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Can someone explain why the underlined inequality is true? Thank you for your time.

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2 Answers 2

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I just realized that this is just a consequence of the Courant-Fischer theorem for symmetric matrices which states that $\substack{\text{max}\\\text{dim}(L)=k}\substack{\text{min}\\x\in L\\x\neq0}\dfrac{x^\text{T}Ax}{x^\text{T}x}=\lambda_k(A)$ with the eigenvalues ordered as $\lambda_1(A)\geq\lambda_2(A)\geq\cdots\geq\lambda_n(A).$

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In the first underlined statement, the minimum is taken over a $k-1$ dimensional space. That is the space selected by the max operator minus $p$. In the second underlined statement, the minimum is again taken over a $k-1$ dimensional space. However, the max operator can now select that space, and it may include a direction not perpendicular to $p$. Therefore, the max operator may be able to achieve a higher value than before. The value is always at least as good, since it can always select the same space as before minus $p$.

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    Why is the minimum taken over a $k-1$ dimensional space in the first underlined statement? It could be $k$ or $k-1$ depending on the relationship between $L$ and $p$ right? For example, $L = \text{span}\left\{\begin{bmatrix}1\\0\\0\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix}\right\}$ and $p = \begin{bmatrix}0\\0\\1\end{bmatrix}.$ – 2017-01-12
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    Good point. The first max operator will include $p$ in $L$, since that reduces the feasible set of the min problem. – 2017-01-12