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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@kotomord but we don't have AB=BA – Idele Nov 16 '16 at 10:45
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1One answer is given here – Ben Grossmann 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.
Ben Grossmann
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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. – Idele Jan 17 '17 at 16:02
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@hctb yes, that's equivalent to the question you asked here. If you want to know more, perhaps you should post a new question. – Ben Grossmann Jan 17 '17 at 16:07
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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.
A.Γ.
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