Suppose $M$ is some $n\times n$ real matrix. Then is it always true that $\det M\neq\det(I_n-M)$, where $I_n$ is the $n\times n$ identity matrix? I have a feeling that it is yes, but I am not sure... Thanks.
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The question has been raised in the comments, given $\det M=\det(I-M)$, what values can $\det M$ take?
Note that $\det M=\prod\lambda$ and $\det(I-M)=\prod(1-\lambda)$, where the products are over all the eigenvalues $\lambda$ of $M$, with multiplicity. In the case $n=2$, we get $ab=(1-a)(1-b)$, which becomes $a+b=1$, so we are looking for the range of $ab$ over pairs such that $a+b=1$. If $a$ and $b$ are real, this implies $ab$ can be any real not exceeding $1/4$. But $a$ and $b$ need not be real. If $a=(1/2)+it$ and $b=(1/2)-it$, then $ab=t^2+(1/4)$, which can take on any real value not less than $1/4$. In short, the range is all of $\bf R$.
For $n=3$, you don't have to go to the complex numbers to get arbitrarily large $\det M$. We want $abc=(1-a)(1-b)(1-c)$. Pick a large number $Q$, let $a=b=Q$, and solve for $c$; $c=(Q-1)^2/(Q^2+(Q-1)^2)$. So $\det M=Q^2(Q-1)^2/(Q^2+(Q-1)^2)$, which is unbounded.