$N^2 = T^*T$, $N$ is non-negative, and $T$ is invertible, how to prove $N$ is also invertible? I'm reading Hoffman's "Linear Algebra", and met this line that I couldn't figure it out.
In $\S9.5$ Spectral Theory, pp 342, he mentioned:

Let $V$ be a finite-dimensional inner product space and let $T$ be any
  linear operator on $V$.  Then $N$ is uniquely determined as the
  non-negative square root of $T^*T$.  If $T$ is invertible, then so is
  $N$ because $$\langle N\alpha, N\alpha\rangle = \langle N^2\alpha, \alpha \rangle = \langle T^*T\alpha, \alpha \rangle = \langle T\alpha, T\alpha\rangle.$$

I'm lost at the last line: I have no problem with the equation, but how would that prove that "If $T$ is invertible, then so is $N$"?

If $T$ is invertible, then
$$
\ker(T^{\ast}) = \text{range}(T^{\ast})^{\perp} = \{0\}
$$
so $T^{\ast}$ is injective, and
$$
\text{range}(T^{\ast})^{\perp} = \ker(T) = \{0\} \Rightarrow \text{range}(T) = V
$$
so $T^{\ast}$ is surjective and hence invertible. So $N^2 = T^{\ast}T$ is invertible. So if $S = (T^{\ast}T)^{-1}$, then
$$
SN^2 = I
$$
and so $N$ is injective. Similarly, $N$ is surjective and you are done.
