary' approach to solve this problem – Spaced Mar 23 '14 at 01:59

2 Answers2

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You already did it. Because you are in $GF(2)$, $-x=x$ for any element. So your $P$ looks like $$ \begin{bmatrix}a&a\\ d&d\end{bmatrix} $$ and now you can use a variety of methods to show that such $P$ is not invertible.

Martin Argerami
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  • That's the most obvious thing to not have come to my mind, thank you very much, $\det P = ad-da =0 \implies $ the Matrix is not invertible. – Spaced Mar 23 '14 at 02:01
  • You are welcome. You could also show that $P$ has non-trivial kernel, or simply write $PT=I$ for some matrix $T$ and show quickly that the equality is impossible. – Martin Argerami Mar 23 '14 at 02:04
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The fact we're in $\Bbb F_2$ is very relevant. The problem with the characteristic polynomial approach is that $+1=-1$ are the same element of $\Bbb F_2$. Now consider the invertibility of your $P$ in $\Bbb F_2$....

anon
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  • Thanks a lot to you too for your answer, at first glance I wasn't considering the $+1=-1$ statement for $\mathbb{F}_2$, I was too focused on solving the equations and spotting an error but it all makes sense now. – Spaced Mar 23 '14 at 02:09