# suppose $A=[a_{ij}]$ is symmetric and positive definite. How can I show that $B=[|a_{ij}|]$ is symmetric and positive definite?

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 $$n=3$$?

• $B$ being symmetric should be obvious, since applying a map to every element of a symmetric matrix maintains symmetry. Jul 6 '21 at 16:22
• 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. Jul 6 '21 at 17:17
• 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 $b^2$. Jul 6 '21 at 17:21
• 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...
– J.G
Jul 6 '21 at 17:26

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}