# Determinants of matrices

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? –  t.b. Nov 6 '11 at 17:39
@t.b that will work for all $n\neq1$ Erics example doesn't have this problem so I like it more :-) –  Listing Nov 6 '11 at 17:48
@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... –  t.b. Nov 6 '11 at 17:50
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. –  Eric Naslund Nov 6 '11 at 17:54
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 $n=1, t=-1$. –  André Nicolas Nov 6 '11 at 17:56

Let $M$ be half of the identity, $$M=\frac{1}{2} I_n.$$
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.