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Iosif Pinelis
Iosif Pinelis
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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:=-\frac1{n-1}$$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 (!).

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 $c:=-\frac1{n-1}$.

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 (!).

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 (!).

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Iosif Pinelis
Iosif Pinelis
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LetConsider the family of matrices $M:=N^2$$M:=M_t:=N^2$, where $N:=P+tI$, $P$ is the $n\times n$ matrix with all entries equal $1$, and $I$ is the $n\times n$ identity matrix, andso that $M^{1/2}=N$. Let $t\downarrow0$, so that $M^{1/2}=N\succ0$$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 $c:=-\frac1{n-1}$.

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 (!).

Let $M:=N^2$, where $N:=P+tI$, $P$ is the $n\times n$ matrix with all entries equal $1$, $I$ is the $n\times n$ identity matrix, and $t\downarrow0$, so that $M^{1/2}=N\succ0$ (but barely), $P^2=nP$, 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 $c:=-\frac1{n-1}$.

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 (!).

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 $c:=-\frac1{n-1}$.

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 (!).

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Iosif Pinelis
Iosif Pinelis
  • 128k
  • 8
  • 107
  • 229

Let $M:=N^2$, where $N:=P+tI$, $P$ is the $n\times n$ matrix with all entries equal $1$, $I$ is the $n\times n$ identity matrix, and $t\downarrow0$, so that $M^{1/2}=N\succ0$ (but barely), $P^2=nP$, 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 $c:=-\frac1{n-1}$.

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 (!).