Prove the following equality:
$$\text{Tr}\{A\}= \langle \Gamma \vert_{RS} I_R \otimes A_S \vert \Gamma \rangle_{RS}.$$
Where $A$ is a square operator acting on a Hilbert space $\mathbf{H}_S$, $I_R$ is the identity operator acting on Hilbert space $\mathbf{H}_R$. Here,
$$\vert \Gamma \rangle_{RS} \equiv \sum_{i=0}^{d-1} \vert i \rangle_R \vert i \rangle_S.$$
Also, trace is given as
$$\text{Tr}\{A\} \equiv\sum_i \langle i \vert A \vert i \rangle.$$
I am a pure math major and am very weak with physics and am stuck computing the RHS of the equality. Thanks in advance for any hints or tips!! I guess what I am trying to say is I am new with all of this notation and how to decipher it all.
We have $$\langle \Gamma|_{RS} = \sum_{i=0}^{d-1}\langle i|_S \langle i|_R$$
and $$(I_R\otimes A_S)|\Gamma\rangle_{RS} = \sum_{i=0}^{d-1}|i\rangle_R\otimes A|i\rangle_S$$
Putting everything together and using the properties of inner products:
$$\langle \Gamma|_{RS}(I_R\otimes A_S)|\Gamma\rangle_{RS} = \sum_{i,j=0}^{d-1} \langle i|j \rangle_R\langle i|A|j\rangle_S$$
Assuming $\{|i\rangle_S\}_{i=0}^{d-1}$ is an orthonormal basis for $H_S$ and that $\{|i\rangle_R\}_{i=0}^{d-1}$ is an orthonormal family then $\langle i | j\rangle = \delta_{ij}$ and the result follows.