# Similarity of matrix over field extension [duplicate]

Let $E/F$ be a field extension and let $A, B$ be $n\times n$ matrices over $F$. Assume that $A, B$ are similar over $E$, i.e. there exists $P\in \mathrm{GL}_{n}(E)$ s.t. $B = PAP^{-1}$. Are $A, B$ also similar over $F$, i.e. can we find $Q\in \mathrm{GL}_{n}(F)$ s.t $B = QAQ^{-1}$?
This statement is true when $E=\mathbb{C}$ and $F = \mathbb{Q}$. More generally, the statement is true for characteristic 0 fields, and the proof can be found in here. However, the proof doesn't work for the char $p$ case, since there exists nonzero polynomials over a characteristic $p$ field (for example, over $\mathbb{F}_{p}$) which became zero for any evaluation. (For example, $f(x) = x^{p}-x\in \mathbb{F}_{p}[x]$ is a nonzero polynomial, but $f(a)=0$ for any $a\in \mathbb{F}_{p}$.) So I want to know whether the theorem is also true for characteristic $p$ case of there's a counterexample. Thanks in advance.
## marked as duplicate by José Carlos Santos linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 20 '18 at 7:32
Two square matrices over a field $F$ have the same rational canonical form. The rational canonical form always has entries in the same field $F$. Therefore if $A$ and $B$ are similar over the extension field $E$ of $F$ then they are similar over $F$.