I am investigation spectrum of perturbed operator (discretized from a PDE):
$P=A+ B$, where $A$ is block tri-diagonal (not necessarily self-adjoint). B only has off-diagonal entries, and is only non-zero for first few rows.
Numerical results indicate that the spectrum is actually independent of $B$. I am trying to gain understanding of the reason behind this numerical result.
For some block matrices, certain such results are known. For example, analytical expressions of eigenvalues are available for some classes of such matrices.
Additionally, we know for example that eigenvalues of upper diagonal matrix are same as diagonal entries.
Are there some results which let us deduce whether or not the spectrum depends on certain off diagonal perturbations ?
If the spectrum of $A + t B$ (including algebraic multiplicity) is independent of $t$, then the characteristic polynomial is also independent of $t$. This can occur if the nonzero entries of $A$ and $B$ are placed so that the first minor of $A+B$ for each nonzero entry of $B$ is $0$. For example, this occurs for $A$ and $B$ of the form $$ A = \pmatrix{* & * & * & * & *\cr * & * & * & * & *\cr 0 & 0 & * & * & *\cr 0 & 0 & 0 & * & *\cr 0 & 0 & 0 & * & *\cr},\ B = \pmatrix{0 & 0 & * & * & *\cr 0 & 0 & * & * & *\cr 0 & 0 & 0 & * & *\cr 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0\cr }$$ (where $*$ denotes an entry that may be nonzero). Thus, corresponding to the $(1,j)$ entry with $j \ge 3$, when we delete the first row and $j$'th column of $A+tB$ the first two columns of the resulting matrix have nonzeros only in the first remaining row, resulting in a determinant of $0$.
Similarly, if the $n \times n$ matrix $A$ has an $m \times m$ block at the top left, such that all entries below this block are $0$, and the only nonzero entries of $B$ are in the first $m$ rows to the right of this block, then the spectrum of $A+B$ won't depend on $B$.