Trace and eigenvalues under multiplication by a matrix with determinant $1$ Let $A,B$ be a $n \times n$ matrices such that $\det(B) = 1$. Will the spectrum
(set of eigenvalues) $AB$ be same as that of $A$. Or, at least is $\mbox{Trace}(A) = \mbox{Trace}(AB)$ ? If not, what can we say about the change in spectrum and trace in $A$ and $AB$.
 A: If two matrices $A,B$ have the same nonzero determinant, then $\det(A^{-1}B)=1$ so $A$ and $B$ are in the same class under multiplication by matrices of determinant $1$. Hence among invertible matrices the only invariant quantity under such multiplication is the determinant itself (of course any function of the determinant is also invariant; this is not very interesting). So there is no chance that the characteristic polynomial, spectrum, trace or whatever is invariant.
A: No. Consider $n=2$, $A=I$, $B$ a diagonal matrix with entries $2, 1/2$ on diagonal. $\det B=1$, but both trace and spectrum of $AB$ are different from those of $A$.
You can't really say anything about the spectrum except that it will not change the product (that is, determinant), which is easy to see by generalizing the example. Choose any $A$, wlog $A$ is in Jordan normal form. Then by a suitable diagonal $B$ we can change the spectrum onto any whatsoever, as long as the product of it stays fixed (and there are as many zeroes for singular $A$).
From that we can see that the trace can be any $\sum_j x_j$ with $\prod x_j=\det (A)$. Unless $A$ is singular, in which case there must be as many zero $x_j$'s as there are zero eigenvalues of $A$ (without any restriction on the others), so for nonzero singular $A$ the trace can be any number.
