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Chapter 9: Problem 25

Let \(B=\left[\begin{array}{rrr}11 & -8 & 4 \\ -8 & -1 & -2 \\ 4 & -2 & -4\end{array}\right]\) (b) Find a maximal set \(S\) of nonzero orthogonal eigenvectors of \(B\). (c) Find an orthogonal matrix \(P\) such that \(D=P^{-1} B P\) is diagonal.

Short Answer

Expert verified
Short Answer: The maximal set S of nonzero orthogonal eigenvectors of matrix B is {饾懃1, 饾懃4} where 饾懃1 = \(\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}\) and 饾懃4 = \(\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}\). The orthogonal matrix P such that D=P^{-1}BP is diagonal is: P = \(\begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\).

Step by step solution

01

Calculate the eigenvalues of matrix B

To calculate the eigenvalues, we need to solve the characteristic equation given by |B - 位I|=0, where 位 is the eigenvalue and I is the identity matrix. For this matrix B, |B - 位I| = |\(\begin{bmatrix} 11-\lambda & -8 & 4 \\ -8 & -1-\lambda & -2 \\ 4 & -2 & -4-\lambda\end{bmatrix}\)| = 0. Now, we compute the determinant to find the characteristic polynomial, |B - 位I| = (11-位)(-1-位)((-4)-位) - (-8)(-2)(4) - (4)(-8)(-2) + (4)^2(11-位) + (4)(-8)(-1-位) + (-2)^2(-8). Expanding and simplifying the polynomial, we get the characteristic polynomial as: 位^3 + (-6)位^2 - 33位 - 36 = 0
02

Find the eigenvalues from the characteristic polynomial

The characteristic polynomial is a cubic equation that might not be easy to solve analytically. However, through inspection or using a numerical methods like Newton's method, we can find that the eigenvalues are: 位1 = -2, 位2 = 3, and 位3 = -6
03

Compute the eigenvectors for each eigenvalue

Now that we have the eigenvalues, we can find the eigenvectors by solving the linear system (B - 位I)饾懃 = 0 for each eigenvalue. For 位1 = -2, we have: \(\begin{bmatrix} 13 & -8 & 4 \\ -8 & 1 & -2 \\ 4 & -2 & -2\end{bmatrix}\) After row reduction: \(\begin{bmatrix} 1 & -\frac{1}{2} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}\) One possible eigenvector for 位1 is 饾懃1 = \(\begin{bmatrix} 1 \\ 2 \\ 0\end{bmatrix}\) For 位2 = 3, we have: \(\begin{bmatrix} 8 & -8 & 4 \\ -8 & 4 & -2 \\ 4 & -2 & -7\end{bmatrix}\) After row reduction: \(\begin{bmatrix} 1 & -1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}\) One possible eigenvector for 位2 is 饾懃2 = \(\begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix}\) For 位3 = -6, we have: \(\begin{bmatrix} 17 & -8 & 4 \\ -8 & -5 & -2 \\ 4 & -2 & 2\end{bmatrix}\) After row reduction: \(\begin{bmatrix} 1 & -\frac{4}{9} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}\) One possible eigenvector for 位3 is 饾懃3 = \(\begin{bmatrix} 9 \\ 4 \\ 0\end{bmatrix}\)
04

Check orthogonality of eigenvectors; find maximal orthogonal set

Now we can check if the eigenvectors we found are orthogonal by taking the dot product between each pair of eigenvectors: 饾懃1 路 饾懃2 = 1 * 1 + 2 * 1 + 0 * 0 = 3 鈮� 0, 饾懃1 路 饾懃3 = 1 * 9 + 2 * 4 + 0 * 0 = 17 鈮� 0, 饾懃2 路 饾懃3 = 1 * 9 + 1 * 4 + 0 * 0 = 13 鈮� 0 Since all the dot products are non-zero, the eigenvectors are not orthogonal to each other. However, the eigenvectors all lie in the xy-plane (since their z-components are all zero). Therefore, 饾懃1 = \(\begin{bmatrix}1 \\ 2 \\ 0\end{bmatrix}\) and any linearly independent vector in the xy-plane can form a maximal orthogonal set for B. For instance, 饾懃1 = \(\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}\) and 饾懃4 = \(\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}\) are orthogonal (\(1*(-2) + 2*1 + 0*0 = 0\)), and make up the maximal orthogonal set. Maximal set S = {饾懃1, 饾懃4}.
05

Construct orthogonal matrix P

Now we can construct an orthogonal matrix P using the orthogonal eigenvectors such that D=P^{-1}BP is diagonal. Since our matrix is 3x3, we need one more orthogonal vector to complete the matrix. Since both 饾懃1 and 饾懃4 lie in the xy-plane, a vector orthogonal to both must have a non-zero z-component and zero x and y components. Thus, 饾懃5 = \(\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\) forms an orthogonal basis with 饾懃1 and 饾懃4. Now construct the orthogonal matrix P: P = \(\begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) Now, D = P^(-1)BP should be diagonal. Check by doing the matrix multiplication.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Eigenvectors
Eigenvectors are special vectors associated with a matrix that do not change direction during a linear transformation represented by the matrix. Identifying an eigenvector involves solving a linear equation system, typically derived from:
  • Subtracting the eigenvalue from the diagonal of the matrix
  • Solving \( (B - \lambda I)x = 0 \)
Here, \( B \) is the matrix, \( \lambda \) is the eigenvalue, and \( x \) is the eigenvector.
The eigenvectors are crucial because they help in understanding the geometric transformations the matrix represents, like scaling, rotating, or reflecting. In the example provided, for each eigenvalue \( \lambda_1 = -2 \), \( \lambda_2 = 3 \), and \( \lambda_3 = -6 \), we computed corresponding eigenvectors. To build the orthogonal matrix \( P \), these vectors are a vital component.
Eigenvalues
Eigenvalues play a crucial role in the analysis of linear transformations. They are scalars corresponding to eigenvectors, which describe the factor by which the associated eigenvector is scaled during the transformation. To find eigenvalues, we solve the characteristic equation obtained from the determinant condition of the matrix subtraction \( |B - \lambda I| = 0 \).
In this specific problem, the characteristic polynomial \( \lambda^3 - 6 \lambda^2 - 33 \lambda - 36 = 0 \) was generated. The solutions to this equation are the eigenvalues which are as follows:
  • \( \lambda_1 = -2 \)
  • \( \lambda_2 = 3 \)
  • \( \lambda_3 = -6 \)
These eigenvalues indicate how each dimension involved in the transformation matrix is scaled.
Characteristic Polynomial
The characteristic polynomial is essential in finding the eigenvalues of a matrix. Derived from the determinant \( |B - \lambda I| = 0 \), it is a polynomial equation where the roots are the eigenvalues of the matrix. Understanding this polynomial allows us to determine important properties of the matrix, like stability and diagonalizability.
For matrix \( B \), the characteristic polynomial was:
  • \( \lambda^3 - 6\lambda^2 - 33\lambda - 36 = 0 \)
Solving this polynomial equation involves finding the roots, which are the eigenvalues. These roots help in understanding how the matrix transforms vectors, which is crucial in the diagonalization process.
Diagonalization
Diagonalization is the process of converting a matrix into a diagonal form, which is easier to work with, as its non-diagonal entries are all zero. This transformation is accomplished using eigenvectors and eigenvalues by finding an orthogonal matrix \( P \) such that:
  • \( D = P^{-1}BP \)
where \( D \) is a diagonal matrix with the eigenvalues on its diagonal.
In the exercise, a set of orthogonal eigenvectors were found, denoted as \( S \), which were used to construct \( P \):
  • \( P = \begin{bmatrix} 1 & -2 & 0 \ 2 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \)
The diagonalization simplifies calculations and provides a clearer understanding of the linear transformation properties represented by matrix \( B \). This process reveals insights into the geometric transformations, like whether they involve scaling or rotations, encoded by the matrix.

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Most popular questions from this chapter

For each of the following matrices, find all eigenvalues and corresponding linearly independent eigenvectors: (a) \(A=\left[\begin{array}{ll}2 & -3 \\ 2 & -5\end{array}\right]\) (b) \(\quad B=\left[\begin{array}{rr}2 & 4 \\ -1 & 6\end{array}\right]\) , (c) \(C=\left[\begin{array}{ll}1 & -4 \\ 3 & -7\end{array}\right]\) When possible, find the nonsingular matrix \(P\) that diagonalizes the matrix.

Find: (a) \(2 \times 2\) matrix \(A\) with eigenvalues \(\lambda_{1}=1\) and \(\lambda_{2}=-2\) and corresponding eigenvectors \(v_{1}=(1,2)\) and \(v_{2}=(3,7)\) (b) \(3 \times 3\) matrix \(A\) with eigenvalues \(\lambda_{1}=1, \lambda_{2}=2, \lambda_{2}=3\) and corresponding eigenvectors \(v_{1}=(1,0,1)\), \(v_{2}=(1,1,2), v_{3}=(1,2,4) .\) [Hint: See Problem 2.19.]

Let \(q(x, y)=x^{2}+6 x y-7 y^{2} .\) Find an orthogonal substitution that diagonalizes \(q\). Find the symmetric matrix \(A\) that represents \(q\) and its characteristic polynomial \(\Delta(t) .\) We have $$ A=\left[\begin{array}{rr} 1 & 3 \\ 3 & -7 \end{array}\right] \quad \text { and } \quad \Delta(t)=t^{2}+6 t-16=(t-2)(t+8) $$ The eigenvalues of \(A\) are \(\lambda=2\) and \(\lambda=-8\). Thus, using \(s\) and \(t\) as new variables, a diagonal form of \(q\) is $$ q(s, t)=2 s^{2}-8 t^{2} $$ The corresponding orthogonal substitution is obtained by finding an orthogonal set of eigenvectors of \(A\). (i) Subtract \(\lambda=2\) down the diagonal of \(A\) to obtain the matrix $$ M=\left[\begin{array}{rr} -1 & 3 \\ 3 & -9 \end{array}\right], \quad \text { corresponding to } \begin{array}{r} -x+3 y=0 \\ 3 x-9 y=0 \end{array} \text { or }-x+3 y=0 $$ A nonzero solution is \(u_{1}=(3,1)\) (ii) Subtract \(\lambda=-8\) (or add 8) down the diagonal of \(A\) to obtain the matrix $$ M=\left[\begin{array}{ll} 9 & 3 \\ 3 & 1 \end{array}\right], \quad \text { corresponding to } \quad \begin{aligned} &9 x+3 y=0 \\ &3 x+y=0 \end{aligned} \text { or } 3 x+y=0 $$ A nonzero solution is \(u_{2}=(-1,3)\). As expected, because \(A\) is symmetric, the eigenvectors \(u_{1}\) and \(u_{2}\) are orthogonal. Now normalize \(u_{1}\) and \(u_{2}\) to obtain, respectively, the unit vectors $$ \hat{u}_{1}=(3 / \sqrt{10}, 1 / \sqrt{10}) \quad \text { and } \quad \hat{u}_{2}=(-1 / \sqrt{10}, 3 / \sqrt{10}) $$ Finally, let \(P\) be the matrix whose columns are the unit vectors \(\hat{u}_{1}\) and \(\hat{u}_{2}\), respectively, and then \([x, y]^{T}=P[s, t]^{T}\) is the required orthogonal change of coordinates. That is, $$ \left.P=\mid \begin{array}{cc} 3 / \sqrt{10} & -1 / \sqrt{10} \\ 1 / \sqrt{10} & 3 / \sqrt{10} \end{array}\right] \quad \text { and } \quad x=\frac{3 s-t}{\sqrt{10}}, \quad y=\frac{s+3 t}{\sqrt{10}} $$ One can also express \(s\) and \(t\) in terms of \(x\) and \(y\) by using \(P^{-1}=P^{T} .\) That is, $$ s=\frac{3 x+y}{\sqrt{10}}, \quad t=\frac{-x+3 t}{\sqrt{10}} $$

Expeat Problem 9.14 for the matrix \(B=\left|\begin{array}{ccc}3 & -1 & 1 \\ 7 & -5 & 1 \\ 6 & -6 & 2\end{array}\right|\)

Suppose \(v\) is an eigenvector of a linear operator \(T\) belonging to the eigenvalue \(\lambda\). Prove (a) For \(n>0, v\) is an eigenvector of \(T^{n}\) belonging to \(\lambda^{n}\). (b) \(f(\lambda)\) is an eigenvalue of \(f(T)\) for any polynomial \(f(t)\)
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