## Convergence of Eigenvalues and Eigenvectors for Uniformly Form-Bounded Operators

Suppose that $$A$$ is an operator on a dense domain $$D(A)\subset L^2$$ with compact resolvent, and with quadratic form $$q(f,g):=\langle f,Ag\rangle$$.

Let $$(r_n)_{n\in\mathbb N}$$ be a sequence of quadratic forms on $$D(A)$$ that are uniformly form-bounded by $$A$$, in the sense that there exists $$0<\alpha<1$$ and $$\beta>0$$ independent of $$n$$ such that $$|r_n(f,f)|\leq \alpha\cdot q(f,f)+\beta\cdot\|f\|_2,\qquad f\in D(A).$$ Since $$\alpha<1$$ the operators $$A_n$$ defined by the form $$q+r_n$$ all have compact resolvent.

Suppose that the $$r_n$$ have a limit $$r_\infty$$, in the sense that $$\lim_{n\to\infty}r_n(f,f)\to r_\infty(f,f)$$ for every $$f\in D(A)$$. Define the operator $$A_\infty$$ through the form $$q+r_\infty$$. Clearly $$|r_\infty(f,f)|\leq \alpha\cdot q(f,f)+\beta\cdot\|f\|_2,\qquad f\in D(A),$$ and thus $$A_\infty$$ has compact resolvent as well.

Question. Does the uniform form-bound of the $$r_n$$ give rise to a dominated convergence-type result for the spectrum of $$A_n$$, that is, the eigenvalues of $$A_n$$ converge to that of $$A_\infty$$, and the eigenfunctions converge in $$L^2$$?

## Rank = # of non-zero eigenvalues of a diagonalizable matrix

Is the rank of a matrix equal to the # of non-zero eigenvalues of a diagonalizable matrix?

## Inaccurate zero eigenvalues for 7*7 matrix

I have symbolic entries for all the elements of a 7*7 matrix. At the symbolic level Eigenvalues gives two zeroes and five others that are extremely complicated. At the same time Det evaluates to exactly zero.

I take this to mean that, no matter what values the symbols within the matrix are assigned, I will have at least two zero eigenvalues. Exactly zero eigenvalues.

However, when I assign values to the symbols by hand, for eg., a=10; b=50, and so on, and evaluate the same codes, Eigenvalues evaluates to give two eigenvalues of the order of 10^-12. For my purpose, this magnitude cannot be treated as a zero. And I am also in need of a different eigenvalue of the same matrix which is very small, of this order or even smaller, but is not exactly zero. So I need the zeroes to show up much more accurately.

I have tried adding \$ MinPrecision=50 to my code before executing the rest of it. It does not help at all.

A similar question was posted, Is there a good way to check, whether a small value produced numerically is a symbolic zero? However I could not decipher anything of use out of it.