Let $\alpha$ be a positive definite automorphism of an inner product space $V$. Is $\alpha^{-1}$ necessarily positive definite?
I know the answer is true for invertible positive definite matrices, but I tried to prove this problem for the general case. However, still, I feel that I'm missing something.
Since $\alpha$ is positive definite, it's self-adjoint and for all nonzero $v\in V$, we have $\langle\alpha(v),v\rangle> 0.$
Let $v\in V$, so $$\langle \alpha^{-1}(v),v\rangle=\langle \alpha^{-1}(v),\alpha\alpha^{-1}(v)\rangle=\langle \alpha\alpha^{-1}(v),\alpha^{-1}(v) \rangle > 0$$
Now for each $v\in V$, we got $\langle \alpha^{-1}(v),v\rangle>0$, but still I need to show $\alpha^{-1}$ is self-adjoint. For finitely generated inner space, I can show that it's true.
Let $v, \:w\in V$. Then,
$$\langle v,w\rangle=\langle \alpha^{-1}\alpha(v),w\rangle=\langle\alpha(v), (\alpha^{-1})^*(v)\rangle=\langle v,\alpha^*(\alpha^{-1})^*(v)\rangle, $$ so $(\alpha^{-1})^*=(\alpha^*)^{-1}. $ Moreover, $$\langle\alpha^{-1}(v),w\rangle=\langle v,(\alpha^{-1})^*(w)\rangle=\langle v,(\alpha^*)^{-1}(w)\rangle=\langle v,\alpha^{-1}(w)\rangle.$$ Hence $\alpha^{-1}$ is self-adjoint.
But how about when $V$ is not finitely generated?
Your argument is correct (though you should mention that you are assuming $v\neq 0$, which implies $\alpha^{-1}(v)\neq 0$ in order to get your inequality). You still need to show that $\alpha^{-1}$ is self-adjoint, though. You can do so by an argument very similar to the one you used.