Does the sum of squares of Hermitian matrices/operators have positive eigenvalues? Let us have the operator $Q=\sum_iA_i^2$ where each $A_i$ is a Hermitian operator. I have found a statement saying that $Q$ must have positive eigenvalues. How can one show this?  
I think it has to do with the fact that Hermitian operators have real eigenvalues and so when we square them (something which we can get from the action of $A_i$ to one of its eigenfuctions/eigenvectors) we get something positive. But, even with this information, I cannot figure a way to prove that $Q$ has positive eigenvalues.
 A: Let's call an operator $T$ positive if $T=T^*$ and $\sigma(T)\subset\mathbb R_{\geq0}$.  Then it's not too difficult to show the following:

Let $T$ be a self-adjoint operator.  Then the following are equivalent:  
  
  
*
  
*$T$ is positive,
  
*$\|\lambda-T\|\leq \lambda$ for all $\lambda\geq \|T\|$,
  
*$\|\lambda-T\|\leq\lambda$ for some $\lambda\geq \|T\|$.
  

Using this, one can show that if $T_1$ and $T_2$ are positive, then $T_1+T_2$ is positive.  Finally, as you've noted, if $T$ is self-adjoint, then $T^2$ is positive.
Thus if $T_1,\ldots, T_n$ are self-adjoint, then $T_1^2,\ldots,T_n^2$ are positive, hence $\sum_kT_k^2$ is positive, and the result follows.
A: You already know that the squares of Hermitian operators are positive; the question concerns their sums. Well, the individual summands $A_i^2$ have the property that $\langle v,A_i^2v\rangle\geq0$ for all vectors $v$. Sum over $i$ and use the fact that the inner product is linear in each factor to get $\langle v,(\sum_iA_i^2)v\rangle\geq0$ for all $v$. In paarticular, if $v$ is an eigenvector of $\sum_iA_i^2$ with eigenvalue $\lambda$, then $\langle v,\lambda v\rangle\geq0$ and therefore $\lambda\geq0$.
