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I know that a matrix is similar to its transpose, if the minimal polynomial over the field splits Now how do I produce a matrix over $\mathbb{R}$ that does not satisfy given property.

Can you give me a hint ?

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    $\begingroup$ Every square matrix $A$ (whose entries are taken from a field) is similar to its transpose, because $A$ and $A^T$ have the same rational canonical form. See this answer by Marc van Leeuwen for more details. $\endgroup$
    – user1551
    Dec 17 '20 at 7:51
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A (square) matrix is always similar to its transponse. One reason (which is not immediate to prove) is that $A,B\in k^{n\times n}$ are similar if and only if they are similar in $\overline k^{n\times n}$: in other words, for two matrices $A,B\in k^{n\times n}$ there is an invertible matrix $P\in k^{n\times n}$ such that $B=PAP^{-1}$ if and only if there is a matrix $Q\in\overline k^{n\times n}$ such that $B=QAQ^{-1}$. In $\overline k^{n\times n}$ all matrices are triangulable, and therefore the result follows.

An intrinsic formulation of this, but not a substantially easier one to prove, is that two matrices $A,B\in k^{n\times n}$ are similar if and only if $\dim\ker (p(A))^s=\dim \ker (p(B))^s$ for all $n\ge1$ and for all irreducible polynomials $p\in k[X]$. With this formulation, you would use identities $\dim\ker M^T=\dim \ker M$ and $p(M^T)=(p(M))^T$.

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