If $A \succ 0$, is $A-cI \succ 0$ for some small $c > 0$? I am studying (strong) convexity, have a question on positive semidefinite matrices.
If $A \succ 0$, i.e., if $A$ is positive definite, is $A - cI \succ 0$ for some small $c>0$? If so, how can I prove it?
 A: A matrix $A$ $\big($over $\mathbb C$ or over $\mathbb R\big)$ is positive-definite if it is Hermitian or symmetric, and its smallest eigenvalue satisfies $\lambda >0\,$.
Then you find some $\,\lambda> c >0\,$ so that $\,A -cI\,$ is still positive-definite.
Weakening to $A$ being positive semi-definite requires the weaker condition $\lambda \geqslant 0\,$.
Thus, if $\lambda = 0$, then each $c>0$ will shift the smallest eigenvalue of $\,A -cI\,$ into the strictly negative, hence its positive semi-definiteness is lost.
A: Since the matrix $\bf A$ is symmetric and positive definite, it has a spectral decomposition ${\bf A} = {\bf Q} {\bf \Lambda} {\bf Q}^\top$, where the entries on the main diagonal of $\bf \Lambda$ (i.e., the eigenvalues of $\bf A$) are positive. Thus,
$$ {\bf A} -  c {\bf I} = {\bf Q} {\bf \Lambda} {\bf Q}^\top - c {\bf Q} {\bf Q}^\top = {\bf Q} \left( {\bf \Lambda} - c {\bf I} \right) {\bf Q}^\top \succ {\bf O} $$
implies that $\lambda_{\min} ({\bf A}) - c > 0$. Hence, $\color{blue}{0 < c < \lambda_{\min} ({\bf A})}$.
