How to find the largest possible minimum eigenvalue for a matrix? Let $\mathcal{X}$ be an arbitrary set and let $\phi: \mathcal{X} \to \mathbb{R}^d$ and $\lVert \phi(x)\rVert_2 \leq 1, \ \forall x \in \mathcal{X}$. Suppose $x$ is drawn from $\mathcal{X}$ according to $P$ and define a matrix $A$ as $$A = \underset{{x \sim P}}{\mathbb{E}}\left[ \, \,\phi(x) \phi(x)^\top \right]$$ I watched a talk in which it was said that the largest possible minimum eigenvalue is $\frac1d$.  Can anybody help me to figure out why the following holds?
$$\max_{P \in \Delta(\mathcal{X})} \sigma_{\min}(A) = \frac{1}{d}$$

Hint: What's the trace of $\phi(x)\phi(x)^T$. With the assumption given, you can bound it by $1$. The trace of a diagonalizable matrix is the sum of its eigenvalues. You can show that the sum of the eigenvalues is less than or equal to 1. You have d eigenvalues whose sum is less than or equal to 1. Can the smallest be strictly bigger than $1/d$?
