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Let $M_1$ and $M_2$ be symmetric positive definite matrices and $M_2 > M_1$ in the Loewner ordering, i.e., $M_2 - M_1$ is positive definite. Does this imply that $M_1^{-1} > M_2^{-1}$?

Rodrigo de Azevedo
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user2792830
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1 Answers1

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The answer is yes. Two facts first:

(1) The statement $M_2>M_1$ is equivalent to $x^TM_2x>x^TM_1x$ for any $x\neq 0$;

(2) For any symmetric positive definite matrix $M$, there exist a positive definition matrix $L$ such that $M=L^2$ (called the square root of $M$).

We can show it is true when $M_1$ is the identity matrix $I$: for $M_2=L_2^2$, $$ x^TM_2^{-1}x=x^TL_2^{-T}L_2^{-1}x=(L_2^{-1}x)^T(L_2^{-1}x) \leq (L_2^{-T}x)^TM_2(L_2^{-T}x)=x^Tx. $$

In the general case for $M_1=L_1^2$, the condition $M_2>M_1$ is equivalent to $L_1^{-1}M_2L_1^{-1}>I$, which implies that $ I>(L_1^{-1}M_2L_1^{-1})^{-1}=L_1M_2^{-1}L_1 $ or $M_1^{-1}>M_2^{-1}$.

Community
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yhhuang
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  • how do you conclude that $(L_2^{-1} x)^T (L_2^{-1} x) \leq (L_2^{-T} x)^T M_2 (L_2^{-T} x)$? And how do you conclude that $(L_2^{-T} x)^T M_2 (L_2^{-T} x) = x^T x$? And how do you conclude that $M_2 > I \Leftrightarrow L_1^{-q}M_2L_1^{-1}$? I think it cannot come from multiplying the equation by $M_1 = L_1 L_1^T$ (see http://math.stackexchange.com/questions/1789462/is-the-loewner-ordering-maintained-after-multiplication-with-positive-definite-m). – Qaswed Jun 02 '16 at 11:48
  • @Qaswed $(L_2^{-1}x)^T(L_2^{-1}x) = (L_2^{-1}x)^TM_2(L_2^{-1}x) \leq (L_2^{-1}x)^TI(L_2^{-1}x)$ because $I\leq M$. And $(L_2^{-1}x)^TM_2(L_2^{-1}x) = x^T L_2^{-1} L_2^2 L_2^{-1} x=x^Tx$ because $M=L_2^2$ and $L_2$ is symmetric. And your third question follows since generally $A>B$ implies $LAL>LBL$ for positive $L$ since $\forall x\neq0.\ x^TAx>x^TBx$ implies $x^TLALx=(Lx)^TA(Lx)>(Lx)^TB(Lx)=x^TLBLx$. – Dominique Unruh Apr 13 '18 at 20:00