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We know in linear algebra that if $A$ and $B$ are $n×n$ complex matrices then$$A\sim B\iff\begin{pmatrix}A&\\&A\end{pmatrix}\sim\begin{pmatrix}B&\\&B\end{pmatrix}.$$I wonder whether it still holds when $A$ and $B$ are integer matrices.(In this case the transition matrix should also be an invertible integer matrix.)

If $A \sim B$ it's easy to show $diag(A,A) \sim diag(B,B).$But the opposite side seems more difficult.I think the opposite side may not hold.But cannot prove it or find a counterexample.Any help will be thanked.

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$A \sim B \Longrightarrow \exists C, \; B = CAC^{-1}; \tag 1$

we thus compute

$\begin{bmatrix} C & 0 \\ 0 & C \end{bmatrix}\begin{bmatrix} A & 0 \\ 0 & A \end{bmatrix}\begin{bmatrix} C^{-1} & 0 \\ 0 & C^{-1} \end{bmatrix} = \begin{bmatrix} CA & 0 \\ 0 & CA \end{bmatrix}\begin{bmatrix} C^{-1} & 0 \\ 0 & C^{-1} \end{bmatrix}$ $= \begin{bmatrix} CAC^{-1} & 0 \\ 0 & CAC^{-1} \end{bmatrix} = \begin{bmatrix} B & 0 \\ 0 & B \end{bmatrix}; \tag 2$

thus

$\begin{bmatrix} A & 0 \\ 0 & A \end{bmatrix} \sim \begin{bmatrix} B & 0 \\ 0 & B \end{bmatrix}. \tag 3$

Note that if $A$ and $B$ are integer matrices and $C$ is an invertible integer matrix in (1), the corresponding applies to (2) and (3).

The reverse implication, going from (3) to (1), appears to be substantially more difficult and I can't write on it at this time.

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    $\begingroup$ Yes that's true but what about the opposite site? $\endgroup$ – Tree23 Apr 11 at 5:30
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    $\begingroup$ @Tree23: what is "the opposite site"? $\endgroup$ – Robert Lewis Apr 11 at 5:33
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    $\begingroup$ Can we have $A \sim B$ when $diag(A,A) \sim diag(B,B)$? $\endgroup$ – Tree23 Apr 11 at 5:36
  • $\begingroup$ @Tree23: need to think a little more about that one! $\endgroup$ – Robert Lewis Apr 11 at 5:38
  • $\begingroup$ @Tree23: well, sorry to say, I don't have an answer to that. But at least I edited my post to reflect that! Cheers! $\endgroup$ – Robert Lewis Apr 11 at 5:46
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If $A = MBM^{-1}$ then

$$\left( \begin{array} { l l } { A } & { } \\ { } & { A } \end{array} \right) = \left( \begin{array} { l l } { M } & { } \\ { } & { M } \end{array} \right) \left( \begin{array} { l l } { B } & { } \\ { } & { B } \end{array} \right) \left( \begin{array} { l l } { M^{-1} } & { } \\ { } & { M^{-1} } \end{array} \right) $$

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