Let $x$ be a vector in $\mathbb{R}^d$. Let $A^*=\sum_1^{i=n} x_ix_i^T$ and $A=\sum_1^{i=n} a_ix_ix_i^T$ for $a_1>a_2>\cdots>a_n$ for $i<j$ where the second summand is called $A^*$.
Is it true that $$a_d\lambda_k(A^*) \le \lambda_k(A)\le a_1 \lambda_k(A^*),$$ where $\lambda_k(\cdot)$ is the $k$-th nonzero eigenvalue?
Under the additional assumption that $d\ge n$ and the vectors $x_i$ are non-zero and mutually orthogonal, then the claim is true. I.e. $x_i\ne0$ for all $i$, $x_i^Tx_j=0$ for all $i\ne j$.
Let $A^*=\sum_{i=1}^n x_ix_i^T$ and $A=\sum_{i=1}^n a_ix_ix_i^T$.
The vector $x_i$ is a eigenvector to both $A$ and $A^*$ with eigenvalue $\lambda_i(A^*) = \|x_i\|^{-2}$ and $\lambda_i(A) = a_i\|x_i\|^{-2}$. Hence $$ a_d\lambda_i(A^*)\le \lambda_i(A)\le a_1\lambda_i(A^*), $$ where $a_1$ and $a_d$ are the largest and smallest of the coefficients $a_i$.
The eigenvalues $n+1\dots d$ are all zero, $\lambda_i(A)=\lambda_i(A^*)=0$ for $i=n+1\dots d$.