Principal Directions & Quadratic Forms - Linear Algebra

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?
Answer: A hyperboloid of one sheet because the quadratic form has one negative and two positive eigenvalues 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: $A = \begin{pmatrix} 2 & 2 & -1 \\ 2 & 3 & 3 \\ -1 & 3 & 1 \end{pmatrix}$ 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.
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 $Q'(\mathbf{y})$ ?
Answer: $Q'(\mathbf{y}) = 3y_1^2 - y_2^2$ 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.
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?
Answer: The principal directions of $Q_1$ and $Q_2$ are related by some orthogonal transformation 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.
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 $P$ ?
Answer: $\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ 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.
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?
Answer: The principal directions are orthogonal and the transformed quadratic form is positive definite with eigenvalues $2$ and $7$ 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.
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 $x_1^2 + x_2^2 = 1$ ?
Answer: 4 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.
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 $Q(\mathbf{x})=0$ ?
Answer: A cone 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.
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 $A$ ?
Answer: The determinant of $A$ is negative. 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.
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 $A$ ?
Answer: The longest semi-axis of the ellipsoid is in the direction of $\mathbf{v}_3$ with length $\frac{1}{\sqrt{\lambda_3}}$ 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.
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?
Answer: The direction is parallel to the vector $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ . The symmetric matrix associated with $Q(\mathbf{x})$ is $A = \begin{pmatrix} 5 & 3 \\ 3 & 5 \end{pmatrix}$ . 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 $(y_1, y_2)$ is $8y_1^2 + 2y_2^2 = 1$ . 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.
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 $Q(\mathbf{x}) = 12$ ?
Answer: $4u^2 - v^2 + 2w^2 = 12$ , representing a hyperboloid of one sheet 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.
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?
Answer: $k > 9$ 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 $k > 9$ .
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 $A$ ?
Answer: $\frac{1}{5}\begin{pmatrix} 36 & -12 \\ -12 & 29 \end{pmatrix}$ 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.
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?
Answer: The principal axes are mutually orthogonal to each other. 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.
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?
Answer: The maximum value of $Q(\mathbf{x})$ on the unit circle is $-5$ . 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.
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?
Answer: The level surfaces are hyperboloids of one sheet with the $\mathbf{v}_2$ direction as the axis of the hyperboloid 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.
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?
Answer: The maximum value is $\frac{7 + 3\sqrt{5}}{2}$ , achieved along the principal direction corresponding to the larger eigenvalue 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.
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 $Q(x,y) = c$ ?
Answer: The matrix has eigenvalues $0$ and $5$ , so the level curves are parallel lines for $c > 0$ 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.

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