$ tr(TS)=0$ and $T,S$ are positive, compact implies $TS=0$

Question

Suppose $T$ and $S$ are positive compact operators and Tr$(TS)=0$ , then show $TS=ST=0$.

Attempt: Apply the spectral decomposition for $T$ to obtain $\lambda_n(T)$ and corresponding orthonormal sequence ${e_n}$. $0=\sum_n \langle TS e_n,e_n\rangle = \sum_n \langle S e_n,\lambda_n(T) e_n\rangle $. Since $\lambda_n(T)$ is positive and $S$ is positive, if $\lambda(T)$ has infinite many non-zeros then $S=0$. So assume finitely nonzero up to N, then $\langle S e_n, e_n\rangle =0$ for $n=1,..N$.

I'm stuck here, its still not enough to show $ST=0$

Answer 1

Notice that $(Se_n, e_n) = || \sqrt{S}e_n ||^2 =0$. Hence, $\sqrt{S}e_n =0$ and thus $Se_n = 0$. Therefore, $S = 0$ on the image of $T$ and thus $ST = 0$

Answer 2

Note that given the spectral decomposition $T(x) = \sum_{n=1}^\infty \lambda_n(T)\langle x,e_n\rangle e_n$, there is a unique positive operator $$T^{1/2}(x)=\sum_{n=1}^\infty \sqrt{\lambda_n(T)}\langle x,e_n\rangle e_n$$ such that $T^{1/2}T^{1/2}=T$. As was already proven in the OP, we have for each $n\ge 1$, $$ \langle Se_n,\lambda_n(T)e_n\rangle = \langle S^{1/2}\sqrt{\lambda_n(T)}e_n, S^{1/2}\sqrt{\lambda_n(T)}e_n\rangle = \|S^{1/2}\sqrt{\lambda_n(T)}e_n\|^2 =0. $$ Since $T^{1/2}e_n = \sqrt{\lambda_n(T)}e_n$, we obtain for each $n\ge 1$, $$ S^{1/2}T^{1/2}e_n=0, $$ i.e. $S^{1/2}T^{1/2}=O$. This implies $$ ST = S^{1/2}(S^{1/2}T^{1/2})T^{1/2} = O $$ and $$TS = (ST)^*=O$$ as wanted.