🔬 Tutorial problems zeta \(\zeta\)

Note

This problems are designed to help you practice the concepts covered in the lectures. Not all problems may be covered in the tutorial, those left out are for additional practice on your own.

\(\zeta\).1

Formulation

Derive canonical equations for the three conic sections in Euclidean space by using the properties stated in the definitions for:

• ellipse

• parabola

• hyperbola

For example, for ellipse, assume that the focal points are located at \((c,0)\) and \((-c,0)\), and consider an arbitrary point \((x,y) \in \mathbb{R}^2\) which satisfies the defining property of the ellipse. Then, simplify the expression to arrive at the canonical equation of the ellipse, and give interpretations for the parameters \((a,b)\).

Perform additional task for each curve:

1. Ellipse: derive expressions for the coordinates of the focal points using standard parameters \(a\) and \(b\)

2. Parabola: explain the geometric meaning of parameter \(p\)

3. Hyperbola: verify what change of variables leads to the equation \(xy=1\) for the same curve

Hint

Start with the definitions of the conic section, and formulate the defining geometric properties of each curve.

Example of what needs to be done, for a circle. Using the definition that a circle is the set of points equidistant from a fixed point (center), we derive the canonical equation of the circle \(x^2 + y^2 = r^2\).

Let a center be situated in the origin. Then, for an arbitrary point \((x,y) \in \mathbb{R}^2\), the distance from the center is given by the Euclidean norm \(\sqrt{x^2 + y^2}\). The defining property of the circle is that this distance is constant, and equal to the radius \(r\). Therefore, the canonical equation of the circle is

\[ \sqrt{x^2 + y^2} = r \quad \iff \quad x^2 + y^2 = r^2 \]

Solution

Ellipse

Let focal points of ellipse be located at \((c,0)\) and \((-c,0)\), and consider an arbitrary point \((x,y) \in \mathbb{R}^2\) which satisfies the defining property of the ellipse, namely the sum of distances from \((x,y)\) to the focal points is constant denoted \(2d\).

\[\begin{split} \begin{array}{rcl} \sqrt{(x-c)^2 + y^2} + \sqrt{(x+c)^2 + y^2} &=& 2d \\ \sqrt{(x-c)^2 + y^2} &=& 2d - \sqrt{(x+c)^2 + y^2} \\ (x-c)^2 + y^2 &=& 4d^2 - 4d\sqrt{(x+c)^2 + y^2} + (x+c)^2 + y^2 \\ 4d\sqrt{(x+c)^2 + y^2} &=& 4d^2 +4xc \\ d^2(x+c)^2 + d^2y^2 &=& d^4 +2xcd^2 +x^2c^2 \\ x^2d^2 + 2xcd^2 + c^2d^2 + d^2y^2 &=& d^4 +2xcd^2 +x^2c^2 \\ x^2 (d^2-c^2) + y^2d^2 &=& d^4 - c^2d^2 \end{array} \end{split}\]

\[ \frac{x^2}{d^2} + \frac{y^2}{d^2-c^2} = 1 \]

That is, the canonical parameters \(a^2 = d^2\) and \(b^2 = d^2-c^2\). Ellipse intersects the \(x\)-axis at \(x = \pm a\) and \(y\)-axis at \(y = \pm b\).

The focal points in canonical parameters are located at \((\sqrt{a^2-b^2},0)\) and \((-\sqrt{a^2-b^2},0)\).

Parabola

Let the focal point of parabola be located at \((0,c)\) and the dirextrix at \(y=-c\). Consider an arbitrary point \((x,y) \in \mathbb{R}^2\) which satisfies the defining property of the parabola, namely the distance from \((x,y)\) to the focal point is equal to the distance from \((x,y)\) to the directrix \(y=-c\). The latter is equal to the distance from the projection of the point \((x,y)\) to point \((0,-c)\) where the directrix crosses \(y\)-axis.

\[\begin{split} \begin{array}{rcl} \sqrt{x^2+(y-c)^2} &=& y+c \\ x^2+(y-c)^2 &=& y^2+2yc+c^2 \\ x^2+y^2-2yc+c^2 &=& y^2+2yc+c^2 \\ x^2 &=& 4yc \end{array} \end{split}\]

That is, the canonical parameter for parabola is \(p = 2c\), the distance from the focal point to the directrix. The vertex of the parabola is in the origin, half way between.

Hyperbola

Again, let the focal points be located at \((c,0)\) and \((-c,0)\), and consider an arbitrary point \((x,y) \in \mathbb{R}^2\) which satisfies the defining property of the hyperbola, namely the difference of distances from \((x,y)\) to the focal points is constant denoted \(2d\).

\[\begin{split} \begin{array}{rcl} \sqrt{(x+c)^2 + y^2} - \sqrt{(x-c)^2 + y^2} &=& 2d \\ \sqrt{(x+c)^2 + y^2} &=& 2d + \sqrt{(x-c)^2 + y^2} \\ (x+c)^2 + y^2 &=& 4d^2 + 4d\sqrt{(x-c)^2 + y^2} + (x-c)^2 + y^2 \\ 4d\sqrt{(x-c)^2 + y^2} &=& 4xc - 4d^2 \\ d^2(x-c)^2 + d^2y^2 &=& x^2c^2-2xcd^2+d^4 \\ x^2d^2 -2xcd^2 +c^2d^2 + d^2y^2 &=& x^2c^2-2xcd^2+d^4 \\ x^2 (c^2-d^2) - y^2d^2 &=& d^2(c^2 - d^2) \end{array} \end{split}\]

\[ \frac{x^2}{d^2} - \frac{y^2}{c^2-d^2} = 1 \]

That is, again the canonical parameters \(a^2 = d^2\) and \(b^2 = c^2-d^2\).

\(\zeta\).2

Formulation

Determine definiteness of the quadratic forms defined with the following matrixes either by Silvester’s criterion or eigenvalue criterion. For the asymmetric matrices use their symmetric part \(\frac{1}{2}(A+A^{T})\) when constructing a quadratic form (see exercise \(\gamma\).2)

\[\begin{split} A_1 = \begin{pmatrix} 5 & 0 & 1 \\ 1 & 1 & 0 \\ -7 & 1 & 0 \end{pmatrix} \end{split}\]

\[\begin{split} A_2 = \begin{pmatrix} 5 & -2 & 3 \\ 0 & 4 & 0 \\ 0 & -1 & 3 \end{pmatrix} \end{split}\]

\[\begin{split} A_3 = \begin{pmatrix} 1 & 0 & 12 \\ 2 & -5 & 0 \\ 1 & 0 & 2 \end{pmatrix} \end{split}\]

\[\begin{split} A_4 = \begin{pmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{pmatrix} \end{split}\]

\[\begin{split} A_5 = \begin{pmatrix} -4 & 2 & -6 \\ 2 & -1 & 3 \\ -6 & 3 & -9 \end{pmatrix} \end{split}\]

Solution

\(A_1\)

Matrix \(A_1\) is not symmetric, we consider its symmetric part

\[\begin{split} \frac{1}{2}(A_1+A_1^{T}) = \begin{pmatrix} 5 & \tfrac{1}{2} & -3 \\ \tfrac{1}{2} & 1 & \tfrac{1}{2} \\ -3 & \tfrac{1}{2} & 0 \end{pmatrix} \end{split}\]

Applying the Silvester’s criterion, first we compute the leading principle minors:

\[ M_1 = \det(5) >0 \]

\[\begin{split} M_2 = \det \begin{pmatrix} 5 & \tfrac{1}{2} \\ \tfrac{1}{2} & 1 \end{pmatrix} = 5 - \tfrac{1}{4} >0 \end{split}\]

\[\begin{split} M_2 = \det \begin{pmatrix} 5 & \tfrac{1}{2} & -3 \\ \tfrac{1}{2} & 1 & \tfrac{1}{2} \\ -3 & \tfrac{1}{2} & 0 \end{pmatrix} = 0-\tfrac{3}{4}-\tfrac{3}{4}-9-\tfrac{5}{4} <0 \end{split}\]

This pattern does not fit neither definite nor semi-definite patters of the Silvester’s criterion, therefore \(A_1\) is indefinite.

You can also verify numerically that eigenvalues have varying signs.

\(A_2\)

Matrix \(A_2\) is not symmetric, we consider its symmetric part

\[\begin{split} \frac{1}{2}(A_2+A_2^{T}) = \begin{pmatrix} 5 & -1 & \tfrac{3}{2} \\ -1 & 4 & -\tfrac{1}{2} \\ \tfrac{3}{2} & -\tfrac{1}{2} & 3 \end{pmatrix} \end{split}\]

Applying the Silvester’s criterion, first we compute the leading principle minors:

\[ M_1 = \det(5) >0 \]

\[\begin{split} M_2 = \det \begin{pmatrix} 5 & -1 \\ -1 & 4 \end{pmatrix} = 20 - 1 >0 \end{split}\]

\[\begin{split} M_2 = \det \begin{pmatrix} 5 & -1 & \tfrac{3}{2} \\ -1 & 4 & -\tfrac{1}{2} \\ \tfrac{3}{2} & -\tfrac{1}{2} & 3 \end{pmatrix} = 60 + \tfrac{3}{4} + \tfrac{3}{4} - 9 - 3 - \tfrac{5}{4} >0 \end{split}\]

Hence, by Silvester’s criterion \(A_2\) is positive definite.

You can also verify numerically that all eigenvalues are strictly positive.

\(A_3\)

Matrix \(A_3\) is not symmetric, we consider its symmetric part

\[\begin{split} \frac{1}{2}(A_3+A_3^{T}) = \begin{pmatrix} 1 & 1 & \tfrac{13}{2} \\ 1 & -5 & 0 \\ \tfrac{13}{2} & 0 & 2 \end{pmatrix} \end{split}\]

Applying the Silvester’s criterion, first we compute the leading principle minors:

\[ M_1 = \det(1) >0 \]

\[\begin{split} M_2 = \det \begin{pmatrix} 1 & 1 \\ 1 & -5 \end{pmatrix} = -5 - 1 < 0 \end{split}\]

This pattern does not fit neither definite nor semi-definite patters of the Silvester’s criterion, therefore \(A_3\) is indefinite.

You can also verify numerically that eigenvalues have varying signs.

\(A_4\)

Matrix \(A_4\) is symmetric, let us assess definiteness after computing eigenvalues

\[\begin{split} \det \begin{pmatrix} 2-\lambda & 2 & 2 \\ 2 & 2-\lambda & 2 \\ 2 & 2 & 2-\lambda \end{pmatrix} = \det \begin{pmatrix} -\lambda & 2 & 2 \\ \lambda & 2-\lambda & 2 \\ 0 & 2 & 2-\lambda \end{pmatrix} = \end{split}\]

(subtracting second column from the first, see the section on the properties of the determinants)

\[ = -\lambda \big( (2-\lambda)^2 -4 \big) -\lambda(4 -2\lambda -4) = \]

\[ = -\lambda \big(\lambda^2 - 4\lambda + 4-4 -2\lambda \big) = -\lambda^2 (\lambda - 6) \]

Hence, the eigenvalues are \(\{0,6\}\), therefore \(A_4\) is positive semi-definite.

Silvester’s criterion agrees with this conclusion:

• all principle minors of order 1 are positive, \(\det(2)>0\)

• all other principle minors are zero

This is consistent with the positive semi-definite definiteness pattern of the Silvester’s criterion.

\(A_5\)

Matrix \(A_5\) is symmetric, let us again assess definiteness after computing eigenvalues

\[\begin{split} \det \begin{pmatrix} -4-\lambda & 2 & -6 \\ 2 & -1-\lambda & 3 \\ -6 & 3 & -9-\lambda \end{pmatrix} = \end{split}\]

\[\begin{split} = -(\lambda+4)(\lambda+1)(\lambda+9) -36 -36 +\\ + 36(\lambda+1) + 4(\lambda+9) +9(\lambda+4) = \end{split}\]

\[\begin{split} = -\lambda^3 -14\lambda^2 - 49\lambda - 36 -72 +\\ +36\lambda + 36 + 4\lambda + 36 + 9\lambda + 36 = \end{split}\]

\[ = -\lambda^3 -14\lambda^2 = -\lambda^2(\lambda + 14) \]

Eigenvalues are \(\{0,-14\}\), hence the \(A_5\) is negative semi-definite.

Silvester’s criterion again agrees with this conclusion:

• all principle minors of order 1 are negative:

\[\begin{split} \det(-4)<0\\ \det(-1)<0\\ \det(-9)<0 \end{split}\]

• all other principle minors are zero:

\[\begin{split} \det \begin{pmatrix} -4 & 2 \\ 2 & -1 \end{pmatrix} =0\\ \det \begin{pmatrix} -4 & -6 \\ -6 & -9 \end{pmatrix} =0\\ \det \begin{pmatrix} -1 & 3 \\ 3 & -9 \end{pmatrix} =0\\ \det \begin{pmatrix} -4 & 2 & -6 \\ 2 & -1 & 3 \\ -6 & 3 & -9 \end{pmatrix} =-36 \cdot 3 +36 \cdot 3 =0 \end{split}\]

This is consistent with the negative semi-definite definiteness pattern of the Silvester’s criterion.

\(\zeta\).3

Formulation

Determine definiteness of the quadratic forms defined with the following matrixes either by Silvester’s criterion or eigenvalue criterion.

\[\begin{split} A_1 = \begin{bmatrix} 4 & 1 & 1 & 1 \\ 1 & 3 & 1 & 1 \\ 1 & 1 & 2 & 1 \\ 1 & 1 & 1 & 1 \end{bmatrix} \end{split}\]

\[\begin{split} A_2 = \begin{bmatrix} -2 & 0 & 0 & 0 \\ 0 & -3 & 1 & 1 \\ 0 & 1 & -2 & 1 \\ 0 & 1 & 1 & -1 \end{bmatrix} \end{split}\]

\[\begin{split} A_3 = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 4 \end{bmatrix} \end{split}\]

\[\begin{split} A_4 = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \\ 3 & 4 & 1 & 2 \\ 4 & 3 & 2 & 1 \end{bmatrix} \end{split}\]

\[\begin{split} A_5 = \begin{bmatrix} 1 & -1 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \end{split}\]

Solution

\(A_1\)

We use leading principal minors to test definiteness:

• \(D_1 = 4 > 0\)

• \(D_2 = \begin{vmatrix} 4 & 1 \\ 1 & 3 \end{vmatrix} = 11 > 0\)

• \(D_3 > 0\)

\[\begin{split} D_3 = \det\begin{pmatrix} 4 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 2 \end{pmatrix} = 4 \cdot \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} = \end{split}\]

\[ = 4 \cdot (5) - 1 \cdot (1) + 1 \cdot (-2) = 20 - 1 - 2 = 1 > 0 \]

• \(D_4 > 0\)

Expanding along the bottom row, and triangle rule for \(3 \times 3\) determinants:

\[\begin{split} D_4 = \det\begin{pmatrix} 4 & 1 & 1 & 1 \\ 1 & 3 & 1 & 1 \\ 1 & 1 & 2 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} = D_3 - 1 \cdot \begin{vmatrix} 4 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 4 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 1 & 1 \\ 3 & 1 & 1 \\ 1 & 2 & 1 \end{vmatrix} = \end{split}\]

\[ = 17 - (12+1+1-3-4-1) + (4+1+2-1-8-1) - (1+6+1-1-3-2) = \]

\[ = 17 - 6 - 3 - 2 >0 \]

All minors positive → Positive definite

\(A_2\)

Leading minors:

• \(D_1 = -2 < 0\)

• \(D_2 = 6 > 0\)

• \(D_3 = -2 (6-2)< 0\)

• \(D_4 = -2 (-6+1+1+2+1+3) = -4 < 0 \)

• Alternate signs, but wrong pattern

→ Indefinite

\(A_3\)

Eigenvalues: \(\{0, 2, 3, 4\}\)
Contains zero and positives → Positive semi-definite

\(A_4\)

Matrix \(D\) is symmetric. Let’s examine its leading principal minors:

• \(D_1 = 1 > 0\)

• \(D_2 = \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = -3 < 0\)

The sign flips, but the first principle minor is not negative or zero, so \(A_4\) is indefinite.

\(A_5\)

Matrix \(E\) is block diagonal, therefore its eigenvalues are simply the union of the eigenvalues of each diagonal block.

• The top-left \(2 \times 2\) block has eigenvalues \(\{0, 2\}\)

• The bottom-right \(2 \times 2\) block is all zeros, so eigenvalues \(\{0, 0\}\)

Total eigenvalues: \(\{0, 0, 0, 2\}\)
→ All eigenvalues non-negative and at least one is zero.

Eigenvalues: \(\{2, 0, 0, 0\}\) → Positive semi-definite

\(\zeta\).4

Formulation

Determine whether each of the following quadratic forms in three variables is positive or negative definite or semidefinite, or indefinite:

1. \(-x^2 - y^2 - 2z^2 + 2xy\)

2. \(x^2 - 2xy + xz + 2yz + 2z^2 + 3zx\)

3. \(-4x^2 - y^2 + 4xz - 2z^2 + 2yz\)

4. \(-x^2 - y^2 + 2xz + 4yz + 2z^2\)

5. \(-x^2 + 2xy - 2y^2 + 2xz - 5z^2 + 2yz\)

6. \(y^2 + xy + 2xz\)

7. \(-3x^2 + 2xy - y^2 + 4yz - 8z^2\)

8. \(2x^2 + 2xy + 2y^2 + 4z^2\)

Solution

1. Quadratic form: \(-x^2 - y^2 - 2z^2 + 2xy\)

Matrix: $$

(1)\[\begin{pmatrix} -1 & 1 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & -2 \end{pmatrix}\]

\[ \begin{align}\begin{aligned}- First-order minors: $-1$, $-1$, $-2$ - Second-order minors: $0$, $2$, $2$ - Determinant: $0$ \\**Conclusion:** Negative semidefinite.\\ ### 2. Quadratic form: $x^2 - 2xy + xz + 2yz + 2z^2 + 3zx$\\Matrix: \end{aligned}\end{align} \]

(2)\[\begin{pmatrix} 1 & -1 & 2 \\ -1 & 0 & 1 \\ 2 & 1 & 2 \end{pmatrix}\]

\[ \begin{align}\begin{aligned}- First-order minors: $1$, $0$, $2$ - First second-order minor: $-1$ \\**Conclusion:** Indefinite.\\ ### 3. Quadratic form: $-4x^2 - y^2 + 4xz - 2z^2 + 2yz$\\Matrix: \end{aligned}\end{align} \]

(3)\[\begin{pmatrix} -4 & 0 & 2 \\ 0 & -1 & 1 \\ 2 & 1 & -2 \end{pmatrix}\]

\[ \begin{align}\begin{aligned}- First-order minors: $-4$, $-1$, $-2$ - Second-order minors: $4$, $4$, $1$ - Determinant: $0$ \\**Conclusion:** Negative semidefinite.\\ ### 4. Quadratic form: $-x^2 - y^2 + 2xz + 4yz + 2z^2$\\Matrix: \end{aligned}\end{align} \]

(4)\[\begin{pmatrix} -1 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & 2 & 2 \end{pmatrix}\]

\[ \begin{align}\begin{aligned}- Leading principal minors: $-1$, $1$, $7$ \\**Conclusion:** Indefinite.\\ ### 5. Quadratic form: $-x^2 + 2xy - 2y^2 + 2xz - 5z^2 + 2yz$\\Matrix: \end{aligned}\end{align} \]

(5)\[\begin{pmatrix} -1 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -5 \end{pmatrix}\]

\[ \begin{align}\begin{aligned}- Leading principal minors: $-1$, $1$, $0$ - First-order minors: $-1$, $-2$, $-5$ - Second-order minors: $1$, $4$, $9$ - Third-order minor: $0$ \\**Conclusion:** Negative semidefinite.\\ ### 6. Quadratic form: $y^2 + xy + 2xz$\\Matrix: \end{aligned}\end{align} \]

(6)\[\begin{pmatrix} 0 & \frac{1}{2} & 1 \\ \frac{1}{2} & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}\]

\[ \begin{align}\begin{aligned}- One first-order minor is positive - A second-order minor (excluding third row and column) is negative \\**Conclusion:** Indefinite.\\ ### 7. Quadratic form: $-3x^2 + 2xy - y^2 + 4yz - 8z^2$\\Matrix: \end{aligned}\end{align} \]

(7)\[\begin{pmatrix} -3 & 1 & 0 \\ 1 & -1 & 2 \\ 0 & 2 & -8 \end{pmatrix}\]

\[ \begin{align}\begin{aligned}- Leading principal minors: $-3$, $2$, $-4$ \\**Conclusion:** Negative definite.\\ ### 8. Quadratic form: $2x^2 + 2xy + 2y^2 + 4z^2$\\Matrix: \end{aligned}\end{align} \]

(8)\[\begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 4 \end{pmatrix}\]

$$

• Leading principal minors: \(2\), \(3\), \(12 > 0\)

Conclusion: Positive definite.

This problem is by Martin J. Osborne of the University of Toronto (reference)

\(\zeta\).5

Formulation

Find conditions on \(a\) and \(b\) under which the following matrix

\[\begin{split} \begin{bmatrix} a & 1 & b \\ 1 & -1 & 0 \\ b & 0 & -2 \end{bmatrix} \end{split}\]

is negative definite, negative semidefinite, positive definite, positive semidefinite, and indefinite.

Hint

There may be no values of \(a\) and \(b\) for which the matrix satisfies some of these conditions.

Solution

First let’s find all principle minors:

\[ D_{1,1} = a,\; D_{1,2} = -1,\; D_{1,3} = -2 \]

\[ D_{2,1} = -a-1,\; D_{2,2} = -2a - b^2,\; D_{2,3} = (-2)(-1) - 0 = 2 \]

\[ D_3 = 2a + b^2 + 2 \]

The matrix can not be positive definite for any value of \(a\) as it would require both \(D_{1,1} = a>0\) and \(D_{2,1} = -a-1 > 0\). It can not be positive semidefinite either because two of the first-order principal minors are negative, namely \(D_{1,2}\) and \(D_{1,3}\).

Necessary and sufficient conditions for it to be negative definite are given by the strict inequalities for the principal leading minors:

• \(D_{1,1} = a < 0\)

• \(D_{2,1} = -a-1 > 0\)

• \(D_3 = 2a + b^2 + 2 <0\)

Thus, it is negative definite if and only if \(a < -1\) and \(b^2 < -2a - 2\).

It is negative semidefinite if and only if all the principal minors have the required signs:

1. \(D_{1,1} = a \leqslant 0\)

2. \(D_{1,2} = -1 \leqslant 0\)

3. \(D_{1,3} = -2 \leqslant 0\)

4. \(D_{2,1} = -a-1 \geqslant 0\)

5. \(D_{2,2} = -2a-b^2 \geqslant 0\)

6. \(D_{2,3} = 2 \geqslant 0\)

7. \(D_{3} = 2a + b^2 + 2 \leqslant 0\)

The first and the fourth conditions give \(a \leqslant -1\), the last inequality implies the fifth, therefore the matrix is negative semidefinite if and only if \(a \leqslant -1\) and \(b^2 \leqslant -2a - 2\).

In other cases, the matrix is indefinite.

This problem is by Martin J. Osborne of the University of Toronto (reference)