# Principal minors and similarity

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.

• 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 $P$). – Nathaniel Johnston Dec 9 '20 at 19:16
• The statement is false for $A=1$, $B=\bigl( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \bigr)$. – Christian Remling Dec 9 '20 at 19:18
• @NathanielJohnston You're right, I will correct this. – Sebastian Schlecht Dec 9 '20 at 19:20
• Thanks for the comments. Sorry, I missed stating that A and B are irreducible. – Sebastian Schlecht Dec 9 '20 at 19:22
• @ChristianRemling I clarified the definitions. – Sebastian Schlecht Dec 9 '20 at 19:59

## 1 Answer

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

$$P^{-1}=P$$ 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, $$DAD^{-1}=\begin{bmatrix} 1& 0& a/c\\b/a&1&b/c\\c/a&c/b&1 \end{bmatrix}\neq B$$.

• 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 $P A P^{-1} = \bigl( \begin{smallmatrix} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{smallmatrix} \bigr)$. – Sebastian Schlecht Dec 11 '20 at 13:31
• 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. – Edwin Franks Dec 11 '20 at 13:38