Approximating a determinant by the matrix diagonal

Question

I asked a related question on Physics SE, but it hasn't received much attention and I think it may be more of a mathematical problem than a physical one.

Suppose I wish to solve $$det(x\mathbb{I}-\mathbb{A})=0$$ for $x$ assuming the matrix $\mathbb{A}$ is not diagonal in general and $\mathbb{I}$ is the identity. What error do I incur by ignoring the off-diagonal elements of $\mathbb{A}$ to approximate the solution? Is there a general way to know, based on properties of the matrix, how good/poor this approximation is (for example, some sort of error bound)?

Answer 1

This is related to the derivative of the determinant: Wikipedia: Jacobi's formula

This is how I interpret your post in Physic SE:

To find the lowest order correction to these Koopman's theorem results, let us ignore the off-diagonal elements of Σ(E).

If your matrix $A=D+hB$ where $D$ is diagonal, and $h\ll 1$.

To simplify, let assume $A$ contains no zero diagonal element, so $D$ is invertible.

$$\det(A)=\det(D)+h\det(D)\operatorname{tr}(D^{-1}B)+o(h)$$

So when the diagonal is dominant and the rest of the matrix $hB$ is of lower order (i.e. negligible compared to the diagonal) then the determinant is approximable by $\det(D)$.

When $D$ is not invertible, the formula is a little bit more complicated (see Wiki's page), but I think you get the principle.

Answer 2

If $\mathbb{A}$ is of the form $\Lambda+\epsilon \mathbb{B}$ where $\Lambda$ is diagonal and $\epsilon$ "small", then basically you use standard perturbation theory with all the caveats therein. This will allow you to approximate the eigenvalues as a series in $\epsilon$. The determinant is then the product of the eigenvalues. After all, perturbation theory is just approximate diagonalization order by order.