Let $K$ be a field and $A$ a square matrix with entries in $K$. Then A and $A^\top$ have the same characteristic polynomial. What do we know about similarity? Do you have an example where $A$ and $A^\top$ are not similar?
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1They are similar, however to prove it (at least how I know to prove it) using the Jordan form of $A$ and $A^t$. See http://math.stackexchange.com/questions/94599/a-matrix-is-similar-to-its-transpose – Ofir Schnabel Jul 28 '15 at 11:24
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The easiest proof is perhaps via Smith normal form. If $P(\lambda)(\lambda I-A)Q(\lambda)=D(\lambda)$ then $Q(\lambda)^T(\lambda I-A^T)P(\lambda)^T=D(\lambda)^T=D(\lambda)$ so $A$ and $A^T$ share the invariant factors (on the diagonal of $D(\lambda)$), thus, are similar. – A.Γ. Jul 28 '15 at 15:56
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$A$ and $A^T$ are necessarily similar. To show this, it suffices to note that $$ \dim \ker [(A - \lambda I)^k] = \dim \ker [(A^T - \lambda I)^k] $$ for all $\lambda$ taken from the algebraic closure of $K$ and all $k \in \Bbb N$.
Ben Grossmann
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This shows the fact for algebraic closed fields. For arbitrary fields $K$ it's left to show that the matrix $S$ with $A=S^{-1}A^\top S$ has its entries in $K$ instead of $\bar K$. – principal-ideal-domain Jul 28 '15 at 13:46
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You mean that a suitable $S$ with entries in $K$ exists. Yes, it is necessary to make that observation. – Ben Grossmann Jul 28 '15 at 14:43
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Perhaps it helps to note that $$ {S: SA - A^TS = 0} $$ is a linear subspace – Ben Grossmann Jul 28 '15 at 14:45
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