Let $A = (a_{ij})$ and $B = (b_{ij})$ be two positive-definite matrices such that $$a_{ij} \geq b_{ij} \text{ for all $i,j$ and for all $i$, } a_{ii} > b_{ii}\,.$$

That is, $A \geq B$ elementwise and the diagonals of $A$ are greater than the diagonals of $B$. In that case, can we say that $\lambda_{A,i} > \lambda_{B,i}$ for $i = 1, \dots, p$?

Intuitively, I think this should be true, but I am not able to show it mainly because I don't know if $A-B$ is positive definite.

  • 1
    $\begingroup$ Check [1, 2 ; 2, 1]... $\endgroup$ – user251257 Jul 16 '18 at 19:01

Counter-example: $$ A=\begin{pmatrix}5 & 0\\ 0 & 5 \end{pmatrix}\qquad B=\begin{pmatrix}\phantom{+}4 & -3\\ -3 & \phantom{+}4 \end{pmatrix} $$ Addendum: In case you are interested, it is worthwhile to point out that this disproves the claim in a slightly stronger setting since $A$ and $B$ are Stieltjes matrices (a subset of the SPD matrices).


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