problem of a positive definite matrix Let $H$ be a positive definite matrix and $I$ the identity matrix. If $k_1,k_2>0$, can we conclude that $k_1H-k_2I$ is positive definite if $k_1\gg k_2$?
 A: The answer is positive.
Note a matrix $A$ is positive definite if and only if its eigenvalues are all strictly greater than $0$. If we use $\lambda(A)$ to denote the eigenvalue of a matrix $A$, it can be easily verified that 
$$\lambda(p(A)) = p(\lambda(A)),$$
where $p$ is any polynomial.
Therefore, $\lambda(k_1 H - k_2 I) = k_1 \lambda(H) - k_2$. In order to have them strictly greater than $0$, it is sufficient to have
$$k_1 > \frac{k_2}{\min\{\lambda(H)\}}.$$
The denominator is always positive since $H$ is assumed to be positive definite. 
A: Yes. Since $H$ is positive definite, we get a norm $\|\cdot\|$ on $\Bbb R^n$ defined by
$$\|x\|^2=x^THx.$$
Let $|\cdot|$ be the standard norm on $\Bbb R^n$. Since on a finite dimensional vector space, every norm are equivalent, there exists $c>0$ such that
$$c|x|^2\leq\|x\|^2,\quad\forall x\in\Bbb R^n.$$
Then, for all $x\neq 0$,
$$x^T(k_1H-k_2I)x=k_1(x^THx)-k_2|x|^2\geq k_1c|x|^2-k_2|x|^2=(k_1c-k_2)|x|^2.$$
Now, it is easy to see that we just have to take $k_1$ large enough so that $k_1c-k_2>0$.
A: $$z^T (k_1H-k_2I) z = k_1z^THz-k_2\|z\|^2=\|z\|^2(k_1v^THv-k_2),$$
where $v=z/\|z\|$. The form $v^THv$ attains it's minimum when $v$ corresponds to the eigenvector of $H$ with smallest eigenvalue. Since all the eigenvalues are positive, your assertion is correct for $k_1>>k_2$. 
