Let $T$ be a linear operator on a finite dim I.P.S and $T=S^*S$, where $S$ is an invertible I.P on $V$. Then $T$ is a positive operator on $V$.
My proof is: we need to show $T=T^*$ and $\langle T(u),u\rangle \ge 0 \,\forall u \in V, T^*=(S^*S)^*=S^*S^{**}$
$$\langle T(u),u\rangle =\langle S^*S(u),u\rangle= \langle S(u),S^{**}(u)\rangle=\langle S(u),S(u)\rangle \ge 0. $$
My doubt is that I do not use the invertibilty or the finite dimensionality in the proof.
For any bounded operator $S$ on any inner product space $T=S^{*}S$ satisfies $\langle Tx,x \rangle \geq 0$ for all $x$. You need not have any doubts about it.