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$$A=\begin{bmatrix} 1 & -1 & 1 & 1 & -1\\ -1 & 1 & -1 & 1 & 1\\ 1 & -1 & 1 & -1 & 1\\ 1 & 1 & -1 & 1 & -1\\ -1 & 1 & 1 & -1 & 1\\ \end{bmatrix}$$
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:
$$x = \begin{pmatrix} 0.675999805464875\4\\ 0.066489151565243\6\\ 0.045909786113699\3\\ 0.509241928539921\0\\ 0.966306531258645 0\\ \end{pmatrix}$$ because with this choice of $x$, we obtain $x^TA^3x = -0.054001591255556$.11$1 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: $$x = \begin{pmatrix} 0.675999805464875\\ 0.066489151565243\\ 0.045909786113699\\ 0.509241928539921\\ 0.966306531258645 \end{pmatrix}$$ because with this choice of$x$, we obtain$x^TA^3x = -0.054001591255556\$.