# If two different symmetric positive definite matrices have equal determinants then $A-B$ is neither positive semidefinite nor negative definite

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)

## 1 Answer

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