Effective Rank

Def Effective Rank
The effective rank of a matrix $A\in {\mathbb{R}}^{n\times n}$ with eigenvalues ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m$ in descending order is defined as:$r_k(A)=\frac{{\sum }_{i>k}{\lambda }_i}{{\lambda }_{k+1}}$Prop For all matrix $A\in {\mathbb{R}}^{n\times n}$, $r_k(A)\ge 1$.
Proof $r_k(A)=\frac{{\sum }_{i>k}{\lambda }_i}{{\lambda }_{k+1}}\ge \frac{{\lambda }_{k+1}}{{\lambda }_{k+1}}=1$

Thrm Suppose $A\in {\mathbb{R}}^{n\times n}$, then${\lambda }_k={\lambda }_1(r_0(A)-1)r_{k-1}(A{)}^{-1}{\prod }_{j=1}^{k-2}(1-r_j^{-1})$Proof Since $r_k(A)=\frac{{\sum }_{i>k}{\lambda }_k}{{\lambda }_{k+1}}$,${\lambda }_{k+1}{r}_k(A)={\sum }_{i>k}{\lambda }_k⟹{\lambda }_{k+1}(r_k(A)-1)={\sum }_{i>k+1}{\lambda }_i$Thus ${\lambda }_k(r_{k-1}(A)-1)={\sum }_{i>k}{\lambda }_i$. Subtract the above two equations,${\lambda }_{k+1}(r_k(A)-1)-{\lambda }_k(r_{k-1}(A)-1)=-{\lambda }_{k+1}$It implies $\frac{{\lambda }_{k+1}}{{\lambda }_k}=\frac{r_{k-1}(A)-1}{r_k(A)}$. Hence we have$\begin{array}{rl}{\lambda }_k& =\frac{{\lambda }_k}{{\lambda }_{k-1}}\cdot \frac{{\lambda }_k}{{\lambda }_{k-1}}\dots \frac{{\lambda }_2}{{\lambda }_1}\cdot {\lambda }_1\\ & ={\lambda }_1\cdot \prod _{j=2}^k\frac{r_{j-2}(A)-1}{r_{j-1}(A)}\\ & ={\lambda }_1(r_0(A)-1)r_{k-1}(A{)}^{-1}\prod _{j=1}^{k-2}(1-r_j(A{)}^{-1})\end{array}$