If $A$ is a symmetric matrix with exactly 3 distinct eigenvalues $\lambda, \mu$ and $\rho$, where $\rho$ is simple eigenvalue and $x$ is its eigenvector, then under which conditions the matrix identity from the title is valid?
It should hold if $A$ is non-negative and $\rho$ is the largest eigenvalue, but it seems that some modification can hold in other cases.
Best Answer
Consider the eigenvalue decomposition, then
\begin{align}(A-\lambda I)(A-\mu I)&=Q\begin{bmatrix} \rho-\lambda & & \\ & 0 & \\ & & \mu-\lambda \end{bmatrix} \begin{bmatrix} \rho-\mu & & \\ & \lambda - \mu & \\ & & 0\end{bmatrix} Q^T \\ &=Q\begin{bmatrix} (\rho-\lambda)(\rho-\mu) & & \\ &0 & \\ & & 0\end{bmatrix} Q^T \\ &= (\rho-\mu)(\rho-\mu)q_1q_1^T\end{align}
We require $(\rho - \mu)(\rho - \mu) > 0$, hence $\rho$ is either the largest or the smallest eigenvalue.