Why is $Q^\frac{1}{2}$ a symmetric nonnegative definite?

Question

Im new in Studying stochastic and currently reading to gain something. This question is utilized in The following link

If $Q$ is an operator on a Hilbert space $U$, $(e_n)$ is an ONB of $U$ consisting of eigenvectors of $Q$, then $(Q^{1/2}e_n)$ is an ONB of $Q^{1/2}U$

Suppose $Q$ is a bounded, linear, symmetric nonnegative definite trace class operator on a separable Hilbert space $U$.

Due to lack of knowledge I fail to show why is $Q^\frac{1}{2}$ a symmetric nonnegative definite? Any help is highly appreciated..

Answer 1

Every symmetric non-negative definite operator $Q$ has a unique symmetric non-negative definite square root and the notation for this square root is $Q^{1/2}$.