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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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    How about a matrix having exactly one $1$ on the diagonal and which is zero elsewhere?2011-11-06
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    @t.b that will work for all $n\neq1$ Erics example doesn't have this problem so I like it more :-)2011-11-06
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    @Listing: right. That's why this is a comment :) Eric's example also has the virtue that both $M$ and $I-M$ are invertible. (slightly) more interesting were a non-diagonal example...2011-11-06
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    Another question that comes up, is given $\det M = \det(I_n-M)$ what values can $\det M$ take? For $n=1$ it can only be $\frac{1}{2}$ and for $n\geq 2$ I am not so sure. We can prove that it can be anything in $[0,\frac{1}{2^n}]$ by letting $M=\frac{1}{2}I_n$ and changing two elements to $\alpha$ and $1-\alpha$. Then varying $\alpha$ allows us to chose anything. Perhaps the range is longer.2011-11-06
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    There is so much freedom that we can arrange for $\det M$ to be $t\det(I_n-M)$ for any $t$, with the sole exception of $n=1, t=-1$.2011-11-06
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    @Eric Naslund: For $n=2$, one can pin down the range exactly. For larger $n$, I have no intuition.2011-11-06

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