Complex matrices are similar over $R$ iff they and thier conjugates are similar over $C$. Prove that there is a nonsingular $T \in M_n(R)$ such that $A = TBT^{−1}$ if and only if there is a nonsingular $S \in M_n$ such that both $A = SBS^{−1}$ and $\bar{A} = S\bar{B}S^{−1}$. (Here $M_n$ is the set of $n \times n$ complex matrices).  If direction: We can take $S = T$ i.e. imaginary part being zero. Then, $A = SBS^{-1}$ obviously. Moreover, $\bar{A} = \overline{SBS^{-1}} = \bar{S}\bar{B}\bar{S^{-1}} = S\bar{B}S^{-1}$. The only if direction: Let $S = H+iC$ and $\bar{S} = H-iC$ where $H,C \in M_n(R)$. Then, $A = SBS^{-1} = (H + iC)B(H+iC)^{-1}, \bar{A} = \bar{S}\bar{B}\bar{S^{-1}} = (H-iC)\bar{B}(H-iC)^{-1} = S\bar{B}S^{−1} =(H + iC)\bar{B}(H+iC)^{-1}$. I guess I should show that $S = \bar{S}$ i.e. $H+iC = H - iC$. Because if complex conjugate of $S$ is equal to $S$ then $S \in M_n(R)$ so, we will complete the proof.
Similar matrices and field extensions Given a field $F$ and a subfield $K$ of $F$. Let $A$, $B$ be $n\times n$ matrices such that all the entries of $A$ and $B$ are in $K$. Is it true that if $A$ is similar to $B$ in $F^{n\times n}$ then they are similar in $K^{n\times n}$?   Any help ... thanks!
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View list of files in ZIP archive on Linux Storage order vs Result order This is a spin-off question from .   says this on the subject of storage order (clustered index) and the output order     A lot of people believe that a clustered index guarantees a sort order on output. But that's not what it does; it guarantees a storage order on disk.   See, for example, .   I've read the blog post by Hugo Kornelis and understands that an index doesn't guarantee that the sql server reads the records in a specific order. Yet I have a hard time accepting that I can't assume this for my scenario?  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