Eigenvalue of self-adjoint

Solution 1:

The proposed solution is not well written, but it appears that $\lambda$ is an arbitrary eigenvalue of $T^*T$ and $x$ is a normalized eigenvector associated with $\lambda$. In that case, $$\lambda = \lambda \langle x,x\rangle = \langle \lambda x,x\rangle = \langle T^*T(x),x\rangle$$ The point of the argument was to demonstrate that all eigenvalues of $T^*T$ are nonnegative.

I would, however, suggest a more direct approach. Note $T^*T$ is self-adjoint, and given $x\in V$, $\langle T^*Tx,x\rangle = \langle Tx,Tx\rangle \ge 0$. Hence, $T^*T$ is positive semidefinite. By a similar argument $TT^*$ is positive semidefinite.

Solution 2:

We can assume that the eigenvector $x$ is normalised, i.e. $\langle x,x \rangle =1$. If that is not the case in the beginning, say $\langle x,x \rangle = c>0$ then $\langle \frac{1}{\sqrt c} x, \frac{1}{\sqrt c} x \rangle =1$ and the vector $\frac{1}{\sqrt c}x$ is still an eigenvector. Therefore $\langle T^*Tx,x \rangle = \langle \lambda x,x \rangle =\lambda \langle x,x \rangle = \lambda$.