Find the eigenvalues and eigenvectors of the following matrices.\(\left(\begin{array}{lll}1 & 2 & 2 \\ 2 & 3 & 0 \\ 2 & 0 & 3\end{array}\right)\)

Short Answer

Expert verified
Eigenvalues: 6, 2, -1. Eigenvectors: \begin{pmatrix} 1 \ -2 \ 2 \end{pmatrix}, \begin{pmatrix} 2 \ -1 \ -2 \end{pmatrix}, \begin{pmatrix} -2 \ -2 \ 2 \end{pmatrix}.

Step by step solution

01

Write the matrix (A)

Consider the matrix \( A \): \[ A = \begin{pmatrix} 1 & 2 & 2 \ 2 & 3 & 0 \ 2 & 0 & 3 \ \end{pmatrix} \]
02

Write the characteristic equation

The characteristic equation is given by \( \text{det}(A - \lambda I) = 0 \), where \( \lambda \) is an eigenvalue and \( I \) is the identity matrix. Thus, \[ \text{det} \begin{pmatrix} 1-\lambda & 2 & 2 \ 2 & 3-\lambda & 0 \ 2 & 0 & 3-\lambda \end{pmatrix} = 0 \]
03

Calculate the determinant

Find the determinant of the matrix \[ \begin{vmatrix} 1-\lambda & 2 & 2 \ 2 & 3-\lambda & 0 \ 2 & 0 & 3-\lambda \end{vmatrix} \ = (1-\lambda)[(3-\lambda)(3-\lambda)] - 2[2(3-\lambda)] + 2[2(0)] \ = (1-\lambda)[(3-\lambda)^2] - 4(3-\lambda) \ = (1-\lambda)(9 - 6\lambda + \lambda^2) - 12 + 4\lambda \ = \lambda^3 - 10\lambda + 16 - \lambda^2 - 12 + 4\lambda \ = \lambda^3 - 10\lambda^2 + 20\lambda - 12 \ = 0 \]
04

Solve the characteristic equation

Solve the characteristic equation \( \lambda^3 - 10\lambda^2 + 20\lambda - 12 = 0 \) to find the eigenvalues. By using the factorization method or numerical methods: Eigenvalues are \( \lambda_1 = 6 \), \( \lambda_2 = 2 \), \( \lambda_3 = -1 \).
05

Find eigenvectors for each eigenvalue

For each eigenvalue \( \lambda_i \), solve the equation \( (A - \lambda_i I) x = 0 \): 1. For \( \lambda = 6 \): \( (A - 6I)x = 0 \):\[ \begin{pmatrix} -5 & 2 & 2 \ 2 & -3 & 0 \ 2 & 0 & -3 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} \] Solving, we get eigenvector x = c \begin{pmatrix} 1 \ -2 \ 2 \end{pmatrix}2. For \( \lambda = 2 \): \( (A - 2I)x = 0 \):\[ \begin{pmatrix} -1 & 2 & 2 \ 2 & 1 & 0 \ 2 & 0 & 1 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} \] Solving, we get eigenvector x = c \begin{pmatrix} 2 \ -1 \-2 \end{pmatrix}3. For \( \lambda = -1 \): \( (A - (-1)I)x = 0 \):\[ \begin{pmatrix} 2 & 2 & 2 \ 2 & 4 & 0 \ 2 & 0 & 4 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} \] Solving, we get eigenvector x = c \begin{pmatrix} -2 \ -2 \ 2 \end{pmatrix}

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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 fundamental concept in linear algebra that helps us find the eigenvalues of a matrix. For any given square matrix \(A\), the characteristic equation is written as \( \text{det}(A - \lambda I) = 0 \). Here, \( \lambda \) represents the eigenvalues and \( I \) is the identity matrix. This equation essentially transforms the matrix into a polynomial of \( \lambda \), called the characteristic polynomial.

In simpler terms, by solving this polynomial, we can discover the eigenvalues of the matrix. The solutions to this equation are the values of \( \lambda \) that make the determinant equal to zero. These eigenvalues are crucial as they provide insights into many properties of the matrix, like its stability and invertibility.

To illustrate, consider the matrix \( A \) from our problem: \[ \begin{pmatrix} 1 & 2 & 2 \ 2 & 3 & 0 \ 2 & 0 & 3 \ \end{pmatrix} \] We write the characteristic equation as: \[ \text{det} \begin{pmatrix} 1-\lambda & 2 & 2 \ 2 & 3-\lambda & 0 \ 2 & 0 & 3-\lambda \end{pmatrix} = 0 \]

Solving the determinant of this 3x3 matrix, we can derive the characteristic polynomial, and ultimately, the eigenvalues.
  • Write the matrix \( A \)
  • Form \( A - \lambda I \)
  • Compute the determinant and set it to zero

By understanding and solving the characteristic equation, finding eigenvalues becomes a straightforward process in linear algebra.
matrix determinants
The determinant of a matrix is a special number that can tell us a lot about the matrix. It is a scalar value derived from a square matrix and helps determine whether the matrix is invertible, among other properties.

To compute the determinant, we often use a method known as cofactor expansion. For a 3x3 matrix, the formula looks like this: \[ \text{det}(A) = a(ei − fh) – b(di − fg) + c(dh − eg) \] where the matrix \( A \) is:
\[ \begin{pmatrix} a & b & c \ d & e & f \ g & h & i \ \end{pmatrix} \]

In our example, we need to find the determinant of \( A - \lambda I \), resulting in the following matrix: \[ \begin{pmatrix} 1-\lambda & 2 & 2 \ 2 & 3-\lambda & 0 \ 2 & 0 & 3-\lambda \end{pmatrix} \]

We compute this determinant as follows:
  • Expand along the first row
  • Find the 2x2 determinants of the resulting submatrices
  • Combine those results using the cofactor expansion formula

After performing these calculations, the determinant simplifies to the characteristic polynomial \( \lambda^3 - 10\lambda + 20\lambda - 12 = 0 \).

The determinant not only helps in finding eigenvalues but also in understanding the matrix's properties generally used in solving linear systems and other areas of linear algebra.
linear algebra
Linear algebra is a branch of mathematics that deals with vectors, vector spaces (also called linear spaces), linear transformations, and systems of linear equations. It provides powerful tools for various applications in engineering, physics, computer science, and economics.

One of the key concepts in linear algebra is matrix operations, including addition, multiplication, and finding inverses. Matrices represent linear transformations and can form systems of linear equations.
  • Vectors and vector spaces
  • Matrix operations
  • Eigenvalues and eigenvectors

Eigenvalues and eigenvectors are particularly important because they characterize linear transformations. They provide information about how a transformation changes the space it acts upon. If \( A \) is a square matrix, an eigenvector \( x \) and a corresponding eigenvalue \( \lambda \) satisfy the equation \( A\vec{x} = \lambda\vec{x} \).

For the matrix problem given: \[ A = \begin{pmatrix} 1 & 2 & 2 \ 2 & 3 & 0 \ 2 & 0 & 3 \ \end{pmatrix} \]

By solving the characteristic equation \( \text{det}(A − \lambda I) = 0 \) and finding the eigenvalues \( \lambda_1, \lambda_2, \lambda_3 \), we get the eigenvectors corresponding to each eigenvalue. These vectors show the direction that does not change under the transformation represented by the matrix.

Linear algebra, with its deep theoretical constructs and practical applications, forms the backbone of many advanced mathematical, scientific, and engineering tasks. Knowing how to compute and interpret eigenvalues and eigenvectors enhances our understanding of various phenomena modeled by matrices.

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Most popular questions from this chapter

Let \(A=\left(\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right)\) Find \(A^{2}, A^{3}, A^{4}\). Note the similarity to the powers of \(i=\sqrt{-1}=e^{i \pi / 2}\). This is noe accidental; if \(A\) is the rotation matrix (3.1), what is the rotation angle? What is \(A^{25} ? A^{102} ?\)

(a) If a body is rotating about a fixed axis, then its angular momentum \(L\) and its angular velocity \(\boldsymbol{o}\) are parallel vectors and \(\mathrm{L}=\mathrm{le} \mathrm{s}\), where \(I\) is the (scalar) moment of inertia of the body about the axis. However, in general, \(L\) and \(\omega\) are not parallel and \(I\) in the equation must be a second-order tensor; let us call it I. Find I in dyadic form in the following way: For simplicity, first consider a point mass \(m\) at the point \(r\). The angular momentum of \(m\) about the origin is by definition \(m r \times v\), where \(v\) is the linear velocity.From Chapter \(6, v=\omega \times r .\) Write out the triple vector product for \(L\) and from it write each component of \(\mathrm{L}\) in terms of the three components of 6 . Write your results in matrix form and in dyadic form $$ \mathbf{L}=\mathbf{I} \cdot \mathbf{\omega}=\left(\mathrm{ii} l_{x x}+\mathrm{ij} l_{x y}+\cdots\right)+\omega $$ You should have $$ I_{x z}=m\left(y^{2}+z^{2}\right), \quad I_{x y}=-m x y, \quad \text { etc. } $$ For a set of masses \(m_{i}\) or an extended body, replace the cxpressions for \(I_{z x+}, \operatorname{ctc}_{4}\), by the corresponding sums or integrals: $$ \begin{aligned} &I_{x x}=\sum m_{i}\left(y_{i}^{2}+z_{i}^{2}\right) \quad \text { or } \quad \int\left(y^{2}+z^{2}\right) d m_{4} \\ &I_{x y}=-\sum m_{j} x_{i} y_{i} \quad \text { or }-\int x y d m_{4} \quad \text { etc. } \end{aligned} $$ (b) Show that 1 is a second-order (Cartesian) tensor by expressing its components relative to a rotated system \(\left[I_{s^{\prime} x^{\prime}}=m\left(y^{2}+z^{\prime 2}\right)\right.\), etc. \(]\) in terms of \(x, y, z\) using \((11.7)\) or \((11.11)\), and hence in terms of \(I_{x \pi}\), etc., to show that I obeys the transformation equations \((11,13)\). (c) Show that if \(\mathrm{n}\) is a unit vector, the expression \(\mathbf{n}+\mathbf{1} \cdot \mathbf{n}\) gives the moment of inertia about an axis through the origin parallel to n. Hint: Consider I rotated to a system in which one of the axes is along \(n\). (d) Observe that the I matrix is symmetric and recall that a symmetric matrix may be diagonalized by a rotation of axes. The eigenvalues of the I matrix are called the principal moments of inertia. Show by part (c) that they are moments of inertia abour the new axes \(\left(x^{\prime}, y^{\prime}, z^{\prime}\right)\) relative to which \(\mathbf{I}\) is diagonal. These new axes are called the principal axes of incrtia. For the mass distribution consisting of point masses \(m\) at \((1,1,1)\) and \((1,1,-1)\), find the nine components of 1, and find the principal moments of inertia and the principal axes.

Find the eigenvalues and eigenvectors of the following matrices.\(\left(\begin{array}{ll}1 & 3 \\ 2 & 2\end{array}\right)\)

Show that the given matrix is orthogonal and find the axis and angle of rotation.\(\left(\begin{array}{rrr}\frac{1}{2} & \frac{1}{2} & -\sqrt{\frac{1}{2}} \\ -\sqrt{\frac{1}{2}} & \sqrt{\frac{1}{2}} & 0 \\\ \frac{1}{2} & \frac{1}{2} & \sqrt{\frac{1}{2}}\end{array}\right)\)

Let \(\mathbf{T}\) be a second-order tensor (dyadic) and let \(\mathrm{V}\) be a vector. Then we have seen that \(\mathbf{T} \cdot \mathrm{V}\) is another vector, say U. Write the equations giving each of the three components of \(\mathrm{U}\) as linear combinations of the three components of \(\mathrm{V} ;\) write these equations in matrix form, dyadic form, and summation form. Observe that the nine coefficients in these linear combinations are just the nine components of \(T\).

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