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Let $V$ be vector space over a field $\mathbb{k}$. I can prove that any matrix is similar to its matrix transpose if $\mathbb{k}$ is an infinite field, but is this still true when $\Bbb k$ is finite? What ideas should I use for finding counterexamples (if they exist)?

P.S. The idea of my proof: I consider the splitting field of characteristical polynomial: $\mathbb{k}[x_1,\dots,x_k]$, therefore(use Jordan decomposition) $(C_0+x_1C_2+\dots+x_kC_{k})A^T = A(C_0+x_1C_2+\dots+x_kC_{k}) \implies C_iA^T=AC_i$. I want $\det(C_0+y_1C_2+\dots+y_kC_{k}) \neq 0$ for $y_i \in \mathbb{k}$ but since it's non zero for $x_{i}$ it's non zero polynomial over $\mathbb{k}$(there i use infiniteness of $\mathbb{k}$).

user26857
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qwenty
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  • Does your proof actually use the fact that $\mathbb k$ is infinite ? – krirkrirk Mar 11 '15 at 17:43
  • Yes (i use $\mathbb{k}[x] \ni f, f(x) \neq 0$ for $x \in $ some extension of $\mathbb{k} \therefore f \neq 0$ like function on $\mathbb{k}$ ) – qwenty Mar 11 '15 at 17:54
  • Just asking, do there exist finite vector spaces? – Pedro Mar 11 '15 at 17:56
  • This doesn't need $\mathbb k$ to be infinite ! @Pedro : well every $\mathbb{R}^{n}$ is one – krirkrirk Mar 11 '15 at 17:56
  • $x(x-1) \neq 0$ in $\mathbb{Z}/_{2}$ but as function it's zero – qwenty Mar 11 '15 at 17:59
  • Some troubles with my first comment, fix: Yes(i use that $f(v) \neq 0$ for $f \in \mathbb{k}[x], v \in $ some extension of $\mathbb{k} $ therefore $f \neq 0$ as function on $\mathbb{k}$ ) – qwenty Mar 11 '15 at 18:01
  • Oh ok. Well the proof I know uses Jordan so I guess you'll need an algebraicallly closed field; but can't help you more sorry. – krirkrirk Mar 11 '15 at 18:18
  • I have added very short version of my proof for infinite fields. – qwenty Mar 11 '15 at 18:33
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    possible duplicate of Matrix is conjugate to its own transpose – Andrew Dudzik Mar 11 '15 at 19:15
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    Short answer: for finite fields, this appears to require something like the rational canonical form or Smith normal form. – Andrew Dudzik Mar 11 '15 at 19:16
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    @Slade i don't understand why $tI - A$ and $tI - A^t$ have same elementary divisors(if understand correctly(Smith normal form is like Frobenius normal form?)). I understand only why $A, A^T$ have the same characteristical polynomials And in first answer they also consider splitting field, but i don't understand how they solve problem with $\det X = 0$;( – qwenty Mar 11 '15 at 19:30
  • @user2715119 Well, for a shortcut you can use your proof for $\overline{k}$, which is infinite. All that's missing is that the elementary divisors (or the rational/Frobenius form) are the same whether considered over $k$ or $\overline{k}$. – Andrew Dudzik Mar 11 '15 at 19:49
  • I'm sorry if i understand incorrectly, but elementary divisors over $k$ and $\bar{k}$ are not same, for example if $\chi_{ A} = t^2+1$ then it's elementary divisor of $A$ over $\mathbb{R}$, but over $\mathbb{C}$ there are two: $t-i, t+i$. – qwenty Mar 11 '15 at 20:09
  • Let us continue this discussion in chat. – qwenty Mar 11 '15 at 22:28
  • @user2715119 Sorry, I'm probably using the wrong term—I think I mean invariant factors, not elementary divisors. In any case, the blocks that appear in the rational canonical form (a.k.a. the Frobenius normal form) are independent of the field extension. – Andrew Dudzik Mar 12 '15 at 06:04

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