$(T^* T)^{-1}$ is self adjoint Suppose $U$, $V$ are ﬁnite-dimensional inner product spaces, and $T \in \mathcal{L}(U, V )$ is invertible. I need to show that $T^* T \in \mathcal{L}(V)$ is positive deﬁnite and invertible. Also show that the inverse of $T^* T$ is self-adjoint and positive deﬁnite.
For the first part I think I'm on the right track but not sure, since $T$ is invertible, I said $T^*$ is also invertible and as a result $T^* T$ is invertible. Further as $\|T^*(u)\|^2$ is greater than or equal to zero for all values of $u$, and we know that the null space of $T^* T$ is $\{0\}$, we can say that $\|T^*(u)\|^2$ is greater than zero for all non-zero values of $u$, and thus we have $\langle T^*(u),T^*(u)\rangle$ is greater than zero which can further be simplified to say $\langle TT^*(u),u \rangle$ is greater than zero i.e $T^*T$ is positive semi definite.
That's what I did for the first part, pretty lost on the second part.

Let $S = (T^{\ast}T)^{-1}$, then for any $v\in V$, let $u = S(v)$, so
$$
0\leq \langle T^{\ast}Tu,u\rangle = \langle v,S(v)\rangle
$$
so $S$ is positive semidefinite. Since it is invertible, it must be positive definite. Also, it follows that all the eigen-values of $S$ are real (in fact, non-negative reals). Hence, $S$ is self-adjoint.
