1 | initial version |

From the examples in the O.P., I assume that we are dealing with homogeneous quadratic polynomials, that is, with quadratic forms.

Let us state the problem: given a quadratic form \(p\) in \(\mathbb{R}^n\), determine the truth value of the proposition

(

P) there exists \(\mathbf{x}\in\mathbb{R}_+^n\) such that \(p(\mathbf{x})>0\).

Here, \(\mathbb{R}_+\) stands for the set of positive real numbers. One can apply, at least, one of the following tests:

**Test 1**. Let \(A\) be the symmetric matrix \(A\) associated to \(p\), i.e., \(p(\mathbf{x})=\mathbf{x}^TA\mathbf{x}\). Compute the eigenvalues of \(A\). If all the eigenvalues are less than or equal to \(0\), then (*P*) is false.

**Test 2**. Compute the maximum of \(p\) on the set \(D=[0,1]^n\). Then (*P*) is true if and only if such a maximum is positive.

Test 1 can be implemented through the `eigenvalues()`

method for matrices. Test 2 can use the `minimize_constrained`

function, already presented in @dan_fulea’s comment. Please note that maximize \(p\) is equivalent to minimize \(-p\).

Let us apply them to the given polynomials.

**Example 1**. Let \(p=-x^2 - y^2 - z^2 + xy + xz + yz\). We apply Test 1:

```
sage: var("x,y,z");
sage: A = matrix([[-1,1/2,1/2],[1/2,-1,1/2],[1/2,1/2,-1]])
sage: A.eigenvalues()
[0, -3/2, -3/2]
```

By Test 1, (*P*) is false. We could also apply Test 2:

```
sage: var("x,y,z");
sage: p = -x^2 - y^2 - z^2 + x*y + x*z + y*z
sage: sol = minimize_constrained(-p, [[0,1]]*3, [1,1,1])
sage: print "Maximum is", p(*sol), "attained at", sol
Maximum is 0.0 attained at (1.0, 1.0, 1.0)
```

Since the maximum is not greater than \(0\), by Test 2, (*P*) is false. The second argument of `minimize_constrained`

is a list of the intervals where \(x\), \(y\) and \(z\) should be, that is, \([0,1]\) for each variable. The last argument is a starting point of the iterative minimization process.

**Example 2**. Let \(q= -x^2 - y^2 - z^2 + (3/2)(xy + xz + yz)\). We first use Test 1:

```
sage: var("x,y,z");
sage: A = matrix([[-1,3/4,3/4],[3/4,-1,3/4],[3/4,3/4,-1]])
sage: A.eigenvalues()
[1/2, -7/4, -7/4]
```

Since \(A\) has one positive eigenvalue, Test 1 fails. We need to apply Test 2:

```
sage: var("x,y,z");
sage: q = -x^2 - y^2 - z^2 + (3/2)*(x*y + x*z + y*z)
sage: sol = minimize_constrained(-q,[[0,1]]*3, [1,1,1])
sage: print "Maximum is", q(*sol), "attained at", sol
Maximum is 1.5 attained at (1.0, 1.0, 1.0)
```

Since the maximum is positive, by Test 2, (*P*) is true.

**Rationale for Test 1**. If \(A\) does not have a positive eigenvalue, then \(A\) is semidefinite negative, and so \(p(\mathbf{x})=\mathbf{x}^TA\mathbf{x}\leq 0\) for all \(\mathbf{x}\in\mathbb{R}^n\). Consequently, (*P*) is false.

**Rationale for Test 2**. Since \(p\) is continuous and \(D\) is compact, there always exists at least one point \(\mathbf{x}_0\in D\) where \(p\) attains a maximum value in \(D\), i.e. \(p(\mathbf{x}_0)\geq p(\mathbf{x})\) for all \(\mathbf{x}\in D\). Now, if (*P*) is true, there exist \(\mathbf{x}_1\in \mathbb{R}_+^n\) such that \(p(\mathbf{x}_1)>0\). Let \(\mathbf{x}_2=\mathbf{x}_1/\lVert\mathbf{x}_1\rVert\), which obviously belongs to \(D\). We deduce that
\[p(\mathbf{x}_0) \geq p(\mathbf{x}_2)=\frac{p(\mathbf{x}_1)}{\lVert\mathbf{x}_1\rVert^2}>0.\]
Conversely, assume that \(p(\mathbf{x}_0)>0\). If \(\mathbf{x}_0\) belongs to the interior of \(D\), then (*P*) holds with \(\mathbf{x}=\mathbf{x}_0\). If \(\mathbf{x}_0\) is in the boundary of \(D\) (so possibly with a null coordinate), by continuity of \(p\), there exists a ball \(B\) centered at \(\mathbf{x}_0\) where the sign of \(p\) is that of \(p(\mathbf{x}_0)\), that is, \(p\) is positive on \(B\). Since \(B\) contains at least one point \(\mathbf{x}_3\) in the interior of \(D\), then (*P*) holds with \(\mathbf{x}=\mathbf{x}_3\).

2 | No.2 Revision |

From the examples in the O.P., I assume that we are dealing with homogeneous quadratic polynomials, that is, with quadratic forms.

Let us state the problem: given a quadratic form \(p\) in \(\mathbb{R}^n\), determine the truth value of the proposition

(

P) there exists \(\mathbf{x}\in\mathbb{R}_+^n\) such that \(p(\mathbf{x})>0\).

Here, \(\mathbb{R}_+\) stands for the set of positive real numbers. One can apply, at least, one of the following tests:

**Test 1**. Let \(A\) be the symmetric matrix ~~\(A\) ~~associated to \(p\), i.e., \(p(\mathbf{x})=\mathbf{x}^TA\mathbf{x}\). Compute the eigenvalues of \(A\). If all the eigenvalues are less than or equal to \(0\), then (*P*) is ~~false.~~false. If there exists a positive eigenvalue \(\lambda\) with an associated eigenvector \(\mathbf{v}\in\mathbb{R}_+^n\), then (*P*) is true.

Note that this test does not cover all the possibilities, since it only checks sufficient conditions.

**Test 2**. Compute the maximum of \(p\) on the set \(D=[0,1]^n\). Then (*P*) is true if and only if such a maximum is positive.

Test 1 can be implemented through the `eigenvalues()`

~~method ~~and `eigenvectors_right()`

methods for matrices. Test 2 can use the `minimize_constrained`

function, already presented in @dan_fulea’s ~~comment. ~~answer. Please note that maximize \(p\) is equivalent to minimize \(-p\).

Let us apply them to the given polynomials.

**Example 1**. Let \(p=-x^2 - y^2 - z^2 + xy + xz + yz\). We apply Test 1:

```
sage: var("x,y,z");
sage: A = matrix([[-1,1/2,1/2],[1/2,-1,1/2],[1/2,1/2,-1]])
sage: A.eigenvalues()
[0, -3/2, -3/2]
```

By Test 1, (*P*) is ~~false. ~~false, since all the eigenvalues are less than or equal to \(0\). We could also apply Test 2:

```
sage: var("x,y,z");
sage: p = -x^2 - y^2 - z^2 + x*y + x*z + y*z
sage: sol = minimize_constrained(-p, [[0,1]]*3,
```~~[1,1,1])
~~[0.1,0.9,0.5])
sage: print "Maximum is", p(*sol), "attained at", sol
Maximum is 0.0 attained at ~~(1.0, 1.0, 1.0)
~~(0.4999999998931121, 0.5000000001068878, 0.5)

Since the maximum is not greater than \(0\), by Test 2, (*P*) is ~~false. ~~false.
The second argument of `minimize_constrained`

is a list of the intervals where \(x\), \(y\) and \(z\) should be, that is, \([0,1]\) for each variable. The last argument is a starting point of the iterative minimization ~~process.~~process. In this example, if we take a different starting point, the maximum is also \(0\), but it can be reached at a different point (\(p\) is \(0\) on the line \(x=y=z\)).

**Example 2**. Let \(q= -x^2 - y^2 - z^2 + (3/2)(xy + xz + yz)\). We first use Test 1:

```
sage: var("x,y,z");
sage: A = matrix([[-1,3/4,3/4],[3/4,-1,3/4],[3/4,3/4,-1]])
sage: A.eigenvalues()
[1/2, -7/4, -7/4]
sage: A.eigenvectors_right()
[(1/2, [(1, 1, 1)], 1), (-7/4, [(1, 0, -1), (0, 1, -1)], 2)]
```

~~Since ~~The matrix \(A\) has one positive eigenvalue, ~~Test 1 fails. We need ~~\(\lambda=1/2\), with an associated
eigenvector \(\mathbf{v}=(1,1,1)\) belonging to \(\mathbb{R}_+^n\). Hence, (*P*) is true. Let us now apply Test 2:

```
sage: var("x,y,z");
sage: q = -x^2 - y^2 - z^2 + (3/2)*(x*y + x*z + y*z)
sage: sol = minimize_constrained(-q,[[0,1]]*3,
```~~[1,1,1])
~~[0.1,0.9,0.5])
sage: print "Maximum is", q(*sol), "attained at", sol
Maximum is 1.5 attained at (1.0, 1.0, 1.0)

Since the maximum is positive, by Test 2, (*P*) is true.

**Rationale for Test 1**. If \(A\) does not have a positive eigenvalue, then \(A\) is semidefinite negative, and so \(p(\mathbf{x})=\mathbf{x}^TA\mathbf{x}\leq 0\) for all \(\mathbf{x}\in\mathbb{R}^n\). Consequently, (*P*) is ~~false.~~false. Likewise, if there exists a positive eigenvalue \(\lambda\) with an associated eigenvector \(\mathbf{v}\in\mathbb{R}_+^n\), then \(p(\mathbf{v})=\mathbf{v}^TA\mathbf{v}=\lambda\mathbf{v}^T\mathbf{v}=\lambda\lVert\mathbf{v}\rVert^2>0\). Hence
(*P*) is true.

**Rationale for Test 2**. Since \(p\) is continuous and \(D\) is compact, there always exists at least one point \(\mathbf{x}_0\in D\) where \(p\) attains a maximum value in \(D\), i.e. \(p(\mathbf{x}_0)\geq p(\mathbf{x})\) for all \(\mathbf{x}\in D\). Now, if (*P*) is true, there exist \(\mathbf{x}_1\in \mathbb{R}_+^n\) such that \(p(\mathbf{x}_1)>0\). Let \(\mathbf{x}_2=\mathbf{x}_1/\lVert\mathbf{x}_1\rVert\), which obviously belongs to \(D\). We deduce that
\[p(\mathbf{x}_0) \geq p(\mathbf{x}_2)=\frac{p(\mathbf{x}_1)}{\lVert\mathbf{x}_1\rVert^2}>0.\]
Conversely, assume that \(p(\mathbf{x}_0)>0\). If \(\mathbf{x}_0\) belongs to the interior of \(D\), then (*P*) holds with \(\mathbf{x}=\mathbf{x}_0\). If \(\mathbf{x}_0\) is in the boundary of \(D\) (so possibly with a null coordinate), by continuity of \(p\), there exists a ball \(B\) centered at \(\mathbf{x}_0\) where the sign of \(p\) is that of \(p(\mathbf{x}_0)\), that is, \(p\) is positive on \(B\). Since \(B\) contains at least one point \(\mathbf{x}_3\) in the interior of \(D\), then (*P*) holds with \(\mathbf{x}=\mathbf{x}_3\).

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