Value of the quadratic form as a function of the determinant Suppose I have a symmetric, positive definite matrix $A$ with all the diagonal elements $\sigma^2 > 0$, and its off-diagonal elements are $\rho \in [-1,1]\backslash\{0\}$.   
Consider a quadratic form $y := x^T A^{-1}x$. 
Suppose, $\sigma^2 \to \infty$. It is easy to see that $det(A) \to \infty$. I saw a claim that this also implies that $y \to 0$. 
Is this true? If yes, why? 
Thanks.
 A: It is true. What we want to show is that all the eigenvalues of $A^{-1}$ go to zero in the limit. For this we need to show that all the eigenvalues of $A$ go to infinity. We would like to say that the off-diagonal is "small" and therfore, we can neglect them and see that we already have the eigenvalues on our diagonal. This is essentially true! Your condition on the off-diagonal implies that
$$ \Vert A - diag( \sigma_1^2, \dots, \sigma_n^2) \Vert_{F} \leq n^2$$
where the norm in question is the Frobenius norm. However, the Hoffman-Wielandt Theorem (see Hoffman-Wielandt Theorem Proof) tells us that the if $\lambda_1(A) \leq \dots \leq \lambda_n(A)$ are the eigenvalues of $A$ and $\mu_1 \leq \dots \leq \mu_n$ are the eigenvalues of $diag(\sigma_1^2, \dots, \sigma_n^2)$ then
$$ \sum_{j=1}^n \vert \lambda_j(A) - \mu_j \vert^2 \leq \Vert A - diag(\sigma_1^2, \dots, \sigma_n^2) \Vert_F^2 $$
Thus, all the eigenvalues of $A$ go to infinity and therefore all the eigenvalues of $A^{-1}$ go to zero. But this implies (as $A$ is diagonalizable) that $\Vert A^{-1} \Vert \rightarrow 0$ and hence (by Cauchy-Schwarz)
$$ y = x^T A^{-1} x \leq \Vert A^{-1} \Vert \cdot \Vert x \Vert^2 \rightarrow 0.$$
It is not true if we drop the assumption on the off-diagonal (first I missed that assumption and wrote this answer and now I cannot bring myself to delete what I wrote). If the diagonal of $A$ goes to infinity, you only get that the product over all eigenvalues of $A$ goes to infinity, but you might have, that only one eigenvalue actually goes to infinity. Then some of the eigenvalues of $A^{-1}$ are bounded away from zero and $y$ might not go to zero (if the base change does that diagonalizes the matrix does not depend on the parameter, then we can just take some eigenvector of an eigenvalue that does not go to infinity).
I started with a diagonal matrix where one eigenvalue is fixed and one that is free for me to play. Then I searched for a unitary matrix such that after conjugation it satisfied the condition you wanted, ie that all the elements of the diagonal are positive and go to infinity.
Here is the example. Take
$$ A_\lambda =  \frac{1}{2}\begin{pmatrix} 1 + \lambda & -\lambda + 1 \\ -\lambda + 1  &1+\lambda
\end{pmatrix} $$
This matrix is symmetric and positive definite, in fact it is similar to $diag(1,\lambda)$ as
$$ U^\star \begin{pmatrix} 1 & 0 \\ 0 & \lambda \end{pmatrix}  U =  A_\lambda$$
where $U$ is the following orthogonal matrix
$$ U=2^{-1/2}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}.$$
We have (using that $U^\star = U^{-1}$)
$$ A_{\lambda}^{-1} 
= \left( U^\star \begin{pmatrix} 1 & 0 \\ 0 & \lambda \end{pmatrix}  U \right)^{-1} 
= \left( U^{-1} \begin{pmatrix} 1 & 0 \\ 0 & \lambda \end{pmatrix}  U \right)^{-1} 
= U^{-1} \begin{pmatrix} 1 & 0 \\ 0 & \lambda \end{pmatrix}^{-1}  U
= U^\star \begin{pmatrix} 1 & 0 \\ 0 & \lambda \end{pmatrix}^{-1}  U 
=  A_{1/\lambda} $$
Furthermore, we have
$$ y=\begin{pmatrix} 1 & 1 \end{pmatrix} A_{1/\lambda} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = 2 $$
This does not go to zero and we are done.
