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If $A, B, A^2-B^2$ are real symmetric positive definite matrices, prove that $A-B$ is also positive definite .

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closed as off-topic by Davide Giraudo, E. Joseph, qwr, Jack's wasted life, Michael Albanese Nov 17 '16 at 2:50

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  • $\begingroup$ @kotomord but we don't have AB=BA $\endgroup$ – Idele Nov 16 '16 at 10:45
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    $\begingroup$ One answer is given here $\endgroup$ – Omnomnomnom Nov 16 '16 at 12:58
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Let $M> 0$ mean that $M$ is positive definite. We have $$ A^2 - B^2 > 0 \implies\\ A^{-1}(A^2 - B^2)A^{-1} > 0 \implies\\ I - A^{-1}B^2A^{-1} > 0 $$ Thus, $A^{-1}B^2A^{-1} = (BA^{-1})^T(BA^{-1})$ is positive definite with eigenvalues strictly less than $1$. Since the spectral norm $\|BA^{-1}\|$ is at most $1$, conclude that the eigenvalues of $BA^{-1}$ are all at most $1$ (in absolute value).

Thus, the matrix $A^{-1/2}BA^{-1/2} = A^{-1/2}(BA^{-1})A^{1/2}$ is positive definite with eigenvalues at most $1$. Thus, we have $$ I - A^{-1/2}BA^{-1/2} > 0 \implies\\ A^{1/2}(I - A^{-1/2}BA^{-1/2})A^{1/2} > 0 \implies\\ A - B > 0 $$ which was the desired conclusion.

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  • $\begingroup$ 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 16:02
  • $\begingroup$ @hctb yes, that's equivalent to the question you asked here. If you want to know more, perhaps you should post a new question. $\endgroup$ – Omnomnomnom Jan 17 '17 at 16:07
  • $\begingroup$ math.stackexchange.com/q/2101727/190802 ok i posted it:p $\endgroup$ – Idele Jan 17 '17 at 16:16
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Denote $A^{1/2}$ to be the Hermitian square root of $A$ and $\rho$ to be the spectral radius.

Hint: $A^2-B^2>0$ $\Leftrightarrow$ $I-A^{-1}BBA^{-1}>0$ $\Leftrightarrow$ $\|BA^{-1}\|<1$ $\Rightarrow$ $\rho(BA^{-1})<1$ $\Leftrightarrow$ $\rho(A^{-1/2}BA^{-1/2})<1$ $\Leftrightarrow$ $I-A^{-1/2}BA^{-1/2}>0$ $\Leftrightarrow$ $A-B>0$.

Try to work out all the implications.

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