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Magazine Chapter 8: Problem 5
For each of the following symmetric matrices, identify the shape of the graph \(\vec{x}^{T} A \vec{x}=1\) and the shape of the graph \(\vec{x}^{T} A \vec{x}=-1\). (a) \(A=\left[\begin{array}{ll}4 & 2 \\ 2 & 1\end{array}\right]\) (b) \(A=\left[\begin{array}{rr}5 & 3 \\ 3 & -3\end{array}\right]\) (c) \(A=\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right]\) (d) \(A=\left[\begin{array}{rrr}1 & 0 & -2 \\ 0 & -1 & -2 \\ -2 & -2 & 0\end{array}\right]\) (e) \(A=\left[\begin{array}{rrr}1 & 8 & 4 \\ 8 & 1 & -4 \\ 4 & -4 & 7\end{array}\right]\)
Short Answer
Expert verified
(a) Ellipsoid; (b) Hyperbola; (c) Hyperboloid; (d) Paraboloid; (e) Hyperboloid.
Step by step solution
01
Understand the Problem
To determine the shape of the graph for the quadratic forms \(\vec{x}^{T} A \vec{x} = 1\) and \(\vec{x}^{T} A \vec{x} = -1\), we need to analyze the eigenvalues of the given symmetric matrix \(A\). The number of positive, negative, and zero eigenvalues will indicate if the quadratic form is an ellipsoid, hyperboloid, paraboloid, or a degenerate form.
02
(a): Compute Eigenvalues of Matrix A
For matrix \(A = \begin{bmatrix} 4 & 2 \ 2 & 1 \end{bmatrix}\), find the eigenvalues by solving the characteristic equation: \(\det(\lambda I - A) = 0\). This gives:\[\det \begin{pmatrix} \lambda - 4 & -2 \ -2 & \lambda - 1 \end{pmatrix} = (\lambda - 4)(\lambda - 1) - (-2)(-2) = \lambda^2 - 5\lambda + 2 = 0\]Solving, \(\lambda^2 - 5\lambda + 2 = 0\), gives the eigenvalues \(\lambda_1 = 4.56\) and \(\lambda_2 = 0.44\). Both are positive.
03
(a): Identify the Shape Based on Eigenvalues
Since both eigenvalues are positive, the graph \(\vec{x}^{T} A \vec{x} = 1\) represents an ellipsoid and \(\vec{x}^{T} A \vec{x} = -1\) is not possible as all terms are positive.
04
(b): Compute Eigenvalues of Matrix A
For matrix \(A = \begin{bmatrix} 5 & 3 \ 3 & -3 \end{bmatrix}\), find the eigenvalues by solving the characteristic equation: \[\det \begin{pmatrix} \lambda - 5 & -3 \ -3 & \lambda + 3 \end{pmatrix} = (\lambda - 5)(\lambda + 3) - (-3)(-3) = \lambda^2 - 2\lambda - 24 = 0\]Solving, we get eigenvalues \(\lambda_1 = 8\) and \(\lambda_2 = -3\).
05
(b): Identify the Shape Based on Eigenvalues
With one positive and one negative eigenvalue, the graph is a hyperbola for both \(\vec{x}^{T} A \vec{x} = 1\) and \(\vec{x}^{T} A \vec{x} = -1\).
06
(c): Compute Eigenvalues of Matrix A
For the 3x3 matrix \(A = \begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \end{bmatrix}\), compute the eigenvalues using the determinant of \(\lambda I - A\). Solving the characteristic polynomial, the eigenvalues turn out to be \(2, -1, -1\).
07
(c): Identify the Shape Based on Eigenvalues
With one positive and two negative eigenvalues, the graph for \(\vec{x}^{T} A \vec{x} = 1\) is a hyperboloid of two sheets. For \(\vec{x}^{T} A \vec{x} = -1\), it is a hyperboloid of one sheet.
08
(d): Compute Eigenvalues of Matrix A
For \(A = \begin{bmatrix} 1 & 0 & -2 \ 0 & -1 & -2 \ -2 & -2 & 0 \end{bmatrix}\), solve the characteristic equation to determine the eigenvalues. The eigenvalues are found to be \(3, -3, 0\).
09
(d): Identify the Shape Based on Eigenvalues
One positive, one negative, and one zero eigenvalue imply that each graph represents a paraboloid with elliptic cross-sections.
10
(e): Compute Eigenvalues of Matrix A
For \(A = \begin{bmatrix} 1 & 8 & 4 \ 8 & 1 & -4 \ 4 & -4 & 7 \end{bmatrix}\), solve the characteristic equation for eigenvalues, getting values \(12, -5, 2\).
11
(e): Identify the Shape Based on Eigenvalues
With two positive and one negative eigenvalues, the graph for \(\vec{x}^{T} A \vec{x} = 1\) is a hyperboloid of one sheet, and \(\vec{x}^{T} A \vec{x} = -1\) is a hyperboloid of two sheets.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Symmetric Matrices
A symmetric matrix is one where the transpose of the matrix is equal to the matrix itself. In simpler terms, it means the elements across the main diagonal are mirrored. This can be visually represented as \[ A = \begin{bmatrix} a & b \ b & d \end{bmatrix} \] where the elements at positions \((1,2)\) and \((2,1)\) are equal.
Symmetric matrices have several important properties:
Symmetric matrices have several important properties:
- The eigenvalues of a symmetric matrix are always real.
- The matrix can be diagonalized using an orthogonal matrix. This means it can be expressed in the form \( PDP^T \), where \(P\) is an orthogonal matrix and \(D\) is a diagonal matrix.
- They often arise in quadratic forms, which are expressions involving terms that are squared and of the form \( \vec{x}^{T} A \vec{x} \).
Eigenvalues
Eigenvalues are key in unlocking the properties of matrices, especially when working with symmetric matrices. They describe how much vectors are stretched or squished along particular directions when multiplied by the matrix.
To find eigenvalues, we solve the characteristic equation, which is given as \[det(\lambda I - A) = 0\] where \(I\) is the identity matrix and \(det\) signifies the determinant. The solutions \(\lambda\) here are the eigenvalues.
To find eigenvalues, we solve the characteristic equation, which is given as \[det(\lambda I - A) = 0\] where \(I\) is the identity matrix and \(det\) signifies the determinant. The solutions \(\lambda\) here are the eigenvalues.
- If all eigenvalues are positive, the quadratic form represents an ellipsoid.
- If there is a mix of positive and negative eigenvalues, the form represents a hyperboloid.
- The number of positive, negative, and zero eigenvalues can help indicate different geometric shapes like cylinders and paraboloids.
Ellipsoids
Ellipsoids are three-dimensional shapes that can be thought of as elongated spheres. These shapes are represented by quadratic forms where all eigenvalues are positive.
An ellipsoid takes on the general equation \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \] where \(a\), \(b\), and \(c\) are the semi-principal axes lengths. These axes are determined by the magnitude of eigenvalues, which dictate the shape鈥檚 dimensions and orientation.
An ellipsoid takes on the general equation \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \] where \(a\), \(b\), and \(c\) are the semi-principal axes lengths. These axes are determined by the magnitude of eigenvalues, which dictate the shape鈥檚 dimensions and orientation.
- All axes are real and distinct for different values of \(a, b,and c\).
- If two axes are of equal length, the ellipsoid exhibits more spherical symmetry.
- The matrix associated with this quadratic form helps determine its principal directions, which align with its eigenvectors.
Hyperboloids
Hyperboloids are intriguing three-dimensional shapes manifesting from quadratic forms that possess both positive and negative eigenvalues. This results in a shape that contrasts with ellipsoids, introducing complex geometry.
A hyperboloid can be visualized through two main categories based on its equation:
A hyperboloid can be visualized through two main categories based on its equation:
- One-sheet Hyperboloid: Takes the form \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1 \] It resembles a saddle and involves a mixture of positive and negative curves.
- Two-sheet Hyperboloid: Expressed as \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1 \] Divided into two disconnected parts, these tend to look like two bowls facing each other.
