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!



Given $\mathbf{x}_i^\top A\mathbf{x}_i$ for a SPD matrix $A$ and orthonormal bases $\mathbf{x}_i$, what is the bound of its eigenvalues?
You are just comparing the diagonal $\vec a$ and the spectrum $\vec\lambda$ of a symmetric matrix (positive semi-definiteness is not an issue here). The key result is a theorem due to R. Horn & C. Johnson : $\vec\lambda$ and $\vec a$ are the spectrum and the diagonal of a symmetric matrix if and only if $\vec\lambda\succ\vec a$ ($\vec\lambda$ *majorizes* $\vec a$). This means that if you list the items in the non-decreasing way,
$$a\_1\le\cdots\le a\_n,\qquad\lambda\_1\le\cdots\le\lambda\_n,$$
then
$$\lambda\_1+\cdots+\lambda\_k\le a\_1+\cdots+a\_k,$$
for every $k=1,\ldots,n$, with equality for $k=n$.


