http://www.eecs.berkeley.edu/~wainwrig/ee227a/SDP_Duality.pdf
On the first page of this lecture note on semidefinite programming (13.3) the following is stated:
$ \lambda_{\min}=\min_{Y}\, \mathrm{tr}(YA)\quad \mathrm{such}\; \mathrm{that}:\;\; \mathrm{tr}Y=1\quad Y\succeq 0\;$ (i.e. is positive semidefinite)
The proof of the above representation of the minimum eigenvalue can be obtained by first showing that we can without loss of generality assume that $A$ is diagonal, and noticing that we can then restrict $Y$ to be diagonal as well.
First, I do not see how assumption that $A$ is diagonal does not result in loss of generality.
Second, I don't see how proof evolves further
Any clarifications?
(Note that in the derivation below I make multiple uses of the trace identity $\mathop{\textrm{Tr}}(AB)=\mathop{\textrm{Tr}}(BA)$, which is valid whenever both matrix multiplications $AB$ and $BA$ are well-posed.)
Since $A$ is symmetric, we know that it can be diagonalized by a unitary matrix $Q$: that is, $$A=Q\Lambda Q^T, \quad Q^TQ=QQ^T=I, \quad \Lambda\triangleq\mathop{\textrm{diag}}(\lambda_1,\lambda_2,\dots,\lambda_n)$$ For a fixed value of $Y$, then, $$\mathop{\textrm{Tr}}(YA)=\mathop{\textrm{Tr}}(YQ\Lambda Q^T)=\mathop{\textrm{Tr}}(Q^TYQ\Lambda)=\mathop{\textrm{Tr}}(\tilde{Y}\Lambda) = \sum_{i=1}^n \tilde{Y}_{ii} \lambda_i$$ where we have defined $\tilde{Y}\triangleq Q^TYQ$. Note that only the diagonal elements of $\tilde{Y}$ are significant, and that $$\mathop{\textrm{Tr}}(Y) = \mathop{\textrm{Tr}}(YQQ^T) = \mathop{\textrm{Tr}}(Q^TYQ) = \mathop{\textrm{Tr}}(\tilde{Y})$$ Thus any solution $Y$ of the original problem corresponds to a specific solution $\tilde{Y}$ to the diagonal version of the problem.
But we really didn't need to assume that $A$ is diagonal, because we're almost done anyway. Again, note that $\mathop{\textrm{Tr}}(YA)=\sum_{i=1}^n \tilde{Y}_{ii} \lambda_i$. Since $\sum_{i=1}^n \tilde{Y}_{ii}=1$, we observe that $$\lambda_n = \left(\sum_i \tilde{Y}_{ii}\right) \lambda_n = \sum_i \tilde{Y}_{ii}\lambda_n \leq \sum_i \tilde{Y}_{ii} \lambda_i.$$ So the minimum value is bounded below by $\lambda_n$. And it is evident that if we choose $$\tilde{Y}=\mathop{\textrm{diag}}(0,\dots,0,1),$$ then $\tilde{Y}$ satisfies our constraints and the objective is indeed $\lambda_n$.
So now let $Y=Q\tilde{Y}Q^T$, where $\tilde{Y}$ is our minimizing matrix determined above. We can confirm that this $Y$ is the minimizing value of our original problem: $$\mathop{\textrm{Tr}}(YA)=\mathop{\textrm{Tr}}(YQ\Lambda Q^T)=\mathop{\textrm{Tr}}(Q^TYQ\Lambda)=\mathop{\textrm{Tr}}(\tilde{Y}\Lambda) = \lambda_n.$$
(EDIT: If you really do want to see why we can assume w.l.o.g that $A$ and $Y$ are diagonal, there is one additional technical detail you need to accept: that if $\tilde{Y}$ is positive semidefinite, then so is the matrix obtained by zeroing out all its off-diagonal entries. Since they affect neither feasibility or the objective, we are free to require that the off-diagonal entries of $\tilde{Y}$ are zero.)