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I believe this variant of the Rayleigh quotient inequality must be well known but I could not find references for it online. It's proof is straightforward.

Let $\mathbf{A}\in\mathbb{R}^{n\times n}$ having non negative eigenvalues. Assume the spectral decomposition of $\mathbf{A}$ is given by

\begin{equation} \mathbf{A}=\mathbf{U}\cdot\left(\begin{array}{ccc} \lambda_{0} & & 0\\ & \ddots\\ 0 & & \lambda_{n-1} \end{array}\right)\cdot\mathbf{V}^{T},\,\mbox{s. t.}\,\,\mathbf{I}_{n}=\mathbf{U}\cdot\mathbf{V}^{T}, \end{equation}

where $0\le\lambda_{0}\le\cdots\le\lambda_{n-1}$, then for all linear combinations of columns of $\mathbf{V}$ and $\mathbf{U}$ respectively of the form \begin{equation} \mathbf{x}=\sum_{0\le k<n}\alpha_{k}\,\mathbf{V}\left[:,\,k\right]\;\mbox{ and }\;\mathbf{y}=\sum_{0\le k<n}\beta_{k}\,\mathbf{U}\left[:,\,k\right], \end{equation} such that $\left\{ \alpha_{k}\cdot\beta_{k}\right\} _{0\le k<n}\subset\mathbb{R}_{\ge0}$ (but not all simultaneously zero), we have \begin{equation} \lambda_{0}\le\frac{\mathbf{x}^{T}\cdot\mathbf{A}\cdot\mathbf{y}}{\mathbf{x}^{T}\cdot\mathbf{y}}\le\lambda_{n-1}. \end{equation}

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    $\begingroup$ Do you have a proof for that claim? I see this for the first time (but I am not an expert on this). $\endgroup$
    – Sebastian Goette
    Commented Dec 11, 2015 at 15:01
  • $\begingroup$ Note that such a spectral decomposition may not exists e.g. when $A$ is the zero matrix. $\endgroup$
    – Surb
    Commented Dec 11, 2015 at 17:16
  • $\begingroup$ Are $U,V$ orthonormal? If $U,V$ are orthonormal and $UV^T = I_n$ then isn't $U = V$? In that case $A$ is symmetric and $x,y$ originate from the same subspace right? In that case isn't this the basic Rayleigh inequality? $\endgroup$
    – Pushpendre
    Commented Dec 11, 2015 at 22:47

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