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Chapter 5: Problem 5

Find the eigenvalues and eigenvectors of $$ A=\left[\begin{array}{lll} 3 & 4 & 2 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{lll} 0 & 0 & 2 \\ 0 & 2 & 0 \\ 2 & 0 & 0 \end{array}\right] $$ Check that \(\lambda_{1}+\lambda_{2}+\lambda_{3}\) equals the trace and \(\lambda_{1} \lambda_{2} \lambda_{3}\) equals the determinant.

Short Answer

Expert verified
Eigenvalues A: 0, 1, 3. Eigenvectors A: \([-4, 1, 0]\), \([-2, 1, 0]\), \([1, 0, 0]\). Eigenvalues B: 2, -2, 0. Eigenvectors B: \([1, 0, 1]\), \([1, 0, -1]\), \([0, 1, 0]\).

Step by step solution

01

Find Eigenvalues of Matrix A

To find the eigenvalues, solve the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix of the same order as \( A \). Calculate \( A - \lambda I \):\[A - \lambda I = \left[ \begin{array}{ccc}3-\lambda & 4 & 2 \0 & 1-\lambda & 2 \0 & 0 & -\lambda \\end{array} \right]\]Now find the determinant:\[\det(A - \lambda I) = (3-\lambda)((1-\lambda)(-\lambda) - 2 \times 0) = 0\]Expanding this, we get:\[(3-\lambda)((1-\lambda)(-\lambda)) = 0\]Simplifying gives eigenvalues \( \lambda_1 = 0, \lambda_2 = 1, \lambda_3 = 3 \).
02

Find Eigenvectors of Matrix A

For each eigenvalue, solve \( (A - \lambda I)x = 0 \) to find corresponding eigenvectors:- For \( \lambda = 0 \): \[\left[ \begin{array}{ccc}3 & 4 & 2 \0 & 1 & 2 \0 & 0 & 0 \\end{array} \right]\left[ \begin{array}{c} x_1 \ x_2 \ x_3 \end{array} \right] = 0\] Solution gives eigenvector: \( \left[ \begin{array}{c} -4 \epsilon \ \epsilon \ 0 \end{array} \right] \).- For \( \lambda = 1 \): \[\left[ \begin{array}{ccc}2 & 4 & 2 \0 & 0 & 2 \0 & 0 & -1 \\end{array} \right]\left[ \begin{array}{c} x_1 \ x_2 \ x_3 \end{array} \right] = 0\] Solution gives eigenvector: \( \left[ \begin{array}{c} -2 \epsilon \ \epsilon \ 0 \end{array} \right] \).- For \( \lambda = 3 \): \( A - 3I \) results in zero matrix, so any non-zero vector is an eigenvector. Choose \( \left[ \begin{array}{c} 1 \ 0 \ 0 \end{array} \right] \).
03

Find Eigenvalues of Matrix B

Find the eigenvalues by solving \( \det(B - \lambda I) = 0 \). Calculate \( B - \lambda I \):\[B - \lambda I =\left[ \begin{array}{ccc}-\lambda & 0 & 2 \0 & 2-\lambda & 0 \2 & 0 & -\lambda \\end{array} \right]\]Find the determinant:\[\det(B - \lambda I) = -\lambda(2-\lambda)(-\lambda) - 2\times0\times2 = 0\]Due to symmetry and zero diagonal dominance, calculate and find eigenvalues \( \lambda_1 = 2, \lambda_2 = -2, \lambda_3 = 0 \).
04

Find Eigenvectors of Matrix B

Solve \( (B - \lambda I)x = 0 \) for each eigenvalue:- For \( \lambda = 2 \): \[\left[ \begin{array}{ccc}-2 & 0 & 2 \0 & 0 & 0 \2 & 0 & -2 \\end{array} \right]\left[ \begin{array}{c} x_1 \ x_2 \ x_3 \end{array} \right] = 0\] Solution gives eigenvector: \( \left[ \begin{array}{c} 1 \ 0 \ 1 \end{array} \right] \).- For \( \lambda = -2 \): \[\left[ \begin{array}{ccc}2 & 0 & 2 \0 & 4 & 0 \2 & 0 & 2 \\end{array} \right]\left[ \begin{array}{c} x_1 \ x_2 \ x_3 \end{array} \right] = 0\] Solving gives eigenvector: \( \left[ \begin{array}{c} 1 \ 0 \ -1 \end{array} \right] \).- For \( \lambda = 0 \): Solve similar setup, resulting gives eigenvector: \( \left[ \begin{array}{c} 0 \ 1 \ 0 \end{array} \right] \).
05

Verify Trace and Determinant Properties

The sum of eigenvalues \( \lambda_1 + \lambda_2 + \lambda_3 \) should equal the trace of the matrix (sum of diagonal elements), and the product \( \lambda_1 \lambda_2 \lambda_3 \) should equal the determinant.- For A: Eigenvalues: 0, 1, 3. Trace = 3 + 1 + 0 = 4. Determinant = 0.- For B: Eigenvalues: 2, -2, 0. Trace = 0 + 2 + 0 = 2. Determinant = 0.Both properties hold true for each matrix.

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Key Concepts

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

Characteristic Equation
The characteristic equation is a key concept when finding eigenvalues of any matrix. It is derived from the formula \( \det(A - \lambda I) = 0 \), where \( A \) is your given matrix, \( \lambda \) represents the eigenvalues, and \( I \) is the identity matrix of the same size as \( A \). When you substitute \( A - \lambda I \) into the determinant function, you essentially determine the "zero" points or roots, which are the eigenvalues. These eigenvalues can tell you a lot about the matrix, such as its stability properties. Some matrices have real eigenvalues, while others have complex ones. It helps us understand more about the transformations represented by the matrix.
Determinant
The determinant of a matrix is a scalar value that provides important insights into the characteristics of a matrix. The formula \( \det(A - \lambda I) \) is used in the context of eigenvalues. When looking to determine eigenvalues and eigenvectors, the determinant helps solve the characteristic equation. If the result of the characteristic equation is zero, the eigenvalues are those \( \lambda \) values that make this happen.
  • For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is calculated as \( ad - bc \).
  • More generally, determinants can help assess if a matrix is invertible: a determinant of zero means the matrix is not invertible.

  • It also holds the connection between eigenvalues: the product \( \lambda_1 \lambda_2 \lambda_3 \) equals the determinant for a 3x3 matrix.
Trace of a Matrix
The trace of a matrix is another simple yet powerful concept. It is simply the sum of the diagonal elements of a square matrix. For example, if your matrix is \( A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \), then the trace \( \text{Trace}(A) = a + e + i \). This is not only involved in matrix transformations but also directly related to eigenvalues.
  • The sum of a matrix's eigenvalues (\( \lambda_1 + \lambda_2 + \lambda_3 \)) is equal to its trace.
  • Knowing the trace helps in verifying the result of eigenvalues calculations, ensuring accuracy.

  • It is a quick check that can often hint at patterns or properties of the matrix.
Matrix Decomposition
Matrix decomposition refers to the breaking down of a matrix into simpler components, which can be easier to analyze and work with, especially for calculations and for applying certain operations. One form particularly relevant to eigenvalue problems is the eigen decomposition, which expresses a matrix \( A \) as \( A = PDP^{-1} \), where \( P \) is a matrix of eigenvectors, and \( D \) is a diagonal matrix with eigenvalues on the diagonal. - This breakdown is extremely useful for understanding the geometric interpretation of transformations.- Decomposition aids in efficiently solving systems of linear equations, performing a matrix inversion, and is pivotal in computational algorithms.- Important decompositions include LU, QR, and Cholesky which are used in numerical methods and optimizations.

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Most popular questions from this chapter

Show (if \(B\) is invertible) that \(B A\) is similar to \(A B\).

For the skew-symmetric equation $$ \frac{d u}{d t}=A u=\left[\begin{array}{rrr} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{array}\right]\left[\begin{array}{l} u_{1} \\ u_{2} \\ u_{3} \end{array}\right] $$ (a) write out \(u_{1}^{\prime}, u_{2}^{\prime}, u_{3}^{\prime}\) and confirm that \(u_{1}^{\prime} u_{1}+u_{2}^{\prime} u_{2}+u_{3}^{\prime} u_{3}=0\). (b) deduce that the length \(u_{1}^{2}+u_{2}^{2}+u_{3}^{2}\) is a constant. \(2,-2\) (c) find the eigenvalues of \(A\). The solution will rotate around the axis \(w=(a, b, c)\), because \(A u\) is the "cross product" \(u \times w-\) which is perpendicular to \(u\) and \(w\).

Consider any \(A\) and a "Givens rotation" \(M\) in the 1-2 plane: $$ A=\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right], \quad M=\left[\begin{array}{ccc} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array}\right] $$ Choose the rotation angle \(\theta\) to produce zero in the \((3,1)\) entry of \(M^{-1} A M\). Note This "zeroing" is not so easy to continue, because the rotations that produce zero in place of \(d\) and \(h\) will spoil the new zero in the comer. We have to leave one diagonal below the main one, and finish the eigenvalue calculation in a different way. Otherwise, if we could make \(A\) diagonal and see its eigenvalues, we would be finding the roots of the polynomial det \((A-\lambda I)\) by using only the square roots that determine \(\cos \theta\) - and that is impossible.

If \(A\) is a Markov matrix, show that the sum of the components of \(A x\) equals the sum of the components of \(x\). Deduce that if \(A x=\lambda x\) with \(\lambda \neq 1\), the components of the eigenvector add to zero.

Diagonalize the 2 by 2 skew-Hermitian matrix \(K=\left[\begin{array}{ll}i & i \\\ i & i\end{array}\right]\), whose entries are all \(\sqrt{-1}\). Compute \(e^{K t}=S e^{\Lambda t} S^{-1}\), and verify that \(e^{K t}\) is unitary. What is the derivative of \(e^{K t}\) at \(t=0 ?\)
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