Assume that $A_{d\times d}$ is a symmetric positive semi-definite matrix, and $\{\mathbf{x}_1,\ldots,\mathbf{x}_d\}$ composes a group of orthogonal bases of $\mathbb{R}^d$ where $\mathbf{x}_i\bot\mathbf{x}_j,\forall i\neq j$ and $\|\mathbf{x}_i\|=1$.

Then, my question is that given $$ \mathbf{x}_i^\top A\mathbf{x}_i=a_i,\forall i $$ how to estimate the bound of the eigenvalues of $A$?

Using eigenvalue decomposition $A=U\Lambda U^\top$ and let $\tilde{\mathbf{x}}_i=U^\top\mathbf{x}_i$, this question can be simplified as to estimate the bound of the elements of the diagonal matrix $\Lambda$ given $$ \tilde{\mathbf{x}}_i^\top\Lambda\tilde{\mathbf{x}}_i=a_i,\forall i. $$ Since $U$ is orthogonal, we also have $\tilde{\mathbf{x}}_i\bot\tilde{\mathbf{x}}_j,\forall i\neq j$ and $\|\tilde{\mathbf{x}}_i\|=1$.

Denote $\Lambda=\mathrm{diag}[\lambda_1\;\ldots\;\lambda_d]^\top$ and $\tilde{\mathbf{x}}\_i=[\tilde{x}\_{i1}\;\ldots\;\tilde{x}\_{id}]$, the equations above can be represented as $$ \sum_j\lambda_j\tilde{x}_{ij}^2=a_i,\forall i $$

However, this question seems to be not so simple yet.

In the case of $d=2$, I found the following method to solve it.

Denote $\tilde{\mathbf{x}}_1=[\cos\theta\;\sin\theta]^\top$ and $\tilde{\mathbf{x}}_2=[-\sin\theta\;\cos\theta]^\top$, we have \begin{eqnarray} \lambda_1\cos^2\theta+\lambda_2\sin^2\theta&=&a_1\\\ \lambda_1\sin^2\theta+\lambda_2\cos^2\theta&=&a_2 \end{eqnarray} Solve the equation system and we get \begin{eqnarray} \lambda_1&=&\frac{1}{\Delta}(a_1\cos^2\theta-a_2\sin^2\theta) =\frac{a_1-a_2\tan^2\theta}{1-\tan^2\theta}=a_2+\frac{a_1-a_2}{1-\tan^2\theta}\\\ \lambda_2&=&\frac{1}{\Delta}(-a_1\sin^2\theta+a_2\cos^2\theta) =\frac{-a_1\tan^2\theta+a_2}{1-\tan^2\theta}=a_1-\frac{a_1-a_2}{1-\tan^2\theta} \end{eqnarray} where $\Delta=\cos^4\theta-\sin^4\theta=\cos^2\theta-\sin^2\theta$.

Without loss of generality, assume that $a_1\geq a_2$ and $\lambda_1>\lambda_2$, we can deduce that $1-\tan^2\theta>0$, then from $\lambda_i\geq 0$, we have $$\tan^2\theta\leq\min\{\frac{a_2}{a_1},\frac{a_1}{a_2}\}=\frac{a_2}{a_1}$$ Therefore it is obvious that when $\tan^2\theta=\frac{a_2}{a_1}$, $\lambda_1$ and $\lambda_2$ take its maximum and minimum respectively as \begin{eqnarray} \lambda_1&=&a_1+a_2\\\ \lambda_2&=&0 \end{eqnarray} and we get the bound of the eigenvalues of $A$.

However, how to solve the problem if $d\geq3$?

Any suggestion will be appreciated! Thank you very much!