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Given two real and irreducible matrices $A$ and $B$ of size $n \times n$. A matrix $A$ is irreducible if there is no permutation matrix $Q$ so that $$ Q^{-1} A Q = \begin{bmatrix} E & G \\ 0 & F \end{bmatrix} $$ where $E$ and $F$ are square.

Also, $A$ and $B$ have equal principal minors, i.e, $$ det A(\alpha) = det B(\alpha) \textrm{ for non-empty } \alpha \subset \{1, \dots, n\}, $$ where $A(\alpha)$ is the submatrix with columns and rows indices in $\alpha$.

There is also symmetric $P$ such that $$ P A P^{-1} = B. $$ Are then $A$ and $B$ necessarily diagonally similar? The opposite direction is clear, i.e., for $A$ and $B$ being diagonally similar, $A$ and $B$ have equal principal minors.

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  • $\begingroup$ As-stated, the answer is "no" for uninteresting reasons: $A = B = I$, $P$ doesn't have to be diagonal. Perhaps you mean to ask "if these conditions hold then does there necessarily exist a diagonal $Q$ such that $QAQ^{-1} = B$?" (but $Q$ might not equal $P$). $\endgroup$ Dec 9, 2020 at 19:16
  • $\begingroup$ The statement is false for $A=1$, $B=\bigl( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \bigr)$. $\endgroup$ Dec 9, 2020 at 19:18
  • $\begingroup$ @NathanielJohnston You're right, I will correct this. $\endgroup$ Dec 9, 2020 at 19:20
  • $\begingroup$ Thanks for the comments. Sorry, I missed stating that A and B are irreducible. $\endgroup$ Dec 9, 2020 at 19:22
  • $\begingroup$ @ChristianRemling I clarified the definitions. $\endgroup$ Dec 9, 2020 at 19:59

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$$A=\begin{bmatrix} 1& 0& 1\\1&1&1\\1&1&1 \end{bmatrix} \quad B=\begin{bmatrix} 1& 1& 1\\0&1&1\\1&1&1 \end{bmatrix} \quad P=\begin{bmatrix} 0& 1& 0\\1&0&0\\0&0&1 \end{bmatrix}$$

$P^{-1}=P$ and $B=PAP^{-1}$. $A$ is irreducible since conjugation by a permutation matrix simply moves around the single zero in $A$ (all of the entries of $A^2$ are positive). The principle minors of $A$ and $B$ are equal. However, if $D=\begin{bmatrix} a& 0& 0\\0&b&0\\0&0&c \end{bmatrix}$ is any invertible diagonal matrix, $$DAD^{-1}=\begin{bmatrix} 1& 0& a/c\\b/a&1&b/c\\c/a&c/b&1 \end{bmatrix}\neq B$$.

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  • $\begingroup$ I believe $A$ is reducible, as for $P=\bigl( \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{smallmatrix} \bigr)$, we have $P A P^{-1} = \bigl( \begin{smallmatrix} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{smallmatrix} \bigr)$. $\endgroup$ Dec 11, 2020 at 13:31
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    $\begingroup$ The matrices $E$ and $F$ are $1\times 2$ and $2\times 1$? Also, $A^2=\begin{bmatrix} 2& 1& 2\\3&2&3\\3&2&3 \end{bmatrix}$ which has no zero entries, which I thought implied irreducibility. $\endgroup$ Dec 11, 2020 at 13:38

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