Is $A – B$ positive definite

Question

1. If $A$ is positive definite and $B$ is negative semidefinite, then $A−B$ is positive definite.

2. If $A$ is indefinite and $B$ is positive definite, then $A+B$ is indefinite.

The definition I am using is if $(Ax,x) > 0$ for all $x$, then $A$ is positive definite. I think (1) is false because $B$ is negative semidefinite so there may be some $x$ for which the negative outweighs the positive from $A$.

For (2), I think the answer is again false. The positive definiteness of $B$ may overweight $A$ so it is possible that $(A+bx,x)$ is positive for all $x$.

Answer 1

1. $x^T(A-B)x = x^TAx - x^TBx$. If $x \ne 0$, then $x^TAx>0$ and $-x^TBx>0$, we have $x^TAx - x^TBx > 0$. $A-B$ is positive definite.

2. Counterexample: If $A = \begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$, we can have $B=I_2$ and $A+B$ is positive semidefinite.