When does a Eigendecomposition result in a Q with det(Q)=1? With Eigendecomposition I can decompose a symmetric real matrix $A$ into  $Q\Lambda Q^T$, where $Q$ is orthogonal. If $det(Q)=1$, $Q$ is a rotation matrix and if $det(Q)=-1$, $Q$ is a rotation matrix with reflection. 
Is there a way to know beforehand if $det(Q)=1$ or $-1$?

You can always arrange that $\det Q = 1$. If $q_1, \ldots, q_n$ is your eigenbasis (the columns of $Q$), note that 
$\det(q_1, \ldots, q_n) = -\det(q_2,q_1,\ldots, q_n)$ and $(q_2,q_1, \ldots, q_n)$ is also an orthonormal eigenbasis.
