Sum of positive definite and symmetric matrix Let $A$ be a real, symmetric, positive definite $n\times n$ matrix. Let $B$ be a real, symmetric $n\times n$ matrix. 
Does there exist an $\epsilon>0$ such that $A+\epsilon B$ is positive definite?
My attempt: According to Wikipedia we can simultaneously diagonalize $A$ and $B$. Since $A$ is positive definite, its eigenvalues are all positive. Call them $\lambda_i$. Let $\rho_i$ be the eigenvalues of $B$. Choose a basis which simultaneously diagonalizes $A$ and $B$. Then $$u^T(A+\epsilon B)u=u^TAu+\epsilon u^TBu=\sum_i(\lambda_i+\epsilon \rho_{i})u_i^2.$$ So if $$\epsilon<\frac{\lambda_{\min}}{|\rho_{\max}|}$$ (assume $B\neq 0$), then $$\lambda_i+\epsilon\rho_i>\lambda_i-\lambda_{\min}\geq0$$ for all $i$, which implies $A+\epsilon B$ is positive definite.
Is this valid? Is there a basis-independent way to show this?
 A: You can make this line of reasoning work, but you should be careful about what "diagonalized" means in this context.  In particular, your wiki page is talking about diagonalization via a congruence rather than a similarity.  
In particular, the theorem is that there exists an invertible matrix $P$ (not necessarily orthogonal) such that both $PAP^T$ and $PBP^T$ are diagonal.  Note that the diagonal entries will not generally be the eigenvalues of $A$ and $B$.

Note that we can't necessarily diagonalize $A$ and $B$ simultaneously in the sense of similarity.  In fact, we can do so if and only if $AB = BA$.  
It is generally true that if $A$ is symmetric and positive definite, then such an $\epsilon$ exists.  One proof is as follows:
If $A$ is positive definite, then for some $r > 0$ (e.g. $r = \lambda_{min}$) we can write
$$
A = rI + (A - rI)
$$
where $I$ denotes the identity matrix and $A - rI$ is positive semidefinite.  With that, we have
$$
A + B\epsilon = [rI + \epsilon B] + (A - rI)
$$
It suffices to choose an $\epsilon$ such that $rI + \epsilon B$ is positive definite.

Another approach that I like: let $A^{-1/2}$ denote the positive definite square root of $A^{-1}$.  Then $A + \epsilon B$ is positive definite if and only if the matrix
$$
A^{-1/2}(A + \epsilon B)A^{-1/2} = I + \epsilon A^{-1/2}B(A^{-1/2})^T
$$
is positive definite.  It therefore suffices to consider the case with $A = I$.
If you choose to diagonalize $I + \epsilon A^{-1/2}B(A^{-1/2})^T$ by diagonalizing the symmetric matrix $A^{-1/2}B(A^{-1/2})^T$, then you are essentially deriving the "simultaneous diagonalizability" of quadratic forms in our specific case.
A: The sought for conclusion binds since the eigenvalues of a matrix depend continuously on its entries; then as $A + \epsilon B$ is continuous in $\epsilon$, and agrees with $A$ for $\epsilon = 0$, all eigenvalues of the symmetric matrix $A + \epsilon B$ are positive for $\vert \epsilon \vert$ sufficiently small, so $A + \epsilon B$ is positive definite for such $\epsilon$.
