If $A, B, A^2-B^2$ are positive definite prove that $A-B$ is positive definite [closed]

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

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, E. Joseph, qwr, Jack's wasted life, Michael Albanese
If this question can be reworded to fit the rules in the help center, please edit the question.

• @kotomord but we don't have AB=BA – Idele Nov 16 '16 at 10:45
• One answer is given here – Omnomnomnom Nov 16 '16 at 12:58

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.

• 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
• @hctb yes, that's equivalent to the question you asked here. If you want to know more, perhaps you should post a new question. – Omnomnomnom Jan 17 '17 at 16:07
• math.stackexchange.com/q/2101727/190802 ok i posted it:p – Idele Jan 17 '17 at 16:16

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.