# Conjugate Matrices: Proof

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.

• 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") – vrugtehagel Oct 5 '17 at 9:28
• 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. – reflexive Oct 5 '17 at 9:57

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 $B\neq I_n$.