# Symmetric positive definite matrix inequality

Hi could you help me with the following:

Show that for a symmetric positive definite matrix $B$, $$b_{ij} + b_{jk} + b_{ki} \leqslant b_{ii} + b_{jj} + b_{kk}$$ holds for any $1 \leqslant i,j,k \leqslant n$ with $b_{ij}$ being the entry at $(i,j)$ of matrix $B$.

Thanks a lot.

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This reduces to the fact that $b_{ii} + b_{jj} \geq 2b_{ij},$ applied three times: specifically, $b_{ii} + b_{jj} \geq 2b_{ij}$,$b_{ii} + b_{kk} \geq 2b_{ik}$ and $b_{jj} + b_{kk} \geq 2b_{jk}$. Adding the three inequalities, and dividing by $2$ gives the desired inequality. I leave you to verify that $b_{ii} + b_{jj} \geq 2b_{ij}.$

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+1. Nice hint. ${}$ – user17762 Jan 4 '13 at 5:09

This is just user Geoff's Answer expanded. I saw it after typing up this whole thing.

Consider the $2 \times 2$ principal minors \begin{align} \begin{bmatrix}b_{ii} & b_{ij} \\b_{ji} & b_{jj} \end{bmatrix}~ and ~\begin{bmatrix}b_{ii} & b_{ik} \\b_{ki} & b_{kk} \end{bmatrix}~ and~ \begin{bmatrix}b_{jj} & b_{jk} \\b_{kj} & b_{kk} \end{bmatrix} \end{align} Determinant of all of them should be positive (related to sylvester's criterion). This gives you $b_{ii}b_{jj}\geq b_{ij}^2$ (note $b_{ij}=b_{ji}$) and other two conditions. From $b_{ii}b_{jj}\geq b_{ij}^2$, you get $b_{ii}+b_{jj}\geq 2*b_{ij}$. Similarily for other minors also, add them all together and you have your inequality.

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Does Sylvester criterion include those as well? it may be the case that j is not i+1 ?? – Salih Ucan Jan 6 '13 at 0:22
Sorry, I just checked, for sylvester's criterion, only leading principal minors need to be positive. But all principal minors being positive is also a equivalent condition. See this link docs.google.com/… – dineshdileep Jan 6 '13 at 9:51
One reasoning for this is , you can find permutation matrices (which do not change positive definiteness) such that the above mentioned principal minors will be rearranged to become the leading principal minors. See also "user1551" answer. It also gives something in this light. – dineshdileep Jan 6 '13 at 9:53

Here is a slightly more general perspective. For any positive diagonal matrix $D$ and real orthogonal matrix $U$, we have $\operatorname{tr}D\ge\operatorname{tr}DU$. Put $U=Q^TPQ$ with a real orthogonal matrix $Q$ and a permutation matrix $P$ and make use of the identity $\operatorname{tr}AB=\operatorname{tr}BA$, we get $\operatorname{tr}QDQ^T\ge\operatorname{tr}QDQ^TP$ and in turn $\operatorname{tr}\tilde{B}\ge\operatorname{tr}\tilde{B}P$ for every positive definite matrix $\tilde{B}$. Now, let $\tilde{B}$ be a 3-by-3 principal submatrix of $B$ and the result follows.

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