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For $n\le 3$ suppose $A=[a_{ij}]$ is symmetric and postive definite. How can I show that $B=[|a_{ij}|]$ is symmetric and positive definite?

I know these statements:

  1. A symmetric $n \times n$ matrix is positive definite if and only if all its eigenvalues are positive
  2. A symmetric matrix $A$ is positive definite if and only if all its leading principal minors are positive; that is $\det A (1 : i, 1 : i) > 0, 1 ≤ i ≤ n$. This called Sylvester’s criterion.
  3. If $A = (a_{ij})$ is positive definite, then $a_{ii}> 0$ for all $i$
  4. If $A = (a_{ij})$ is positive definite, then the largest element in magnitude of all matrix entries must lie on the diagonal.
  5. The sum of two positive definite matrices is positive definite

I guess for $n=1$ I can use statement number $2$. But what about $n=2$ and $n=3$?

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    $\begingroup$ $B$ being symmetric should be obvious, since applying a map to every element of a symmetric matrix maintains symmetry. $\endgroup$ Jul 6 '21 at 16:22
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    $\begingroup$ Notably, the statement fails for $n \geq 4$. For instance: $$ A = \pmatrix{10 & 3 & -2 & 1\\ 3 & 10 & 0 & 9\\ -2 & 0 & 10 & 4\\ 1 & 9 & 4 & 10}, \qquad B = \pmatrix{10 & 3 & 2 & 1\\ 3 & 10 & 0 & 9\\ 2 & 0 & 10 & 4\\ 1 & 9 & 4 & 10} $$ You may verify that $A$ is strictly positive definite, while $B$ fails to be positive semidefinite (since it has a negative determinant: $\det B= -364$). cf. Horn and Johnson's "Matrix Analysis", 2nd edition, problem 7.5.P6. $\endgroup$ Jul 6 '21 at 17:17
  • $\begingroup$ The statement is easy to show for $n \geq 2$: the matrix $$ \pmatrix{a&b\\b&c} $$ is positive definite iff $a > 0$ and $ac > b^2$. Taking the absolute values of the entries might change $b$, but it cannot change $b^2$. $\endgroup$ Jul 6 '21 at 17:21
  • $\begingroup$ It's not super illuminating, but it seems $n=3$ can be shown in a similar way using Sylvester's criterion--you can search for this same question on this site and find the proof. That said, it would be nice if there were a more elegant argument... $\endgroup$
    – J.G
    Jul 6 '21 at 17:26
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This follows easily from Sylvester's criterion. I will consider only $n=3$, as this is the only interesting case. As $A$ and $B$ share identical diagonal elements as well as principal $2\times2$ minors, it suffices to prove that $\det(B)\ge\det(A)$ (so that $\det(B)>0$). Let $$ A=\pmatrix{p&s_1a&s_2b\\ s_1a&q&s_3c\\ s_2b&s_3c&r} \ \text{ and }\ B=\pmatrix{p&a&b\\ a&q&c\\ b&c&r} $$ where $p,q,r>0,\,a,b,c\ge0$ and $s_1,s_2,s_3\in\{1,-1\}$. Then \begin{aligned} \det(A) &=pqr+2s_1s_2s_3abc-(pc^2+qb^2+ra^2)\\ &\le pqr+2abc-(pc^2+qb^2+ra^2)\\ &=\det(B). \end{aligned}

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