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Suppose that $A$ is a positive definite symmetric matrix (P.D.S.). Now consider the matrix $|A|$, the matrix arrived at by taking the absolute value of all the entries of $A$. Is $|A|$ also P.D.S.? I have been trying to construct a counter example, but I can't seem to get one. Can someone proved a proof or counterexample?

Wintermute
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    I've proven that there are no $2 \times 2$ counterexamples, but my proof doesn't adapt to any higher dimensions, since it uses the fact that a $2 \times 2$ matrix with positive trace and positive determinant has positive eigenvalues. By contrast this doesn't hold for $3 \times 3$ (consider eigenvalues of $4,-2,-1$). Interesting problem, +1. – Ian Jul 09 '15 at 14:10
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    Turns out the statement is true in dimensions 2 and 3 but fails for dimension 4 or greater. Gershgorin's circle theorem provides the insite. As a side note this problem arose from a suggestion to modify a correlation matrix in an economic model. – Wintermute Jul 09 '15 at 17:12
  • Could you present the proof for the $3 \times 3$ case as an additional answer? I think the $2 \times 2$ case is simple enough; the $4 \times 4$ case is in the accepted answer; and I don't think it's hard to adapt the $4 \times 4$ case to build higher examples (certainly not if you're willing to let $A$ be semidefinite). But I don't see a super-simple proof of the $3 \times 3$ case. – Ian Jul 09 '15 at 18:14
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    To prove $3\times 3$ case one could use Sylvester criterion: $$ A=\left[\matrix{a & d & f\d & b & e\f & e & c}\right] $$ is pos.def. iff $a>0$, $ab-d^2>0$ and $abc+2def-ae^2-bf^2-cd^2>0$. Since the diagonal elements are necessarily positive, the only difference between $A$ and $|A|$ is in off-diagonal ones. It means that the only term that depends on signs is $def$ in the third condition that becomes easier to satisfy for $|A|$ since $def>0$. – A.Γ. Jul 10 '15 at 13:14

2 Answers2

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The statement is not true. Here is the simulation to find a counterexample. Basically, I generate a random matrix $A$ by using normal distribution (in matlab randn(4,4)) and get a psd matrix $B$ by its gram matrix (A'A). I check the eigenvalue of absolute value of $B$.

If you do it many times, you will eventually find a counterexample. Here is one example I found and you could verify. 

A =
   -1.8429    0.9184   -0.4437    1.2861
   -1.5589    0.7211    0.6427   -0.1817
    1.4980   -0.9764   -0.6835    0.9816
    1.1269    1.7022    1.0411   -1.0071

B =
    9.3402   -2.3612   -0.0349   -1.7512
   -2.3612    5.2145    2.4954   -1.6227
   -0.0349    2.4954    2.1610   -2.4070
   -1.7512   -1.6227   -2.4070    3.6650

The eigenvalues of $B$,

d = 
    0.0865
    1.7064
    7.9794
   10.6082

The eigenvalues of $|B|$,

d_abs =
   -0.1351
    2.7554
    6.2533
   11.5070

I'm interested if there are more elegant ways to disprove the statement.

eigenchris
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Sisi
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  • How many attempts did you need to find this? I have just tried maybe 40 times and I get positive definite results each time. – Ian Jul 09 '15 at 14:18
  • It only took me 11 times (it depends, experimentally on average less than 100 times). Make sure you use normal distribution but not rand if you use matlab. – Sisi Jul 09 '15 at 14:20
  • Never mind, I forgot to abs. Silly mistake; now I found it in 56 iterations, and then in 18 after that. – Ian Jul 09 '15 at 14:26
  • It seems that the important feature is that there is a pair of relatively large negative off-diagonal elements in $B$. I have a three significant digit example: $B=\begin{bmatrix} 1.75 & 0.59 & -0.92 & 0.67 \ 0.59 & 10.7 & -0.06 & -3.97 \ -0.92 & -0.06 & 0.52 & -0.42 \ 0.67 & -3.97 & -0.42 & 2.28 \end{bmatrix}$. Here the effect appears to be driven by the $-3.97$, in the sense that if you flip only those signs the result is already indefinite. This makes a degree of sense from the perspective of continuity of eigenvalues. – Ian Jul 09 '15 at 14:42
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    BTW using B=randi([-1,1],4,4);A=B'*B gives "nicer" examples. – Algebraic Pavel Jul 09 '15 at 15:01
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One counterexample could be $$ A=\left[\matrix{10 & 3 & -2 & 1\\3 & 10 & 0 & 9\\-2 & 0 & 10 & 4\\1 & 9 & 4 & 10}\right]. $$ $A$ is positive-definite, however $|A|$ is not positive-semidefinite.

Another one: $$ A=\left[\matrix{1.5 & 1 & 0 & -1\\1 & 1.5 & 1 & 0\\0 & 1 & 1.5 & 1\\-1 & 0 & 1 & 1.5}\right]. $$

A.Γ.
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