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Suppose $A_1,A_2$ are positive definite matrices such that $\lambda_{min}(A_1) \leq \lambda_{min}(A_2)$ where $\lambda_{min}(.)$ denotes the minimum eigenvalue. Let $B$ be a positive semi-definite matrix. Is it true that $\lambda_{min}(A_1+B) \leq \lambda_{min}(A_2+B)$?

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    $\begingroup$ Yes. Hint: for a Hermitian matrix $A$, the minimum eigenvalue is the minimum of $\langle Av, v \rangle$ over all unit vectors $v$. $\endgroup$ – keej May 2 '17 at 7:58
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No. Counterexample: consider $A_1=\pmatrix{1\\ &2}$ and $A_2=B=\pmatrix{2\\ &1}$.

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  • $\begingroup$ thanks, what if $\lambda_{min}(A_1)< \lambda_{min}(A_2)$? $\endgroup$ – Sanand May 2 '17 at 8:43
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    $\begingroup$ @Sanand It doesn't matter. The counterexample still works if you replace the $1$ in $A_1$ by $1-\epsilon$ for some small $\epsilon>0$. $\endgroup$ – user1551 May 2 '17 at 8:49
  • $\begingroup$ yes, thanks I found a counterexample already inspired by your previous one!! $\endgroup$ – Sanand May 2 '17 at 9:08
  • $\begingroup$ @user1551 In addition to OP's conditions, if we have $A_2 - A_1 > 0$, does that imply what he needs? $\endgroup$ – dineshdileep May 2 '17 at 12:39
  • $\begingroup$ @dineshdileep If $A_2-A_1>0$, then $(A_2+B)-(A_1+B)>0$, so the minimum eigenvalue of $A_2+B$ dominates that of $A_1+B$. $\endgroup$ – user1551 May 2 '17 at 13:30

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