Eigenvalues of $\sum_{i=1}^n \frac{(x_i - x_{i-1})^2}{\lambda_i}$

Question

Consider the cuadratic form $$ \mathbf{x}^{\intercal}Q\mathbf{x} = \frac{x_1^2}{\lambda_1} + \sum_{i=2}^n \frac{(x_i - x_{i-1})^2}{\lambda_i}\ . $$ Is it true that the eigenvalues of $Q$ are $\lambda_i^{-1}$ ?

Answer 1

The answer to your question is no. $Q$ can be written in the form $$\begin{bmatrix}1 & -1 \\ & 1 & -1 \\ & & 1 & \ddots \\ & & & \ddots & -1 \\ & & & & 1 \end{bmatrix} \begin{bmatrix}\lambda_1^{-1} \\ & \lambda_2^{-1} \\ & & \lambda_3^{-1} \\ & & & \ddots \\ & & & & \lambda_n^{-1} \end{bmatrix} \begin{bmatrix}1 \\ -1 & 1 \\ & -1 & 1 \\ & & \ddots & \ddots \\ & & & -1 & 1 \end{bmatrix} $$ Call that right-hand matrix $Z$. The eigenvalues would be $\lambda_i^{-1}$ if $Z^TZ=ZZ^T=I$. But $Z^TZ\neq I$. The eigenvalues are not $\lambda_i^{-1}$, and it's not hard to see this by trying a couple of numerical examples.