Change in eigenvalues when rank 1 matrix is added to Positive definite matrix Let $v \in \mathbb{R}^n $ be a column vector and $ A, B$ are positive definite matrices with eigen values $ \lambda_A^1 \geq ... \geq \lambda_A^n$ and $ \lambda_B^1 \geq... \geq \lambda_B^n$ respectively. 
We also know that a) $ \lambda_A^1 \leq \lambda_B^1 $ and $ \lambda_A^n \geq \lambda_B^n $, b) $Trace(A) = Trace(B)$ , c)  $ \det(A) \leq \det(B) $
What can we say about eigenvalues (or change in eigenvalues) of following matrices (apart from Cauchy interlacing theorem):
1) $ \hat{A} = A + vv^T$
2) $ \hat{B} = B + vv^T$
Is there any relationship between $ \frac{\det(\hat{A})}{\det(A)} $ and $  \frac{\det(\hat{B})}{\det(B)}  $?
 A: Yep, we can say things. The two ingredients you need are the following:


*

*the matrix determinant lemma: $\det(M+xy^T) = (1+y^TM^{-1}x)\det(M)$;

*the Sherman-Morrison identity: $(M+xy^T)^{-1}= M^{-1} - \frac{M^{-1}xy^TM^{-1}}{1+y^TM^{-1}x}$.


Equipped with these things, you can find out more quantitative relationships between the old and new matrices.
Edit: The question has been updated to ask about relationships between the ratios of the determinants, which according to the matrix determinant lemma are the quantities 
$$1+v^TA^{-1}v\quad \text{and}\quad 1+v^TB^{-1}v$$
The short general answer is no. Even though we have some relationships between the spectra of $A$ and $B$, we can't in general say much about $v^TA^{-1}v$ compared to $v^TB^{-1}v$, since it may happen that $v$ is well-aligned with an eigenvector of $A^{-1}$ of large eigenvalue, and an eigenvector of $B^{-1}$ of small eigenvalue. 
An extreme example meeting all conditions in the question would be to take a positive definite matrix $A$ which has large variation in its spectrum, and make $B$ be a rotated copy of $A$, i.e. obtained by conjugating with an orthogonal transformation. Then it may well happen that for some $v$ the ratio of determinants is huge, while for others it's tiny.
