# Given a square matrix where $a_{11}=c\neq 0$ and $a_{ij}=0$ otherwise, can we find a matrix B such that B and A+B have no common eigenvalues?

Given a matrix where $a_{11}=c\neq 0$ and $a_{ij}=0$ otherwise, can we find a matrix B such that B and A+B have no common eigenvalues?

If instead the matrix had its nonzero entry component at $a_{1n}$ (or $a_{n1}$) an easy example would be to take the matrix with ones on the subdiagonal (or superdiagonal) and zeros elsewhere. This matrix would be nilpotent and thus its only eigenvalue would be zero, but whereas the sum matrix would be invertible and therefore have no zero eigenvalues.

Returning the original question, for the $n=2$ case, we can take the permutation matrix which has eigenvalues $\pm1$ and the sum matrix would be the Fibonacci matrix which has irrational eigenvalues. I'm not sure how to create a general example and see the result in a simple way. I'd imagine the construction wouldn't be too different from this case, but I also would think there is an easier way to do this. I considered looking at characteristic polynomials but didn't see anything enlightening there.

Thanks!

If $p(\lambda)$ is the characteristic polynomial of $B$ and $q(\lambda)$ is the characteristic polynomial of the matrix $C$ obtained from $B$ by removing the first row and column, then the characteristic polynomial of $A+B$ is $p(\lambda) + c q(\lambda)$. So all you need to do is ensure that $q$ and $p$ share no roots. That will almost always be the case. For a simple example, you might take $A$ to be the companion matrix for a polynomial with nonzero constant term $p(0)$, in which case $p(\lambda) = \lambda q(\lambda) + p(0)$.