Why $\alpha I +A$ can improve the condition nubmer of a SPD matrix $A$?

Question

For Poisson equation with Dirichlet boundary conditions in 2 dimension: $$ -\Delta u=f, $$ using FDM (centered difference) or FEM discretization, we can obtain a SPD system of linear equations as follows: $$ Ax=b. $$ And if the step size is $h$, then the spectral condition number of matrix $A$ is $O(h^{-2})$.

A conclusion goes that "Given a positive constant $\alpha>0$, then matrix $\alpha I+A$ is well-conditioned". Why the addition of a positive term $\alpha$ to the main diagonal of matrix $A$ can improve the condition number of matrix $A$?

Because in my opinion, the new matrix condition number is

$$ \mathrm{cond}_2(\alpha I+A) = \frac{\alpha+\lambda_{\max}(A)}{\alpha+\lambda_{\min}(A)}>\frac{\lambda_{\max}(A)}{\lambda_{\min}(A)}=\mathrm{cond}_2(A). $$

But the numerical results contrast as follows:

clc;clear;
n=10;
A=gallery('poisson',n);
cond(full(A))

n=10;
A=gallery('poisson',n);
cond(full(A)+speye(n^2))

n=20;
A=gallery('poisson',n);
cond(full(A)+speye(n^2))

The numerical results as follows:

ans =

   48.3742


ans =

    7.6056


ans =

    8.5723

As is seen, the condition number of $I+A$ is less than $A$ (48.3742 > 7.6056).

Furthermore, when the system size increases, the condition number almost do not increase (from 7.6056 to 8.5723), which seems that the condition number of matrix $\alpha I+A$ is independent on $h$. Why this happens? Does it really independent on step size $h$?

Answer 1

The inequality $\frac{\alpha + \lambda_{max}(A)}{\alpha+\lambda_{min}(A)} > \frac{\lambda_{max}(A)}{\lambda_{min}(A)}$ doesn't hold because $\lambda_{max}(A) > 1 > \lambda_{min}(A) > 0$ and $\alpha > 0$. In fact, the reverse inequality holds in this case. Hence, there is no contradiction here.

EDIT: From the comments, I realized that the question is not answered completely yet.

Also, $\kappa(A+\alpha I) = \frac{\alpha + \lambda_{max}(A)}{\alpha+\lambda_{min}(A)} = \frac{\frac{\alpha}{\lambda_{max}(A)}+1}{\frac{\alpha}{\lambda_{max}(A)}+\frac{\lambda_{min}(A)}{\lambda_{max}(A)}} = \frac{\frac{\alpha}{\lambda_{max}(A)}+1}{\frac{\alpha}{\lambda_{max}(A)}+O(h^2)}$.

Hence, for small enough $h$ and big enough $\alpha$, we have $\kappa(A) \approx 1+\frac{\lambda_{max}(A)}{\alpha}$. One can see that $\lambda_{max}(A)$ changes very little. For example,

poisson = @(n) full(gallery('poisson',n,n));
max(abs(eig(poisson(10))))

ans =

7.837971894457977

max(abs(eig(poisson(40))))

ans =

7.988263204734964

max(abs(eig(poisson(60))))

ans =

7.994696359539318