I remember reading somewhere about the following properties of non-negative definite matrix. But I don't know how to prove it now.

Let $A$ and $B$ be two non-negative definite matrices. If $A^2\succ B^2$, then it necessarily follows that $A\succ B$, but $A\succ B$ doesn't necessarily leads to $A^2\succ B^2$.

How can you prove it? Thanks!

  • $\begingroup$ Are you sure about the other side? Because I feel that the other direction should also be correct. $\endgroup$ – Arash Oct 1 '13 at 8:47

In general, if $X\succ0$ and $Y\succeq0$, then $X\succ Y$ if and only if $\rho(X^{-1}Y)<1$.

Now suppose $A\succ B\succeq0$ and $A^2\succeq B^2$. Then $\rho\left(A^{-2}B^2\right)<1$ and hence $$ \rho(A^{-1}B)^2 \le\|A^{-1}B\|_2^2 =\rho\left(A^{-1}B(A^{-1}B)^\ast\right) =\rho(A^{-1}B^2A^{-1}) =\rho\left(A^{-2}B^2\right)<1. $$ Thus $\rho(A^{-1}B)<1$ and $A\succ B$.

By a limiting argument, we may also prove that $A\succeq B\succeq0$ if $A^2\succeq B^2\succeq0$. Pick any $t>0$. By unitarily diagonalising $A$, we see that $P=A+tI$ is positive definite and $P^2\succ A^2$. Thus $P^2\succ B^2\succeq0$ and $P\succ B$. Let $t\to0$, we get $A\succeq B$.

However, note that the converses to the above are not true. E.g. when $A\succ B$, it is not necessarily true that $A^2\succ B^2$, as illustrated by the following counterexample where $\epsilon>0$ is small: \begin{align*} A&=\pmatrix{2+\epsilon&1\\ 1&3},\\ B&=\pmatrix{1\\ &2},\\ A-B&=\pmatrix{1+\epsilon&1\\ 1&1}\succ0,\\ A^2&=\pmatrix{(2+\epsilon)^2+1&5+\epsilon\\ 5+\epsilon&10},\\ B^2&=\pmatrix{1\\ &4},\\ A^2-B^2&=\pmatrix{(2+\epsilon)^2&5+\epsilon\\ 5+\epsilon&6}\not\succ0. \end{align*}

  • $\begingroup$ What is meant by the notation $x^H$? $\endgroup$ – Babai Aug 9 '16 at 5:39
  • 1
    $\begingroup$ @Babai It's the conjugate transpose (aka $\color{red}{H}$ermitian transpose) of $x$, i.e. $x^\ast$. The notation $x^\ast$ is more popular in theoretical linear algebra literature, while $x^H$ is more popular in numerical linear algebra literature. $\endgroup$ – user1551 Aug 9 '16 at 9:44
  • $\begingroup$ Hello, what if we have $A,B,A-B$ non-negative definite matrix (which means we may not have $A^{-1}$) , can we have $\sqrt{A}-\sqrt{B}$ also is non-negative definite matrix? Thanks. $\endgroup$ – Idele Jan 17 '17 at 15:52

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