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Recently I've asked question about the set given as $\{x: x^T(A-B)x = 1\}$. In that question I wanted to prove that such set isn't bounded, and according to the answer I can do it if I can prove the fact from the title of this question i.e.:

If $A$ and $B$ are different, symmetric, positive definite $n \times n$ $(n > 1)$ matrices such that $\det A = \det B$ then the matrix $A - B$ is neither positive semidefinite nor negative definite

Could you please help me to start the proof? I have no idea how to do it (namely how to use the equality of determinants)

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Audience request: we can begin with orthogonal $Q$ such that $Q^TB Q = D$ is diagonal with positive diagonal elements. We can then continue with a diagonal matrix $W$ with diagonal elements certain reciprocals of square roots, so that $W Q^T BQW = W^T Q^T B Q W = I,$ and name P = QW.$

there is a matrix $P$ with $\det P \neq 0$ so that $P^T B P = I.$ Then $$ P^T ( A - B) P = C - I, $$ where $$ C = P^T A P $$ is symmetric, positive definite and, since $\det B \cdot \det^2 P = 1,$ so $\det A \cdot \det^2 P = 1,$ we get $$ \det C = 1.$$

Let's see, if all eigenvalues of $C$ are equal to $1$ then it is the identity matrix; that needs a little proof using symmetry and orthogonal matrices. In this case, $A = B$ and $x^T (A-B)x = 1$ is impossible.

Otherwise, $C$ has an eigenvalue larger than $1$ and another eigenvalue smaller than $1,$ which tells us that $C-I$ is indefinite.

You should look up Sylvester's Law of Inertia, which has to do with the "congruence" I am using.

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  • $\begingroup$ Why $\exists \ P: P^TBP = I$? I thought there should be a diagonal matrix, but not necessary an identity... $\endgroup$ – D F Feb 12 '18 at 19:36
  • $\begingroup$ Wow! Extremely beautiful solution. Thanks a lot! $\endgroup$ – D F Feb 12 '18 at 19:45

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