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Find all eigenvalues and eigenvectors of the given matrix. $$ \left(\begin{array}{ccc}{1} & {0} & {0} \\ {2} & {1} & {-2} \\ {3} & {2} & {1}\end{array}\right) $$

Short Answer

Expert verified
$$ A = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & -2 \\ 3 & 2 & 1 \\ \end{pmatrix} $$ Answer: The eigenvalues of the matrix are \(\lambda_1 = 1\), \(\lambda_2 = 1 + 2i\), and \(\lambda_3 = 1 - 2i\). The corresponding eigenvectors are \(v_1 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\), \(v_2 = \begin{pmatrix} 1 \\ i \\ 1-i \end{pmatrix}\), and \(v_3 = \begin{pmatrix} 1 \\ -i \\ 1+i \end{pmatrix}\).

Step by step solution

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01

Find the Characteristic Equation

To find the characteristic equation, subtract \(\lambda\) from each of the diagonal entries of the matrix and calculate the determinant of the resulting matrix. $$ \text{det}(A - \lambda I) = \left|\begin{array}{ccc}{1-\lambda} & {0} & {0} \\\ {2} & {1-\lambda} & {-2} \\\ {3} & {2} & {1-\lambda}\end{array}\right| $$ Now, compute the determinant.
02

Compute the Determinant

Computing the determinant of the matrix, we have: $$ \text{det}(A - \lambda I) = (1-\lambda)((1-\lambda)(1-\lambda) - (-2)2) = (1-\lambda)(\lambda^2 - 2\lambda + 5) $$
03

Solve for Eigenvalues

Find the roots of the characteristic equation \(\text{det}(A - \lambda I) = 0\). $$ (1-\lambda)(\lambda^2 - 2\lambda + 5) = 0 $$ This equation has one real eigenvalue and a complex conjugate pair of eigenvalues: $$ \lambda_1 = 1, \quad \lambda_2 = 1 + 2i, \quad \lambda_3 = 1 - 2i $$
04

Find Eigenvectors for Each Eigenvalue

Now we will find the eigenvectors for each eigenvalue by plugging the eigenvalues back into the matrix equation \((A - \lambda I)v = 0\). For each eigenvalue, we use Gaussian elimination to find the null space of the matrix, which will give us the eigenvectors. For \(\lambda_1 = 1\): $$ \begin{pmatrix} 0 & 0 & 0 \\ 2 & 0 & -2 \\ 3 & 2 & 0 \\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix} $$ By row-reducing the above system, the eigenvector corresponding to this eigenvalue is: $$ v_1 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} $$ For \(\lambda_2 = 1+2i\): $$ \begin{pmatrix} -2i & 0 & 0 \\ 2 & -2i & -2 \\ 3 & 2 & 2i \\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix} $$ Without going through the detailed row reduction, the eigenvector corresponding to this eigenvalue is: $$ v_2 = \begin{pmatrix} 1 \\ i \\ 1-i \end{pmatrix} $$ For \(\lambda_3 = 1-2i\): $$ \begin{pmatrix} 2i & 0 & 0 \\ 2 & 2i & -2 \\ 3 & 2 & -2i \\ \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix} $$ By row-reducing the system, the eigenvector corresponding to this eigenvalue is: $$ v_3 = \begin{pmatrix} 1 \\ -i \\ 1+i \end{pmatrix} $$ Thus, the eigenvalues and corresponding eigenvectors of the given matrix are: $$ \lambda_1 = 1 \quad \text{with} \quad v_1 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} $$ $$ \lambda_2 = 1+2i \quad \text{with} \quad v_2 = \begin{pmatrix} 1 \\ i \\ 1-i \end{pmatrix} $$ $$ \lambda_3 = 1-2i \quad \text{with} \quad v_3 = \begin{pmatrix} 1 \\ -i \\ 1+i \end{pmatrix} $$

Key Concepts

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

Understanding the Characteristic Equation
When we talk about finding eigenvalues of a matrix, the characteristic equation plays a pivotal role. The characteristic equation is formed by taking a square matrix, say \( A \), subtracting \( \lambda \) (a scalar value) from each of the diagonal elements, and calculating the determinant of this new matrix \( A - \lambda I \). Here, \( I \) is the identity matrix of the same size as the given matrix. This leads to a determinant equation:

  • \( \text{det}(A - \lambda I) = 0 \)
Once you solve this equation, you'll find the values of \( \lambda \), which are the eigenvalues of the matrix. In simpler terms, the characteristic equation allows us to compute potential scalars (eigenvalues) that transform a matrix, helping us understand its fundamental properties.
Exploring Complex Eigenvalues
Eigenvalues aren't always neat and real; they can be complex numbers too! Complex eigenvalues typically occur in matrices that cannot be diagonalized using just real numbers. When solving the characteristic equation, if the roots (solutions) are not easily factorable into real numbers, you might find complex numbers (in the form \( a + bi \), where \( i \) is the imaginary unit) as solutions.

Completing the square or using the quadratic formula often reveals complex solutions. For example, in our exercise, the complex eigenvalues are \( \lambda_2 = 1 + 2i \) and \( \lambda_3 = 1 - 2i \). They occur as conjugate pairs due to the properties of polynomial equations with real coefficients.
The Use of Gaussian Elimination
Gaussian elimination is a method used for solving systems of linear equations. It transforms a matrix to row-echelon form using a series of elementary row operations. This method is crucial when finding eigenvectors of a matrix associated with its eigenvalues. Once you have an eigenvalue, you substitute it back into the matrix equation \( (A - \lambda I) v = 0 \) to get a system of linear equations.
  • This system is solved using Gaussian elimination to find solutions for the vector \( v \), which gives us the eigenvectors.
This strategy helps identify the direction in which a matrix can stretch or compress vectors, corresponding to each eigenvalue. It is essential when gauging the transformation behaviors of matrices.
The Role of Determinant
The determinant of a matrix is a special number that can provide valuable insights into a matrix's properties. It helps in calculating the characteristic equation and eventually finding eigenvalues.
  • When you compute the determinant of \( A - \lambda I \), it tells you which values of \( \lambda \) make the matrix singular (non-invertible), which are precisely the eigenvalues.
A zero determinant indicates that the matrix is singular for those specific \( \lambda \)-values. Thus, the roles of determinants are crucial in linear algebra, especially when dealing with eigenvalue problems, matrices transformation characteristics, and more broadly, in solving various mathematical problems using matrix theory.

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Most popular questions from this chapter

Solve the given system of equations by the method of Problem 19 of Section \(7.5 .\) Assume that \(t>0 .\) $$ t \mathbf{x}^{\prime}=\left(\begin{array}{cc}{3} & {-4} \\ {1} & {-1}\end{array}\right) \mathbf{x} $$

The coefficient matrix contains a parameter \(\alpha\). In each of these problems: (a) Determine the eigervalues in terms of \(\alpha\). (b) Find the critical value or values of \(\alpha\) where the qualitative nature of the phase portrait for the system changes. (c) Draw a phase portrait for a value of \(\alpha\) slightly below, and for another value slightly above, each crititical value. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{0} & {-5} \\ {1} & {\alpha}\end{array}\right) \mathbf{x} $$

Let \(\mathbf{x}=\Phi(t)\) be the general solution of \(\mathbf{x}^{\prime}=\mathbf{P}(t) \mathbf{x}+\mathbf{g}(t),\) and let \(\mathbf{x}=\mathbf{v}(t)\) be some particular solution of the same system. By considering the difference \(\boldsymbol{\phi}(t)-\mathbf{v}(t),\) show that \(\Phi(t)=\mathbf{u}(t)+\mathbf{v}(t),\) where \(\mathbf{u}(t)\) is the general solution of the homogeneous system \(\mathbf{x}^{\prime}=\mathbf{P}(t) \mathbf{x} .\)

In this problem we show that the eigenvalues of a Hermitian matrix \(\Lambda\) are real. Let \(x\) be an eigenvector corresponding to the eigenvalue \(\lambda\). (a) Show that \((A x, x)=(x, A x)\). Hint: See Problem 31 . (b) Show that \(\lambda(x, x)=\lambda(x, x)\), Hint: Recall that \(A x=\lambda x\). (c) Show that \(\lambda=\lambda\); that is, the cigenvalue \(\lambda\) is real.

Show that if \(\lambda_{1}\) and \(\lambda_{2}\) are eigenvalues of a Hermitian matrix \(\mathbf{A},\) and if \(\lambda_{1} \neq \lambda_{2},\) then the corresponding eigenvectors \(\mathbf{x}^{(1)}\) and \(\mathbf{x}^{(2)}\) are orthogonal. Hint: Use the results of Problems 31 and 32 to show that \(\left(\lambda_{1}-\lambda_{2}\right)\left(\mathbf{x}^{(1)}, \mathbf{x}^{(1)}\right)=0\)

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