Let $A$,$B \in \mathbb{R}^n$, $C\in\mathbb{R}^{n\times n}$, and $C=D^TD$ where $D$ is a $n\times n$ psd matrix, is it guaranteed that $A^TCA \leq B^TCB \Rightarrow A^TA \leq B^TB$?
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$$ C = \left( \begin{array}{cc} 1 & 0 \\ 0 & 9 \end{array} \right), $$ $$ A = \left( \begin{array}{c} 2 \\ 0 \end{array} \right), $$ $$ B = \left( \begin{array}{c} 0 \\ 1 \end{array} \right), $$
Will Jagy
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Since $C = D^{T}D$,and we have all eigenvalues of $C$ are great than $0$ ,which means $> 0$.
Assume the eigenvector $x$ and $y$of $C$ corresponding to $\lambda$ and $\mu$,respectively,and we have \begin{equation} C x=\lambda x \end{equation} and \begin{equation} C y =\mu y \end{equation} ,which mean
\begin{equation} x^{T} C x \leq y^{T} C y \rightarrow \lambda x^{T}x \leq \mu y^{T}y \end{equation}.
because that we don't know the relations between $ \lambda $ and $\mu$,even more we have known both,and then we can't identify which is bigger between $x^{T}x$ and $y^{T}y$.
askuyue
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