Are the eigenvectors of $DA$ with $D$ diagonal positive definite and $A$ symmetric positive definite orthogonal?

Question

Let $D$ and $A$ be matrices of size $n\times n$, with $D$ diagonal and positive definite, and $A$ symmetric positive definite. Then the eigenvalues of $DA$ are real and positive (see this post on Overflow).

Under which conditions on $D$ do the eigenvectors of $DA$ form an orthnormal basis for $\mathbb R^n$?

Answer 1

If a matrix has an orthonormal eigenbasis, it can be unitarily diagonalised. Hence it must be Hermitian. The converse is also true (and the proof can be found in almost every introductory linear algebra text). So, in your case, $DA$ has an orthonormal eigenbasis if and only if $DA$ is symmetric, i.e., iff $DA=AD$.