$T*T$ Notation and proof Let $T:H\to H$ be compact where $H$ is a Hilbert space and let $T^*$ be the adjoint operator of $T$. Prove that $T^*T$ is compact and self adjoint and that the eigenvalues of $T^*T$ are nonnegative. Find a compact self adjoint operator $S:H\to H$ such that $S^2=T^*T$
What will be a good approach for this question?
 A: While there is a lot of context missing from the question, perhaps it is $T^* \circ T$, where $T^*$ is the adjoint of $T$?  That is easily seen to be self-adjoint and with positive eigenvalues: Since $(AB)^*=B^* A^*$ and $A^{**}=A$, we have that $(T^*T)=T^*T^{**}=T^*T$.  Further, since if $T*Tx=\lambda x$ and $x\neq 0$, then $$\lambda = \langle T^*Tx,x \rangle/ \langle x,x \rangle =\langle Tx,Tx \rangle/\langle x,x \rangle,$$ which is the quotient of two non-negative quantities, with non-zero denominator.
Finding the square root also isn't so bad.  Since the operator is self-adjoint, the spectral theorem gives an orthonormal basis of eigenvectors, and we already have that the eigenvalues are non-negative, and thus have real square roots.  Define $Sx$ on an eigenvector $T^*Tx=\lambda x$ by $Sx=\sqrt{\lambda}x$.  Verify that this is compact and self-adjoint.
It's been too long since I've taken functional analysis to recall the relevant facts about compactness, but hopefully this is enough that you can finish things off (by showing $T^*T$ and $S$ are compact).
