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If $A$ is copositive, what about $A^3$? Is it also copositive? More generally, my question is whether the odd power of a copositive matrix is still copositive.

Any reference is appreciated

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2 Answers 2

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A counter-example is given by $$\pmatrix{.6,0,0,.6,1}\pmatrix{ 1 & -1 & 1 & 1 & -1 \\\ -1 & 1 & -1 & 1 & 1 \\\ 1 & -1 & 1 & -1 & 1\\\ 1 & 1 & -1 & 1 & -1\\\ -1 & 1 & 1 & -1 & 1} ^3 \pmatrix{.6 \\\ 0 \\\ 0\\\ .6 \\ 1}=-0.44.$$

while the matrix in the centre is actually copositive. It is easy to check that there are no counter-examples for $2\times 2$ matrices.

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  • $\begingroup$ ah! you typed in the answer while I was typesetting it. Same matrix ;-) $\endgroup$
    – Suvrit
    Jun 24, 2011 at 2:25
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My preliminary experiments show that the answer is no. Here is why.

In the paper Constructing copositive matrices from interior matrices, the following matrix (from Horn's quadratic form) is mentioned to be copositive:

$$ A=\begin{bmatrix} 1 & -1 & 1 & 1 & -1\\\\ -1 & 1 & -1 & 1 & 1\\\\ 1 & -1 & 1 & -1 & 1\\\\ 1 & 1 & -1 & 1 & -1\\\\ -1 & 1 & 1 & -1 & 1\\\\ \end{bmatrix} $$

Now plug this matrix into Matlab, and search to see if $x^TA^3x < 0$ occurs for a nonnegative $x$. Here is one particular value:

$$x = \begin{pmatrix} 4\\\\ 6\\\\ 3\\\\ 0\\\\ 0\\\\ \end{pmatrix} $$ because with this choice of $x$, we obtain $x^TA^3x = -11$

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  • $\begingroup$ We both gave the same example, though I've seen $A$ in another context, it is given as a counter-example to "every copositive matrix is the sum of a positive semidefinite and nonnegative matrix". $\endgroup$ Jun 24, 2011 at 2:38
  • $\begingroup$ @Gjergji: Nice! In the abovecited paper too, they mention that this matrix is an example of an "exceptional" copositive matrix (i.e., a copos matrix that is not of the semidef + nonneg form) $\endgroup$
    – Suvrit
    Jun 24, 2011 at 2:42
  • $\begingroup$ Thanks for your answer. I would like to tick both answers, but the system don't allow.... I know you don't mind about this. $\endgroup$
    – Sunni
    Jun 24, 2011 at 2:50
  • $\begingroup$ You are welcome; and of course, no problems! $\endgroup$
    – Suvrit
    Jun 24, 2011 at 2:52

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