Determinant of an eigenmatrix of a symmetric positive definite matrix

Question

Let $A$ be a symmetric positive definite matrix with eigenvalue decomposition $A = VDV^\top$. In this case, the eigenmatrix $V$ is orthogonal and thus the determinant of $V$ is either $1$ or $-1$. Is there a case where the determinant of $V$ is always $1$, a pure rotational matrix? Thanks

Answer 1

You can change the sign of $\det V$ by negating one of the eigenvectors, i.e. columns of $V$: $$ \mathbf{v}_i \mapsto -\mathbf{v}_i $$ or by swapping the two columns $$ \mathbf{v}_i \leftrightarrow \mathbf{v}_j $$ Let $U$ denote the resulting matrix after doing one of these operations on $V$.

Claim: $U$ is special orthogonal, i.e. $U^\top U = I$ and $\det U = 1$.

As a result the matrix $D' = U^{-1} A U$ is also diagonal. Hence, $$ A = U D' U^{-1} = U D' U^\top $$ is still a factorization in terms of eigenbasis and eigenvalues.

In the first case, the $D' = D$, and in the latter case, $D'$ has the two diagonal entries at index $i$ and $j$ swapped.