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Is there a possible proof of the statement: Any two $n\times n$ invertible matrices over a field are conjugate?

I've tried thinking of it and think it should be true but really don't know how to prove it.

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  • $\begingroup$ I'm not sure how you define "conjugate". Normally (at least, how I learned it), the conjugate of a matrix is simply obtained by taking the complex conjugate of each element. Conjugation in that sense is an operation though, and you seem to treat it like a property ("any two $n\times n$ invertible matrices are conjugate") $\endgroup$ – vrugtehagel Oct 5 '17 at 9:28
  • $\begingroup$ Welcome to Stack Exchange! You should mention in your question what you mean by "conjugate". And also try to show your attempt, whatever you've thought of the problem. $\endgroup$ – reflexive Oct 5 '17 at 9:57
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No. I think that "conjugate" means "similar".

Similar matrices have the same eigenvalues , determinat, etc ...

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Nope, your claim is wrong. Conjugated matrices share the same determinant, characteristic polynomial and many more properties. So just take two matrices with different determinants, you will not be able to conjugate one to the other. :)

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  • $\begingroup$ Its like a Matrix M which has a set of basis and if we take the matrix with any two basis then they both would be conjugate. $\endgroup$ – ZDD Oct 5 '17 at 9:33
  • $\begingroup$ I know how to conjugate matrices, but still, they will keep their determinant. :) $\endgroup$ – Dirk Oct 5 '17 at 9:39
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Here is an easy example to see that the claim cannot be true in general. Take two matrices $A=I_n$ and $B=I_n+E_{1n}$ for $n\ge 2$. They are certainly not conjugate, i.e., there is no invertible $S$ such that $B=SAS^{-1}$, because $SAS^ {-1}=I_n$, but $B\neq I_n$.

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