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I'm looking at the proof in this question, and I can't convince myself that this is correct (assuming $A$ and $B$ are symmetric and positive definite): $$\forall x\; x^T Ax \ge x^T Bx \iff \lambda_{min}(A) \ge \lambda_{max}(B).$$

Intuitively, it kinda makes sense because every $x$ can be expressed as a linearly combination of the eigenvectors of $A$ or $B$. However, although I can convince myself that the RHS implies the LHS, I can't convince myself of the other direction. It seems possible that the RHS is sufficient but not necessary for the LHS.

So does the statement actually hold?

Zizheng Tai
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1 Answers1

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RHS $\implies$ LHS but not the other way around. A straightforward counterexample is given by $$ A = \begin{pmatrix} 1 & 0 \\ 0 & 1/2 \end{pmatrix} \qquad \text{and} \qquad B = \begin{pmatrix} 3/4 & 0 \\ 0 & 1/4 \end{pmatrix}. $$

The difference $A-B$ is positive definite but $\lambda_{\min}(A) = 1/2 \not\geq 3/4 = \lambda_{\max}(B)$.

Rammus
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