Product of matrices has real eigenvalues?

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Let $A$ be a (symmetric) positive definite matrix and $\hat{n}$ be an arbitrary unit vector. Consider $b,c,d,k$ arbitrary positive integers. I would like to know if the following matrix has real eigenvalues (I would like the more general answer, for more than product of four matrices but even for three I don't know).

$$(I - \hat{n}\hat{n}^T)\cdot A^b\cdot(I - \hat{n}\hat{n}^T)\cdot A^c\cdot (I - \hat{n}\hat{n}^T)\cdot A^d\cdot (I-\hat{n}\hat{n}^T)\cdot A^k\cdot (I-\hat{n}\hat{n}^T).$$

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  1. Let $P=I-\hat{n}\hat{n}^T$. Suppose that $A$ is positive definite, $m_1,m_2,\ldots,m_r$ are some positive integers and $$ M=PA^{m_1}PA^{m_2}\cdots PA^{m_r}P $$ has some non-real eigenvalues. Since $\lim_{k\to\infty}MA^{1/k}P=M$, if we put $B=A^{1/k},n_{r+1}=1$ and $n_i=km_i$ for each $i\le r$, then the product $$ M'=PB^{n_1}PB^{n_2}\cdots PB^{n_r}PB^{n_{r+1}}P $$ also has some non-real eigenvalues when $k$ is sufficiently large. In other words, if the statement in the OP is false when $r$ powers of $A$ are involved, it is in general also false when more than $r$ powers are involved. Since the statement is true when $r=2$, if one wants to disprove the statement, one may consider the case $r=3$ first.
  2. By changing the orthonormal basis, one may assume that $\hat{n}=(0,0,\ldots,0,1)^T$. The problem thus reduces to the following: let $A$ be an $n\times n$ positive definite matrix. For any positive integer $m$, denote by $S_m$ the leading principal $(n-1)\times(n-1)$ submatrix of $A^m$. If $r\ge3$ and $m_1,m_2,\ldots,m_r$ are $r$ positive integers, is it necessarily true that $S_{m_1}S_{m_2}\cdots S_{m_r}$ has a real spectrum?
  3. When $n=2$, the answer is affirmative because every $S_m$ is a positive number. When $n=3$ and $A$ is also entrywise nonnegative, the answer is also affirmative because each $S_m$ is a $2\times2$ entrywise nonnegative matrix. By Perron-Frobenius theorem, $\rho(S_{m_1}\cdots S_{m_r})$ is a real eigenvalue of $S_{m_1}\cdots S_{m_r}$. Hence the other eigenvalue is real too.