Eigenvalue of self-adjoint I am solving an exercise:
Let $T: V \rightarrow W$ be a linear transformation.
$V$ and $W$ are finite-dimensional inner product spaces. Prove $T^*T$ and $TT^*$ are semidefinite.
This is a solution that I don't understand: 
$T^*T$ and $TT^*$ are self-adjoint, then we have $T^*T(x) = \lambda x$. Hence: 
$$\lambda = ⟨T^*T(x),x⟩ = ⟨T(x),T(x)⟩ ≥ 0.$$
$\lambda$ is $≥ 0$, hence $T^*T$ is semidefinite.
I don't understand why the eigenvalue is equal to $⟨T^*T(x),x⟩$.
Thank you for any kind of help!
 A: 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 $\langle T^*Tx,x \rangle = \langle \lambda x,x \rangle =\lambda \langle x,x \rangle = \lambda$.
A: 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.
