Question:

Matrix inequality with arbitrary large ratios

Emily: 2 days ago

Let $M = (m_{ij})$ be $n \times n$ symmetric positive definite matrix. Then it can be proven that $$ M^{1/2}A M^{1/2} \succeq M^{1/2}D M^{1/2}\succ 0 $$ so \begin{equation} \lambda_{\min}(M^{1/2}A M^{1/2}) \geq \lambda_{\min}(M^{1/2}D M^{1/2}), \tag{!} \end{equation} where D is diagonal matrix with diagonal elements equal to $\frac{1}{m_{ii}}$ and A is matrix with diagonal elements equal to $$a_{ii} = \frac{1}{n-1}\sum_{j = 1,i\neq j}^n \frac{m_{jj}}{m_{ii}m_{jj} - m_{ij}m_{ji}} $$ and non-diagonal elements equal to $$a_{ij} = \frac{-1}{(n-1)} \frac{m_{ij}}{m_{ii}m_{jj} - m_{ij}m_{ji}}. $$

But in reality, ratio of these two minimal eigenvalues can be arbitrary large, for instance $$ M = \begin{bmatrix} 1& 0.99 \\[0.3em] 0.99 & 1 \end{bmatrix}$$ adding $9s$ to $0.99$, ratio $\frac{\lambda_{\min}(M^{1/2}A M^{1/2})}{\lambda_{\min}(M^{1/2}D M^{1/2})}$ is not bounded. For $n = 2$, answer is that $M$ should be close to singular. And their ratio is only equal to $1$ only for diagonal matrix. My intuition is that similar structure with $2 \times 2$ submatrices (close to singular) should apply for higher dimensions. Also my experiments suggest so. Does anybody have idea how to proof that for matrix with some specific structure, this ratio can be large (at least 2).

Answer:
Isaiah: 2 days ago

Consider the family of matrices $M:=M_t:=N^2$, where $N:=P+tI$, $P$ is the $n\times n$ matrix with all entries equal $1$ and $I$ is the $n\times n$ identity matrix, so that $M^{1/2}=N$. Let $t\downarrow0$, so that $N\succ0$ (but barely). Note that $P^2=nP$, and the eigenvalues of $P$ are $n$ and $0$. Then $m_{ii}=d:=n+2t+t^2$ and $m_{ij}=e:=n+2t$ if $i\ne j$. So, $A=-cP+(b+c)I$ and $D=\frac1d\,I$, where $b:=\frac{d^2}{d^2-e^2}$ and $c:=-\frac b{n-1}$.

Next, $M^{1/2}D M^{1/2}=NDN=\frac1d\,M=\frac1d\,(eP+t^2I)$, so that the RHS of (!) is $\frac{t^2}d\sim\frac{t^2}n$.

On the other hand, $M^{1/2}A M^{1/2}=NAN=\frac{1+O(t)}{2(n-1)}\,(nI-P)$, so that the LHS of (!) is $\frac{1+O(t)}2\sim\frac12$, which much greater than the RHS of (!).