Compute eigenvalues If we know that $A$ is a symmetric $2$ by $2$ matrix, $\displaystyle \max_{\|x\|=1}\:x^{T}Ax=a$, and $\displaystyle\min_{\|x\|=1}\:x^{T}Ax=b$ for some given $a, b$, can we compute the eigenvalues of $A$?
 A: Since $A$ is a symmetric matrix, the eigenvalues are real and the eigenvectors can be chosen to be orthonormal. Let $x_1$, $x_2$ be the two orthonormal eigenvectors and $\lambda_1$, $\lambda_2$ be the two eigenvalues respectively. Any vector in $\mathbb{R}^2$ can be hence written as $x = c_1 x_1 + c_2 x_2$. Hence, we get that 
$$
\begin{align}
x^TAx & = (c_1 x_1 + c_2 x_2)^T A (c_1 x_1 + c_2 x_2)\\
& = c_1^2 x_1^T A x_1 + c_1c_2 x_1^T A x_2 + c_1c_2 x_2^T A x_1 + c_2^2 x_2^T A x_2\\
& = c_1^2 \lambda_1 + c_2^2 \lambda_2
\end{align}
$$
Hence, $\displaystyle \max_{||x||=1} x^TAx$ is equivalent to $\displaystyle \max_{c_1^2 + c_2^2 = 1} c_1^2 \lambda_1 + c_2^2 \lambda_2$. Hence, $$\displaystyle \max_{||x||=1} x^TAx = \lambda_{max}.$$
By the same argument, $\displaystyle \min_{||x||=1} x^TAx$ is equivalent to $\displaystyle \min_{c_1^2 + c_2^2 = 1} c_1^2 \lambda_1 + c_2^2 \lambda_2$. Hence, $$\displaystyle \min_{||x||=1} x^TAx = \lambda_{min}.$$
The same argument works in any $\mathbb{R}^n$ as well.
