minimum eigenvalue for difference of two matrices Let $A$ a symmetric positive definite matrix, and $B$ a matrix constructed from $A$ by setting all its off-diagonal elements to zero. Then is there a way to see for which values of positive scalars $a$ and $b$ $$C=aA-bB$$ is positive definite?
Do we know something on the smallest eigenvalue of $C$? 

@uranix commented that I could be after some $\epsilon$ such that $\det(A-\epsilon B)=0$ and have $C_{\epsilon}$ be PSD. I looked at $A=[1,1;1,7]$ and found $\epsilon=1.37796$ but then the eigenvalues of $C_{1.37796}$ are $-3.02372$ and $0$ ... 
 A: Let's consider simplified problem about eigenvalues of
$$
C = A - \epsilon B, \quad A = A^\top > 0
$$
where $B = \operatorname{diag}(A)$.
Let's inspect the equation
$$
\operatorname{det}(A - \epsilon B) = 0.
$$
Since $B$ is a diagonal matrix it is easy to construct $B^{-1/2}$ - matrix with diagonal elements raised in $-1/2$ power.
$$
\operatorname{det}(B^{1/2})
\operatorname{det}(B^{-1/2}AB^{-1/2} - \epsilon)
\operatorname{det}(B^{1/2})
 = 0\\
\operatorname{det}(B^{-1/2}AB^{-1/2} - \epsilon) = 0.
$$
The matrix $G = B^{-1/2}AB^{-1/2}$ is positive definite since
$$
x^\top B^{-1/2}AB^{-1/2}x = (AB^{-1/2}x, B^{-1/2}x) > 0, \quad \forall x \neq 0
$$
Thus all eigenvalues of $G$ are positive, and those are precisely the roots $\epsilon_i$ of 
$$
\operatorname{det}(G - \epsilon) = 0.
$$
Assume that $0 < \epsilon_1 \leq \epsilon_2 \leq \dots \leq \epsilon_n$.
Assuming that $\lambda_{\min}(C)$ is a continuous function of $\epsilon$ (that sounds sane) and obviously $\lambda_{\min}(A) > 0$
the function $$\lambda_{\min}(A - \epsilon B)$$ thus is positive when $0 \leq \epsilon < \epsilon_1$. At $\epsilon = \epsilon_1$ the determinant of the $A - \epsilon B$ vanishes, so $\lambda_{\min}$ can not be positive.
A: Notice that $A-\epsilon B$ is positive definite iff $B^{-\frac{1}{2}}(A-\epsilon B)B^{-\frac{1}{2}}$ is positive definite.
Now $B^{-\frac{1}{2}}(A-\epsilon B)B^{-\frac{1}{2}}= B^{-\frac{1}{2}}AB^{-\frac{1}{2}}-\epsilon Id$. 
So $B^{-\frac{1}{2}}(A-\epsilon B)B^{-\frac{1}{2}}$ is positive definite iff $\epsilon<$ smallest eigenvalue of $B^{-\frac{1}{2}}AB^{-\frac{1}{2}}$.
