If $A$ has eigenvalues $\lambda_1,...,\lambda_n$, is there a relationship between the eigenvalues of $A$ and $\hat{A}$ Suppose a square matrix $A$ has eigenvalues $\lambda_1,...,\lambda_n$.
Note that A is a $n\times n$ matrix where $n$ is even.
Let $\widehat{A}$ be a matrix that is obtained from $A$ only by multiplying every second row in $A$ by $-1$. Is there a relationship between the eigenvalues of $A$ and $\widehat{A}$?
I don't think there is, but I am not sure.
 A: Write $D=\text{diag}(1,-1,1,-1,\dots)$, so that $\hat{A} = DA$.  Then we have $$\det(\hat{A}) = \det(D)\det(A) = (-1)^{n/2}\det(A),$$ so the product of the eigenvalues of $A$ is the same as that of $\hat{A}$, up to a sign.
In general, you can't expect strong relationships among the eigenvalues of $A$ and $\hat{A}$, for the following reason: Choose any two sets of eigenvalues $\lambda_1,\dots,\lambda_n$ and $\chi_1,\dots,\chi_n$.  Naively we might expect it should be possible to find a matrix $A$ s.t. $A$ has the first set of eigenvalues and $\hat{A}$ has the second.  This is because $\lambda$ is an eigenvalue of $A$ iff $\det(A-\lambda) = 0$, so we are effectively solving $2n$ equations (one for each of $\lambda_i$ and $\chi_i$) in the $n^2$ coefficients of $A$.  Provided $n\geq 2$ (so that $n^2\geq 2n$) , we expect a solution to exist.
This argument is not rigorous, because the equations involving $A$ and those involving $\hat{A}$ need not be independent.  (Indeed, we know it has to be false because of the determinant relationship above.)  However, some experiments in Wolfram Alpha suggest that e.g. for $n=2$ it is possible to choose all but one of the eigenvalues of $A$ and $\hat{A}$ independently.  Perhaps the determinant relationship is the only one that exists among the eigenvalues of $A$ and $\hat{A}$?
However, for some special classes of matrices we can find further relationships: Note that $\hat{A}^\dagger\hat{A} = A^\dagger D^\dagger D A = A^\dagger A$.  For a normal matrix $M$, if $M$ has eigenvalues $\lambda_i$ then $M^\dagger M$ has eigenvalues $|\lambda_i|^2$.  This means that if $A$ and $\hat{A}$ are both normal, then their eigenvalues must be equal up to complex phases. 
