$N^2 = T^*T$, $N$ 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 $\ker T=\{0\}$. The equation
$$
\|N\alpha\|^2 = \langle N\alpha,N\alpha\rangle = \langle T\alpha,T\alpha\rangle=
\|T\alpha\|^2
$$
implies $\ker N = \ker T$. Hence $\ker N=\{0\}$, and since it is a linear mapping from the finite-dimensional $V$ to $V$, it is invertible.
