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Recently i'm reading a paper,there is a inequality that confuse me. L is a symmetric,irreducible and semi-positive definite matrix with eigenvalues of $0=\lambda_{1}(L)<\lambda_{2}(L)\leq...\leq\lambda_{m}(L)$(counting the multiplicities),P is a positive definite matrix with maximum eigenvalues $\lambda_{m}(P)$,both of them are square matrix,and it satisfied

$$x^{T}(L^{2}\otimes P^{2})x\leq \lambda_{m}(L)\lambda_{m}(P)x^{T}(L\otimes P)x$$

I don't know how to prove this inequality,I have searched many references ,but hardly find any result.Maybe there needs other properties,welcome to point out!

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  • $\begingroup$ What precisely is your question? $\endgroup$
    – Stefan Kohl
    Oct 8, 2015 at 9:03
  • $\begingroup$ sorry ,my question is why this inequality is right,how to prove it? $\endgroup$
    – linzzwz
    Oct 8, 2015 at 9:48
  • $\begingroup$ it would certainly help if you would tell us in which paper you found this inequality. $\endgroup$ Oct 8, 2015 at 9:57
  • $\begingroup$ math.stackexchange.com/questions/1469948/… $\endgroup$ Oct 8, 2015 at 10:28

1 Answer 1

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We use $L^t \otimes P^t = (L\otimes P)^t$ twice below. The proof follows by observing that \begin{eqnarray*} \sup_{x\neq 0}\frac{x^T(L\otimes P)^2 x}{x^T(L\otimes P)x} = \sup_{z=(L\otimes P)^{1/2}x} \frac{z^T(L\otimes P)z}{z^Tz} \le \sup_{z \neq 0}\frac{z^T(L\otimes P)z}{z^Tz} = \|L\otimes P\|. \end{eqnarray*}

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  • $\begingroup$ Also observe that $\|L\otimes P\| = \lambda_m(L)\lambda_m(P)$ as these are psd. $\endgroup$
    – Suvrit
    Oct 8, 2015 at 12:13
  • $\begingroup$ is it obvious that $L^2\otimes P^2=(L\otimes P)^2$ ? $\endgroup$ Oct 8, 2015 at 13:16
  • $\begingroup$ Well, $(AC \otimes BD) = (A \otimes B)(C\otimes D)$, and for fractional powers, we can use this lemma with the eigenvector decomposition of the involved matrices... $\endgroup$
    – Suvrit
    Oct 8, 2015 at 13:26
  • $\begingroup$ i have understood! thank you very much! $\endgroup$
    – linzzwz
    Oct 8, 2015 at 14:51

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