If $L$ is a matrix that represents real physical quantity, why is $L^2$ non-negative real physical quantity? In my textbook, it says that when $L$ is a matrix that represents real($\mathbb{R}$) physical quantity, $L^2$ represents non-negative real physical quantity. What would be the proof of this?
 A: There exists a set of eigenstates, $\{ |\psi_\lambda\rangle  \}$, such that $L|\psi_\lambda\rangle = \lambda|\psi_\lambda\rangle$
where $\lambda$ is a real eigenvalue and $|\psi_\lambda\rangle$ is an eigenstate of $L$.  The $\lambda$ represents the value of a physical observable associated with the eigenstate $|\psi_\lambda\rangle$.
There is currently no stipulation as to the sign of $\lambda$.  It can be positive or negative, but we are given that it is real.  Now we reapply the $L$ operator to our eigenvector equation to arrive at: 
$LL|\psi_\lambda\rangle = \lambda L|\psi_\lambda\rangle = \lambda^2|\psi_\lambda\rangle$ 
Therefore, 
$L^2|\psi_\lambda\rangle = \lambda^2|\psi_\lambda\rangle$ 
$\lambda^2$ must be positive, since $\lambda$ is real.  Therefore the eigenvalues, which represent the values of the physical observables associated with the eigenstates of $L^2$, must be positive and real.
A: If $\psi$ is a normalized eigenvector of $L^2$ and $\lambda$ the corresponding eigenvalue then 
$\lambda=\psi^*\lambda\psi=\psi^*L^2\psi=(L\psi)^*(L\psi)$.
Thus $\lambda$ is manifestly real and nonnegative.
This even holds if $L$ is a vector of noncommuting real (i.e., Hermitian) quantities, such as angular momentum. Then we get in the last step 
$\sum_i (L_i\psi)^*(L_i\psi)$, which is again manifestly real and nonnegative.
A: If $L$ is an operator of a real physical quantity, it means it has its own eigenvectors $|\psi_l\rangle$ and eigenvalues $l$: $$L|\psi_l\rangle=l|\psi_l\rangle$$. If $l$ is real, then $l^2$ (corresponding to the eigenvalue of $L^2$) is positive.
