problem of a positive definite matrix

Question

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 $k_1\gg k_2$?

Answer 1

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.

Answer 2

$$z^T (k_1H-k_2I) z = k_1z^THz-k_2\|z\|^2=\|z\|^2(k_1v^THv-k_2),$$

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 $k_1>>k_2$.

Answer 3

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 $k_1c-k_2>0$.