Linear Algebra Quiz: Principal Directions And Quadratic Forms

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Principal Directions And Quadratic FormsQuestion 1 of 18

A quadratic form in $\mathbb{R}^3$ has the property that its associated matrix $A$ has eigenvalues $\lambda_1 > \lambda_2 > \lambda_3 > 0$ with corresponding eigenvectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ . If we consider the ellipsoid $\mathbf{x}^T A \mathbf{x} = 1$ , which statement correctly describes the relationship between the principal axes of this ellipsoid and the eigenstructure of $A$ ?

The longest semi-axis of the ellipsoid is in the direction of

\mathbf{v}_1

with length

\frac{1}{\sqrt{\lambda_1}}

The principal axes are determined by the eigenvalues but their lengths depend on the determinant of

A

The semi-axes have lengths

\sqrt{\lambda_1}, \sqrt{\lambda_2}, \sqrt{\lambda_3}

in the directions

\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3

respectively

The longest semi-axis of the ellipsoid is in the direction of

\mathbf{v}_3

with length

\frac{1}{\sqrt{\lambda_3}}

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Linear Algebra Quiz

Linear Algebra Quiz: Principal Directions And Quadratic Forms

Practice Principal Directions And Quadratic Forms in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Principal Directions And Quadratic Forms, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The longest semi-axis of the ellipsoid is in the direction of $\mathbf{v}_1$ with length $\frac{1}{\sqrt{\lambda_1}}$
The principal axes are determined by the eigenvalues but their lengths depend on the determinant of $A$
The semi-axes have lengths $\sqrt{\lambda_1}, \sqrt{\lambda_2}, \sqrt{\lambda_3}$ in the directions $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ respectively
The longest semi-axis of the ellipsoid is in the direction of $\mathbf{v}_3$ with length $\frac{1}{\sqrt{\lambda_3}}$ (correct answer)

Explanation: When you encounter quadratic forms and ellipsoids, the key insight is understanding how eigenvalues and eigenvectors of the matrix translate to the geometric properties of the resulting ellipsoid. The equation

\mathbf{x}^T A \mathbf{x} = 1

defines an ellipsoid whose principal axes align with the eigenvectors of

A

. To find the semi-axis lengths, consider what happens when you move along an eigenvector direction. If

\mathbf{x} = t\mathbf{v}_i

where

\mathbf{v}_i

is a unit eigenvector, then

\mathbf{x}^T A \mathbf{x} = t^2 \lambda_i = 1

. Solving for

t

gives

t = \frac{1}{\sqrt{\lambda_i}}

, which is the semi-axis length in the

\mathbf{v}_i

direction. Since

\lambda_1 > \lambda_2 > \lambda_3

, we have

\frac{1}{\sqrt{\lambda_1}} < \frac{1}{\sqrt{\lambda_2}} < \frac{1}{\sqrt{\lambda_3}}

. Therefore, the longest semi-axis corresponds to the smallest eigenvalue

\lambda_3

, making answer D correct. Answer A incorrectly associates the longest axis with the largest eigenvalue. The relationship is actually inverse—larger eigenvalues produce shorter semi-axes. Answer B is vague and incorrect; the determinant doesn't directly determine axis lengths. Answer C has the eigenvalue-to-length relationship backwards; it should be

\frac{1}{\sqrt{\lambda_i}}

, not

\sqrt{\lambda_i}

. Study tip: Remember the inverse relationship in quadratic forms: for ellipsoids defined by

\mathbf{x}^T A \mathbf{x} = 1

, larger eigenvalues create stronger "compression" in those directions, resulting in shorter axes. The smallest eigenvalue always corresponds to the longest axis.

Question 2

Let $Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}$ be a quadratic form where $A$ is a $2 \times 2$ symmetric matrix. If the level sets $Q(\mathbf{x}) = c$ for any non-zero constant $c$ are hyperbolas, which of the following must be true about matrix $A$ ?

The trace of $A$ is zero.
The determinant of $A$ is positive.
The determinant of $A$ is negative. (correct answer)
The matrix $A$ is not invertible.

Explanation: The level sets of a quadratic form are hyperbolas if and only if the form is indefinite. A quadratic form is indefinite if its associated symmetric matrix

A

has both positive and negative eigenvalues. The determinant of a matrix is the product of its eigenvalues. If one eigenvalue

\lambda_1

is positive and the other

\lambda_2

is negative, their product

\det(A) = \lambda_1 \lambda_2

must be negative.

Question 3

A quadratic form $Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}$ is defined by a symmetric matrix $A$ with eigenvalues $\lambda_1 = 3$ and $\lambda_2 = -1$ . The corresponding orthonormal eigenvectors are $\mathbf{v}_1 = \begin{pmatrix} 1/\sqrt{5} \\ 2/\sqrt{5} \end{pmatrix}$ and $\mathbf{v}_2 = \begin{pmatrix} -2/\sqrt{5} \\ 1/\sqrt{5} \end{pmatrix}$ . After an orthogonal change of variables $\mathbf{x} = P\mathbf{y}$ , where $P = [\mathbf{v}_1 \, \mathbf{v}_2]$ , what is the resulting quadratic form $Q'(\mathbf{y})$ ?

$Q'(\mathbf{y}) = \frac{1}{5}y_1^2 - \frac{2}{5}y_2^2$
$Q'(\mathbf{y}) = y_1^2 + y_2^2$
$Q'(\mathbf{y}) = 3y_1^2 - y_2^2$ (correct answer)
$Q'(\mathbf{y}) = 3y_1^2 + 2y_1y_2 - y_2^2$

Explanation: The Principal Axes Theorem states that an orthogonal change of variables

\mathbf{x} = P\mathbf{y}

, where the columns of

P

are orthonormal eigenvectors of

A

, transforms the quadratic form

\mathbf{x}^T A \mathbf{x}

into

\mathbf{y}^T D \mathbf{y}

, where

D

is a diagonal matrix with the eigenvalues of

A

on the diagonal. The new quadratic form has no cross-product terms and is given by

Q'(\mathbf{y}) = \lambda_1 y_1^2 + \lambda_2 y_2^2 + \dots

. In this case, with

\lambda_1 = 3

and

\lambda_2 = -1

, the new form is

Q'(\mathbf{y}) = 3y_1^2 - y_2^2

. The eigenvector information is used to define the transformation

P

but does not appear as coefficients in the new form.

Question 4

Consider the quadratic form $Q(x_1, x_2) = 3x_1^2 + 2x_1x_2 + 3x_2^2$ . What is the maximum value of $Q(x_1, x_2)$ for all points on the unit circle $x_1^2 + x_2^2 = 1$ ?

2
3
4 (correct answer)
6

Explanation: The maximum value of a quadratic form

Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}

subject to the constraint

\|\mathbf{x}\|=1

is the largest eigenvalue of the matrix

A

. The matrix for this quadratic form is

A = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}

. The characteristic equation is

(3-\lambda)^2 - 1 = 0

, which means

3-\lambda = \pm 1

. The eigenvalues are

\lambda = 3 \pm 1

, so

\lambda_1 = 4

and

\lambda_2 = 2

. The largest eigenvalue is 4, which is the maximum value of

Q(\mathbf{x})

on the unit circle.

Question 5

A $2 \times 2$ symmetric matrix $A$ has eigenvalues $\lambda_1 = -2$ and $\lambda_2 = -5$ . Let $Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}$ . Which of the following statements about $Q(\mathbf{x})$ is FALSE?

$Q(\mathbf{x}) < 0$ for all non-zero vectors $\mathbf{x}$ .
The level set $Q(\mathbf{x}) = -1$ is an ellipse.
The matrix $A$ is invertible.
The maximum value of $Q(\mathbf{x})$ on the unit circle is $-5$ . (correct answer)

Explanation: The maximum value of

Q(\mathbf{x})

on the unit circle is the largest eigenvalue of

A

. The eigenvalues are

-2

and

-5

. The larger of these two values is

-2

. Therefore, the statement that the maximum value is

-5

is false. Let's verify the other statements: (A) Since both eigenvalues are negative, the quadratic form is negative definite, meaning

Q(\mathbf{x}) < 0

for all

\mathbf{x} \neq \mathbf{0}

. This is true. (B) For a negative definite form, level sets

Q(\mathbf{x}) = c

for

c<0

are ellipses. This is true. (C) The determinant of

A

is the product of its eigenvalues,

\det(A) = (-2)(-5) = 10 \neq 0

, so

A

is invertible. This is true.

Question 6

An orthogonal transformation $\mathbf{x} = P\mathbf{y}$ converts a quadratic form $Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}$ into the diagonal form $Q'(\mathbf{y}) = 4y_1^2 + 9y_2^2$ . The principal direction corresponding to the $y_1$ -axis is parallel to the vector $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ . What is the matrix $A$ ?

$\begin{pmatrix} 4 & 0 \\ 0 & 9 \end{pmatrix}$
(correct answer)

Explanation: The eigenvalues of

A

are

\lambda_1=4

and

\lambda_2=9

. The eigenvector for

\lambda_1=4

is parallel to

\begin{pmatrix} 3 \\ 4 \end{pmatrix}

. Its normalized form is

\mathbf{v}_1 = \frac{1}{5}\begin{pmatrix} 3 \\ 4 \end{pmatrix}

. The eigenvector for

\lambda_2=9

must be orthogonal to

\mathbf{v}_1

, so it is parallel to

\begin{pmatrix} -4 \\ 3 \end{pmatrix}

. Its normalized form is

\mathbf{v}_2 = \frac{1}{5}\begin{pmatrix} -4 \\ 3 \end{pmatrix}

. We reconstruct

A

using spectral decomposition

A = PDP^T

. With

P = [\mathbf{v}_1 \, \mathbf{v}_2] = \frac{1}{5}\begin{pmatrix} 3 & -4 \\ 4 & 3 \end{pmatrix}

and

D = \begin{pmatrix} 4 & 0 \\ 0 & 9 \end{pmatrix}

, we compute:

A = \frac{1}{25} \begin{pmatrix} 3 & -4 \\ 4 & 3 \end{pmatrix} \begin{pmatrix} 4 & 0 \\ 0 & 9 \end{pmatrix} \begin{pmatrix} 3 & 4 \\ -4 & 3 \end{pmatrix} = \frac{1}{25} \begin{pmatrix} 12 & -36 \\ 16 & 27 \end{pmatrix} \begin{pmatrix} 3 & 4 \\ -4 & 3 \end{pmatrix} = \frac{1}{25} \begin{pmatrix} 36+144 & 48-108 \\ 48-108 & 64+81 \end{pmatrix} = \frac{1}{25} \begin{pmatrix} 180 & -60 \\ -60 & 145 \end{pmatrix} = \begin{pmatrix} 36/5 & -12/5 \\ -12/5 & 29/5 \end{pmatrix}

. Choice A results from incorrectly associating

\lambda=9

with the given eigenvector. Choice B has a sign error in the off-diagonal. Choice D is the correctly simplified matrix.

Question 7

Let $A$ be a $3 \times 3$ symmetric matrix with three distinct eigenvalues. The level surface $Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x} = 1$ is an ellipsoid. Which of the following statements about the principal axes of this ellipsoid is always true?

The lengths of the principal semi-axes are equal to the eigenvalues of $A$ .
The principal axes are aligned with the standard basis vectors $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ .
The principal axes are parallel to the column vectors of the matrix $A$ .
The principal axes are mutually orthogonal to each other. (correct answer)

Explanation: The Principal Axes Theorem states that the principal axes of the level surface are in the directions of the eigenvectors of the matrix

A

. For a symmetric matrix, eigenvectors corresponding to distinct eigenvalues are always orthogonal. Since the three eigenvalues are distinct, the three corresponding eigenvectors (which define the principal axes) must be mutually orthogonal.

Question 8

The quadratic form $Q(\mathbf{x}) = 2x_1^2 + 8x_1x_2 + 2x_2^2$ is diagonalized by an orthogonal change of variables $\mathbf{x} = P\mathbf{y}$ , resulting in a new quadratic form with no cross-product term. Which of the following matrices could be $P$ ?

$\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$
(correct answer)
$\begin{pmatrix} 2 & 4 \\ 4 & 2 \end{pmatrix}$

Explanation: The matrix of the quadratic form is

A = \begin{pmatrix} 2 & 4 \\ 4 & 2 \end{pmatrix}

. The columns of the change-of-variable matrix

P

must be a set of orthonormal eigenvectors of

A

. The eigenvalues of

A

are given by

(2-\lambda)^2 - 16 = 0

, which yields

\lambda_1 = 6

and

\lambda_2 = -2

. The corresponding eigenvectors are proportional to

\begin{pmatrix} 1 \\ 1 \end{pmatrix}

and

\begin{pmatrix} 1 \\ -1 \end{pmatrix}

, respectively. To form the orthogonal matrix

P

, these eigenvectors must be normalized to have a length of 1. The normalized vectors are

\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}

and

\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}

. Placing these as columns gives the matrix in choice B. Choice D is also a valid

P

matrix (the order and sign of eigenvectors can vary), but standard construction often pairs the eigenvectors in the given order. Let's assume the question asks for one possibility. Both B and D are technically correct, but B corresponds to the eigenvalue order 6 then -2. Let's stick with B. A contains non-normalized eigenvectors. C is the matrix

A

itself. D is also a valid matrix, but B is the most standard construction matching the order of eigenvalues found.

Question 9

Let $A$ be a $3 \times 3$ symmetric matrix representing the quadratic form $Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}$ . The eigenvalues of $A$ are $5, 2,$ and $-3$ . Which statement best describes the geometry of the set of points in $\mathbb{R}^3$ where $Q(\mathbf{x})=0$ ?

A cone (correct answer)
A hyperboloid of one sheet
An ellipsoid
Only the origin

Explanation: When you encounter quadratic forms and their zero sets, you're dealing with conic sections in higher dimensions. The key insight is that the eigenvalues of the symmetric matrix determine the geometric shape through the signs and multiplicities of these values. For the quadratic form

Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x} = 0

, we can diagonalize the symmetric matrix

A

using its eigenvalues. In the principal axis coordinate system, this becomes

5y_1^2 + 2y_2^2 - 3y_3^2 = 0

, or equivalently

5y_1^2 + 2y_2^2 = 3y_3^2

. This equation describes a cone because we have two positive eigenvalues and one negative eigenvalue. The positive terms are balanced by the negative term, creating the characteristic double-cone structure where the surface extends infinitely in both directions along the axis corresponding to the negative eigenvalue. Choice A is correct - this mixed signature (two positive, one negative eigenvalue) always produces a cone. Choice B (hyperboloid of one sheet) would require the equation

Q(\mathbf{x}) = k

where

k \neq 0

, not

Q(\mathbf{x}) = 0

. Choice C (ellipsoid) would need all eigenvalues to have the same sign, which isn't the case here. Choice D (only the origin) would occur if all eigenvalues were positive or all were negative, making the quadratic form definite. Remember this pattern: for

Q(\mathbf{x}) = 0

, count the positive and negative eigenvalues. Mixed signs with zero eigenvalues give cylinders, mixed signs without zero eigenvalues give cones, and same signs give only the trivial solution.

Question 10

Consider the quadratic form $Q(\mathbf{x}) = 5x_1^2 + 6x_1x_2 + 5x_2^2$ . The level set $Q(\mathbf{x}) = 1$ defines an ellipse in the $x_1x_2$ -plane. What is the direction of the major axis of this ellipse?

The direction is parallel to the vector $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ .
The direction is parallel to the vector $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ . (correct answer)
The direction is parallel to the vector $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ .
The direction is parallel to the vector $\begin{pmatrix} 5 \\ 3 \end{pmatrix}$ .

Explanation: The symmetric matrix associated with

Q(\mathbf{x})

A = \begin{pmatrix} 5 & 3 \\ 3 & 5 \end{pmatrix}

. The principal axes of the ellipse are along the eigenvectors of

A

. The characteristic equation is

(5-\lambda)^2 - 9 = 0

, which gives eigenvalues

\lambda_1 = 8

and

\lambda_2 = 2

. The equation of the ellipse in the principal axis coordinate system

(y_1, y_2)

8y_1^2 + 2y_2^2 = 1

. This can be written as

\frac{y_1^2}{(1/\sqrt{8})^2} + \frac{y_2^2}{(1/\sqrt{2})^2} = 1

. The semi-major axis has length

1/\sqrt{2}

, which corresponds to the smaller eigenvalue

\lambda_2 = 2

. We find the eigenvector for

\lambda_2=2

(A-2I)\mathbf{v} = \begin{pmatrix} 3 & 3 \\ 3 & 3 \end{pmatrix} \mathbf{v} = \mathbf{0}

, which gives an eigenvector proportional to

\begin{pmatrix} 1 \\ -1 \end{pmatrix}

. This is the direction of the major axis.

Question 11

Consider the quadratic form $Q(x,y) = 3x^2 + 4xy + 6y^2$ . If the principal directions are determined by the eigenvectors of the associated matrix, which of the following statements about the transformation that diagonalizes this quadratic form is correct?

The principal directions are orthogonal and the transformed quadratic form has one positive and one negative eigenvalue
The principal directions are orthogonal and the transformed quadratic form is positive definite with eigenvalues $2$ and $7$ (correct answer)
The principal directions are not orthogonal because the original matrix is not symmetric
The principal directions correspond to the null space and column space of the coefficient matrix

Explanation: The matrix associated with the quadratic form is

A = \begin{pmatrix} 3 & 2 \\ 2 & 6 \end{pmatrix}

. Since

A

is symmetric, its eigenvectors are orthogonal and form the principal directions. The characteristic polynomial is

\det(A - \lambda I) = (3-\lambda)(6-\lambda) - 4 = \lambda^2 - 9\lambda + 14 = (\lambda-2)(\lambda-7)

. Both eigenvalues are positive (2 and 7), making the quadratic form positive definite. Choice A is wrong because both eigenvalues are positive. Choice C is incorrect because the matrix is symmetric. Choice D confuses eigenvectors with fundamental subspaces.

Question 12

Consider the quadratic form $Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}$ where $A$ is a $3 \times 3$ symmetric matrix with eigenvalues $2, -1, 3$ . If $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are the corresponding orthonormal eigenvectors, and we define a new coordinate system using these principal directions, which statement about the level surfaces $Q(\mathbf{x}) = k$ for $k > 0$ is correct?

All level surfaces are ellipsoids because the matrix is positive definite
The level surfaces are hyperboloids of one sheet with the $\mathbf{v}_2$ direction as the axis of the hyperboloid (correct answer)
The level surfaces are hyperboloids of two sheets because two eigenvalues are positive and one is negative
The level surfaces are unbounded in the directions corresponding to negative eigenvalues

Explanation: With eigenvalues

2, -1, 3

, the canonical form is

2u^2 - v^2 + 3w^2 = k

where

(u,v,w)

correspond to directions

\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3

. Since we have two positive and one negative eigenvalue, for

k > 0

, the level surfaces are hyperboloids of one sheet. The negative term corresponds to the

\mathbf{v}_2

direction, which becomes the axis along which the hyperboloid extends. Choice A is wrong because the matrix has a negative eigenvalue. Choice C confuses hyperboloids of one sheet (2 positive, 1 negative eigenvalue) with two sheets (1 positive, 2 negative). Choice D is incorrect because level surfaces are bounded.

Question 13

The quadratic form $Q(x,y) = x^2 + 4xy + 4y^2$ represents a degenerate conic. After diagonalization, what can be concluded about the principal directions and the geometric nature of the level curves $Q(x,y) = c$ ?

The matrix has eigenvalues $0$ and $5$ , so the level curves are parallel lines for $c > 0$ (correct answer)
The matrix has eigenvalues $0$ and $5$ , so the level curves are parabolas opening along the principal direction
The matrix has a repeated eigenvalue $5$ , so the principal directions are not unique
The matrix has eigenvalues $-1$ and $6$ , so the level curves are hyperbolas for all $c > 0$

Explanation: The matrix is

A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}

. The characteristic polynomial is

\det(A - \lambda I) = (1-\lambda)(4-\lambda) - 4 = \lambda^2 - 5\lambda + 4 - 4 = \lambda^2 - 5\lambda = \lambda(\lambda - 5)

. So eigenvalues are

0

and

5

. Since one eigenvalue is zero, the quadratic form is degenerate (determinant of

A

is zero). The level curves

Q(x,y) = c

for

c > 0

are indeed parallel lines because the form can be written as

5v^2 = c

in principal coordinates, where

u

corresponds to the zero eigenvalue. Choice B is wrong because these are lines, not parabolas. Choice C is incorrect about repeated eigenvalues. Choice D has wrong eigenvalues.

Question 14

Consider two quadratic forms $Q_1(\mathbf{x}) = \mathbf{x}^T A_1 \mathbf{x}$ and $Q_2(\mathbf{x}) = \mathbf{x}^T A_2 \mathbf{x}$ where $A_1$ and $A_2$ are $2 \times 2$ symmetric matrices. If $A_1$ and $A_2$ have the same eigenvalues but different eigenvectors, which statement about their principal directions is necessarily true?

The angle between corresponding principal directions depends on the difference between the eigenvalues
The principal directions of $Q_1$ and $Q_2$ are identical because eigenvalues determine eigenvectors uniquely
The principal directions of $Q_1$ and $Q_2$ are related by some orthogonal transformation (correct answer)
The principal directions can be arbitrarily different, and their relative orientation contains no information about the eigenvalue multiplicities

Explanation: When you encounter questions about quadratic forms and their geometric properties, focus on the relationship between eigenvalues, eigenvectors, and the geometric transformations they represent. The key insight here is understanding what "same eigenvalues but different eigenvectors" means geometrically. Since both matrices are symmetric with identical eigenvalues, they represent ellipses (or other conic sections) with the same "shape" - the same ratio of axes lengths. However, these ellipses are oriented differently in space because their eigenvectors (principal directions) differ. The correct answer is C because matrices with the same eigenvalues are related by a similarity transformation. Specifically, if

A_1

and

A_2

have the same eigenvalues, then

A_2 = QA_1Q^T

for some orthogonal matrix

Q

. This orthogonal transformation rotates the principal directions of

A_1

to those of

A_2

while preserving the eigenvalues. Let's examine why the other options fail: A is incorrect because the angle between principal directions depends on the orthogonal transformation relating the matrices, not on eigenvalue differences. B makes a fundamental error - eigenvalues definitely do not determine eigenvectors uniquely, especially when eigenvalues have multiplicity greater than one. D is wrong because the relative orientation is precisely determined by the orthogonal transformation; it's not arbitrary. Remember this principle: for symmetric matrices, having the same eigenvalues means the quadratic forms have the same "intrinsic shape" but potentially different orientations, connected by a rotation. This geometric perspective helps clarify the relationship between algebraic properties and spatial transformations.

Question 15

For the quadratic form $Q(x,y,z) = 2x^2 + 3y^2 + z^2 + 4xy - 2xz + 6yz$ , suppose we want to eliminate all cross-product terms by rotating to principal axes. If the coefficient matrix has eigenvalues $\lambda_1 = 6$ , $\lambda_2 = 3$ , and $\lambda_3 = -3$ , what type of quadric surface does the equation $Q(x,y,z) = 18$ represent?

An ellipsoid because all eigenvalues of the coefficient matrix are positive
An elliptic paraboloid because the trace of the coefficient matrix equals the sum of eigenvalues
A hyperboloid of two sheets because the constant term is positive while one eigenvalue is negative
A hyperboloid of one sheet because the quadratic form has one negative and two positive eigenvalues (correct answer)

Explanation: When you encounter a quadratic form and need to classify the resulting quadric surface, the key insight is that the eigenvalues of the coefficient matrix determine the surface type after rotation to principal axes. First, let's extract the coefficient matrix from

Q(x,y,z) = 2x^2 + 3y^2 + z^2 + 4xy - 2xz + 6yz

. The symmetric matrix is:

After diagonalization using the given eigenvalues

\lambda_1 = 6

\lambda_2 = 3

\lambda_3 = -3

, the equation

Q(x,y,z) = 18

becomes

6u^2 + 3v^2 - 3w^2 = 18

in principal coordinates, or equivalently

\frac{u^2}{3} + \frac{v^2}{6} - \frac{w^2}{6} = 1

. This canonical form has two positive terms and one negative term, which defines a hyperboloid of one sheet. The correct answer is D. Here's why the other options fail: A is wrong because not all eigenvalues are positive (

\lambda_3 = -3

). B incorrectly identifies the surface type—while the trace does equal the sum of eigenvalues, this gives us a hyperboloid, not a paraboloid. C confuses hyperboloid types: having one negative eigenvalue with a positive constant gives a hyperboloid of one sheet, not two sheets. Study tip: For quadric surface classification, count the signs of the eigenvalues in the canonical form. Two positive, one negative = hyperboloid of one sheet. Two negative, one positive = hyperboloid of two sheets. All same sign = ellipsoid.

Question 16

A quadratic form $Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}$ in $\mathbb{R}^3$ has eigenvalues $\lambda_1 = 4$ , $\lambda_2 = -1$ , and $\lambda_3 = 2$ with corresponding orthonormal eigenvectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ . After rotating to principal axes, what is the canonical form of the quadratic surface $Q(\mathbf{x}) = 12$ ?

$4u^2 - v^2 + 2w^2 = 12$ , representing a hyperboloid of one sheet (correct answer)
$\frac{u^2}{3} - v^2 + \frac{w^2}{6} = 1$ , representing a hyperboloid of one sheet
$4u^2 - v^2 + 2w^2 = 12$ , representing a hyperboloid of two sheets
$\frac{u^2}{3} + \frac{v^2}{12} + \frac{w^2}{6} = 1$ , representing an ellipsoid

Explanation: In the principal coordinate system

(u,v,w)

aligned with the eigenvectors, the quadratic form becomes

4u^2 - v^2 + 2w^2 = 12

. Since we have eigenvalues with mixed signs (two positive, one negative), this represents a hyperboloid of one sheet, not two sheets. Choice B incorrectly divides the equation by 12. Choice C has the correct equation but wrong surface type (hyperboloid of two sheets would require two negative eigenvalues). Choice D incorrectly assumes all eigenvalues are positive.

Question 17

Consider the constrained optimization problem: maximize $Q(x,y) = 2x^2 + 6xy + 5y^2$ subject to $x^2 + y^2 = 1$ . Using the relationship between quadratic forms and their principal directions, what is the maximum value of $Q$ on the unit circle?

The maximum value is $\frac{7 - 3\sqrt{5}}{2}$ , which corresponds to the smaller eigenvalue
The maximum value is $8$ , achieved at the point where the gradient of $Q$ is parallel to the constraint gradient
The maximum value is $\frac{7 + 3\sqrt{5}}{2}$ , achieved along the principal direction corresponding to the larger eigenvalue (correct answer)
The maximum value is $7$ , which is the trace of the coefficient matrix divided by 2

Explanation: When you encounter a quadratic form optimization problem on a constraint, you're dealing with finding eigenvalues of a matrix. The key insight is that the maximum and minimum values of a quadratic form on the unit circle are precisely the largest and smallest eigenvalues of its coefficient matrix. First, extract the coefficient matrix from

Q(x,y) = 2x^2 + 6xy + 5y^2

. Since the cross-term

6xy

contributes

3

to both off-diagonal entries, the matrix is $$A = \begin{pmatrix} 2 & 3 \ 3 & 5 \end{pmatrix}

To find eigenvalues, solve $$\det(A - \lambda I) = 0$$: $$\det\begin{pmatrix} 2-\lambda & 3 \\ 3 & 5-\lambda \end{pmatrix} = (2-\lambda)(5-\lambda) - 9 = \lambda^2 - 7\lambda + 1 = 0$$ Using the quadratic formula: $$\lambda = \frac{7 \pm \sqrt{49-4}}{2} = \frac{7 \pm 3\sqrt{5}}{2}$$ The maximum value occurs at the larger eigenvalue: $$\frac{7 + 3\sqrt{5}}{2}$$, confirming answer C. Answer A gives the smaller eigenvalue, which would be the minimum value, not maximum. Answer B incorrectly suggests the maximum is $$8$$ - while the gradient condition is relevant for Lagrange multipliers, this specific value doesn't arise from our eigenvalue calculation. Answer D claims the maximum is $$7$$ (half the trace), but the trace only gives the sum of eigenvalues, not individual extrema. Remember: for quadratic form optimization on the unit circle, always find the eigenvalues of the coefficient matrix - the largest eigenvalue is your maximum value.

Question 18

For which values of the constant $k$ is the quadratic form $Q(x_1, x_2) = kx_1^2 + 6x_1x_2 + x_2^2$ positive definite?

$k > 9$ (correct answer)
$k > 0$
$k < 9$
$k > -1$

Explanation: A quadratic form is positive definite if and only if all eigenvalues of its associated symmetric matrix are positive. The matrix is

A = \begin{pmatrix} k & 3 \\ 3 & 1 \end{pmatrix}

. For a

2 \times 2

matrix, the eigenvalues are both positive if and only if the determinant is positive and the trace is positive. Condition 1:

\det(A) = k(1) - 3(3) = k - 9 > 0

, which implies

k > 9

. Condition 2:

\mathrm{tr}(A) = k + 1 > 0

, which implies

k > -1

. Both conditions must hold. The first condition,

k > 9

, is stricter and automatically implies the second condition. Thus, the quadratic form is positive definite if and only if

k > 9