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Let $M$ be a $n \times n$ symmetric positive definite matrix. Let $u$ and $v$ be vectors of length $n$ with entries consisting $n-m_u$ (or $n-m_v$) $0$'s and $m_u $ (or $m_v$) $1$'s, where $m_u,m_v \in \{1,...,n-1\}$. Sort $u$ so that the first $n-m_u$ entires of $u$ are $0$'s and the last $m_u$ entries are $1$'s. Sort $v$ in the same way. Suppose $m_u>m_v$. Is the following weak inequality true?

(As shown below by @Niki Di Giano, this is not true) $$ u^T M u \geq v^TMv $$

kx526
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    I don't think you need all that, I think it is true in general for all the matrices with positive entries – Lucio Tanzini Aug 12 '19 at 23:03
  • @LucioTanzini That is absolutely true. Sorry, for the actual question, $X$ does not need to have only positive entries. I corrected the question. – kx526 Aug 12 '19 at 23:08

3 Answers3

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Consider : $$ M= \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{bmatrix}, u = \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} ^T, v = \begin{bmatrix} 0 & 1 & 1 \end{bmatrix} ^T $$

M is positive definite and : $$ u^T M u = v^T M v = 2 $$ so it's not a strict inequality (is $m_u \lt n$ ?). I couldn't find an example where the inequality is not working though.

yprobnoob
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  • Thank you for the comment! Indeed, $m_n, m_v <n$. Sorry for the confusion, I edited my question. – kx526 Aug 12 '19 at 23:38
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    I just checked and if you set $m_u =2$ and $m_v = 1$ you also get 2 so it’s still not working for $m_u \lt n$. – yprobnoob Aug 12 '19 at 23:40
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    I see. I will edit the question to a weak inequality which I am still interested in showing. Thank you. – kx526 Aug 12 '19 at 23:44
  • Sorry to bother you again, but would you please help me on this question? – kx526 Aug 13 '19 at 17:23
  • I saw Niki Di Giano’s answer and he proved that the inequality is false. What is left to prove? – yprobnoob Aug 13 '19 at 17:28
  • What I hope to show is $$\frac{1+m_v}{1+m_u} \leq \frac{1+u^\top(M+I_n)^{-1} u}{1+v^\top(M+I_n)^{-1} v} \leq \frac{1+m_u}{1+m_v}$$ where $m_v<m_u$, and $M$ is symmetric PD with positive entries. – kx526 Aug 13 '19 at 17:33
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Another example where the inequality fails:

The matrix $\begin{bmatrix}3&-2\\-2&3\end{bmatrix}$ is positive definite. Also, $\begin{bmatrix}1\\1\end{bmatrix}>\begin{bmatrix}1\\0\end{bmatrix}$. But $$\begin{bmatrix}1&1\end{bmatrix}\begin{bmatrix}3&-2\\-2&3\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=2<3=\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}3&-2\\-2&3\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}$$ which violates the inequality.

subrosar
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The matrix $M$: $$M = \begin{bmatrix} 1 & 0 & 0\\ 0 & 10 & - 9\\ 0 & -9 & 10\\ \end{bmatrix} $$ Yields $u^T M u = 2$ and $v^T M v = 10$ for the vectors $u = (0, 1, 1)$ and $v =(0, 0, 1)$ respectively.

Niki Di Giano
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  • It is required that the number of $1$'s are strictly below $n$. Is it possible to show a counterexample for that case? Thank you. – kx526 Aug 13 '19 at 00:10
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    @kenex I have edited the answer. As you can see it doesn't really matter since it easily generalises. – Niki Di Giano Aug 13 '19 at 00:12