Is $A^2-B^2$ positive definite too when $A-B,B$ is positive definite?

Question

Denote $A,B\in M_n(\mathbb{R})$
If $A-B,B$ is positive definite, it's easy to see $A^2-B^2$ is symmetric.
Now the question is:

Prove or disprove: $A^2-B^2$ is positive definite.

I have checked some easy examples(mostly 2x2), and now I believe this is true.
But we know here $A,B$ do not necessarily commute.
I tried to write $A$ as $A=P^\mathrm{T}P$, and then I know all the eigenvalues of $(P^\mathrm{T})^{-1}BP^{-1}$ all lay in the interval $(0,1)$.
Another question show me that when $A,B$ is positive definite, $AB+BA$ can also be no positive definite. I can't move forward. Hint will also be appreciated.

I don't really see why $A^2-B^2$ is symmetric, can you elaborate? — Stan Tendijck, Apr 23 '19 at 11:14
Here by "positive define", I mean $A^\mathrm{T}=A\land\forall x\in\mathbb{R}^n\quad x^\mathrm{T}Ax\geq 0$, so $(A^2-B^2)^\mathrm{T}=(A^2-B^2)$ — Oolong Milktea, Apr 23 '19 at 11:26

score 4 · Accepted Answer · answered Apr 23 '19 at 11:31

4

For the first question try: $$ A = \pmatrix{2 & 1\cr 1 & 7\cr},\ B = \pmatrix{1 & -1\cr -1 & 2\cr} $$

For the second,

$$ A = \pmatrix{2 & 1\cr 1 & 1\cr},\ B = \pmatrix{ 5 & -3\cr -3 & 2} $$

answered Apr 23 '19 at 11:31

Robert Israel

448,999

Thanks, a good counterexample. – Oolong Milktea Apr 23 '19 at 12:06

Is $A^2-B^2$ positive definite too when $A-B,B$ is positive definite?

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