Proving the existence of a positive definite difference between symmetric matrices Given a real-valued symmetric $m$-by-$m$ matrix $A$ such that $\operatorname{tr}(A)<\gamma$ where $\gamma>0$, prove that there exists another symmetric matrix $Z$ such that $\operatorname{tr}(Z)<\gamma$ and $A<Z$.
I think I need to choose a $Z$ and then prove both that $\operatorname{tr}(Z)<\gamma$ and that every eigenvalue of $Z-A$ is positive.  I've tried using $Z=cI$ and placing $c$ less than $\gamma/m$ and greater than $\lambda_{max}(A)$, but I can never prove both conditions.  I can get $\operatorname{tr}(Z)=mc<\gamma$ and $\operatorname{tr}(Z-A)>0$, but I don't think that's good enough.
What am I missing?

Hint: Take
$$
Z = A + \frac{\gamma - \operatorname{tr}(A)}{2m}I
$$
