# Producing a matrix not similar to its transpose [duplicate]

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 ?

• 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. Dec 17 '20 at 7:51

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 $$p(M^T)=(p(M))^T$$.