Let $\textbf D$ be a square matrix with eigenvalues separated into two or more groups using their magnitude (say the first group of eigenvalues is $\{1,2,\dots,10\}$ and the second group has eigenvalues $\{1010,1021,1051,\dots,10000\}$).
Then consider the eigenvalue relation of transition matrix $\textbf{D}$, $$\textbf{D}\textbf{R}=\textbf{R}\Lambda$$ Partitioning $\Lambda$ gives $$\textbf{D}\textbf{R}=\textbf{R} \begin{bmatrix}\Lambda_{dd}&\textbf{0}_{dm}\\\textbf{0}_{md}&\Lambda_{mm}\end{bmatrix}$$ wheere $\Lambda_{dd}$ has the smallest group and $\Lambda_{mm}$ has the largest group of eigenvalues. Rescale the system using $k$ (a value with magnitude greater than the small group of eigenvalues and less than large group of eigenvalues - say 100 or 1000 in the previous example of eigengroups) as $$\frac{1}{|k|}\textbf{D}\textbf{R}=\frac{1}{|k|}\textbf{R} \begin{bmatrix}\Lambda_{dd}&\textbf{0}_{dm}\\\textbf{0}_{md}&\Lambda_{mm}\end{bmatrix}$$ $$\frac{1}{|k|}\textbf{D}\textbf{R}=\textbf{R} \begin{bmatrix}\frac{\Lambda_{dd}}{|k|}&\textbf{0}_{dm}\\\textbf{0}_{md}&\frac{\Lambda_{mm}}{|k|}\end{bmatrix}$$ Assume that $k$ infinitely large, then $\lim\limits_{|k|\rightarrow \infty}\frac{\Lambda_{dd}}{|k|}=\textbf{0}_{dd}$,
$$ \tilde{\textbf{D}}\textbf{R}=\textbf{R}\begin{bmatrix}\textbf{0}_{dd}&\textbf{0}_{dm}\\\textbf{0}_{md}&\frac{\Lambda_{mm}}{|k|}\end{bmatrix}$$
However, to rescale, I need to know what is the separation between eigenvalues to find a suitable k. This is hard in large dimensional matrices due to the need of computing a large number of eigenvalues. Then instead of checking the gap in eigenvalues, How can confirm that rescaling using the gap in elements of D will rescale the eigenvalues to zero.
A worked out example is given below
In the first figure, elements of D is sorted by their magnitude and second figure shows the eigenvalues. If I rescale D using $10^4$ (A value in the second gap in the element spectrum. first gap is not useful in the case), I will get the first group of eigenvalues rescaled to zero. Then by $10^6$, I may (sometimes there will be the same number of zero eigenvalues by stepping up) rescale the second group to zero and so on.
How can I confirm this is true in general or for some specific cases under some conditions?.
A small observation about infinity.
What is infinity? When we say a value $x$ is infinite, what are we meant by that?.
To my knowledge: if $x$ is defined as $\infty$, then $1/x=0$. But we know for any large number, no $x$ will satisfy this condition. There will be a decimal value at some point in $1/x$. The question is, are we considering this decimal point or not. If we are truncating it, then we can say $x=\infty$ which is an approximation.
If someone says, 100 is infinite for him, then $1/100=0.01$ is a zero for him. By using this fact I trucated the matrix D by removing the decimal point which satisfies the limit condition on $k$.