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Chapter 7: Problem 19

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

Short Answer

Expert verified
A: The eigenvalues can be found by solving the characteristic equation: 位鈧� = 0 and 位鈧� = 2. The corresponding eigenvectors can be found by solving the system of linear equations: For 位鈧� = 0, the eigenvector x鈧� = (1/2)(1, -鈭�3)岬� For 位鈧� = 2, the eigenvector x鈧� = (1/2)(1, 鈭�3)岬�

Step by step solution

01

Write down the given matrix

First, let's write down the given matrix A: $$ A=\left(\begin{array}{cc}{1} & {\sqrt{3}} \\\ {\sqrt{3}} & {-1}\end{array}\right) $$
02

Find the characteristic equation for the eigenvalues

The characteristic equation can be found by evaluating the determinant of \((A-\lambda I)\), denoted by \(|A-\lambda I|\), where \(\lambda\) is an eigenvalue, and \(I\) is the identity matrix. For given matrix A: $$ |A-\lambda I|=| \begin{pmatrix} 1-\lambda & \sqrt{3} \\ \sqrt{3} & -1 - \lambda \end{pmatrix} | = (1-\lambda)(-1-\lambda)-(\sqrt{3})^2 $$
03

Solve the characteristic equation

Next, let's solve the characteristic equation: $$ (1-\lambda)(-1-\lambda)-3=-(\lambda^2-2\lambda) $$ Solving the above quadratic equation, we get two eigenvalues: $$ \lambda_1=0, \: \lambda_2=2 $$
04

Find the eigenvectors for each eigenvalue

For each eigenvalue, let's find the eigenvectors by solving the following system of linear equations \((A-\lambda_i I)x=0\). For \(\lambda_1=0\), \((A-\lambda_1 I)x=0\) becomes: $$ A x = \left(\begin{array}{cc}{1} & {\sqrt{3}} \\\ {\sqrt{3}} & {-1}\end{array}\right) \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} =0 $$ Solving the above system, we get the eigenvector for \(\lambda_1\): $$ x_{1} = \frac{1}{2} \begin{pmatrix} 1 \\ -\sqrt{3} \end{pmatrix} $$ For \(\lambda_2=2\), \((A-\lambda_2 I)x=0\) becomes: $$ (A-\lambda_2 I)x = \left(\begin{array}{cc}{-1} & {\sqrt{3}} \\\ {\sqrt{3}} & {-3}\end{array}\right) \begin{pmatrix} x_3 \\ x_4 \end{pmatrix} =0 $$ Solving this system of equations, we get the eigenvector for \(\lambda_2\): $$ x_{2} = \frac{1}{2} \begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix} $$
05

State the final solution

The eigenvalues and corresponding eigenvectors for the given matrix are: $$ \lambda_{1} = 0, \: x_{1} = \frac{1}{2} \begin{pmatrix} 1 \\ -\sqrt{3} \end{pmatrix} $$ and $$ \lambda_{2} = 2, \: x_{2} = \frac{1}{2} \begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix} $$

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Key Concepts

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

Characteristic Equation
In the realm of linear algebra, the characteristic equation is a crucial tool used to find eigenvalues of a matrix. It involves calculating the determinant of the matrix \((A - \lambda I)\), where \(A\) is the given matrix, \(\lambda\) represents an eigenvalue, and \(I\) is the identity matrix of the same size. For the exercise at hand, the matrix \(A\) was:
  • \(\begin{pmatrix} 1 & \sqrt{3} \ \sqrt{3} & -1 \end{pmatrix}\)
To find the characteristic equation, we compute the determinant:\[|A - \lambda I| = \begin{vmatrix} 1-\lambda & \sqrt{3} \ \sqrt{3} & -1-\lambda \end{vmatrix}\]This results in the equation:\[(1-\lambda)(-1-\lambda) - 3\]Solving this quadratic equation gives the eigenvalues \(\lambda_1 = 0\) and \(\lambda_2 = 2\). The roots of this equation are key in understanding the behavior of the matrix in various transformations.
Matrix Algebra
Matrix algebra is a branch of mathematics dealing with matrices and operations like addition, multiplication, and finding the determinant. In our case, matrix algebra helps in deriving the characteristic equation and solving for eigenvectors. When the characteristic equation gives us eigenvalues, we use them in the expression \((A - \lambda I)x = 0\) to find eigenvectors. During the solution:
  • For \(\lambda_1 = 0\), solving \(Ax = 0\) led to the eigenvector \(\begin{pmatrix} 1/2 \ -\sqrt{3}/2 \end{pmatrix}\).
  • For \(\lambda_2 = 2\), it resulted in the eigenvector \(\begin{pmatrix} 1/2 \ \sqrt{3}/2 \end{pmatrix}\).
These eigenvectors provide insight into the directions along which the matrix transformation acts as simple scaling.
Linear Algebra
Linear algebra is the foundation of vector spaces and linear transformations. It allows us to study systems of linear equations, which is essential in finding eigenvalues and eigenvectors. In this context, understanding linear spaces, basis, and dimension plays a significant role. The aim is to find vectors \(x\) such that transforming them by the matrix \(A\) merely stretches or compresses them by a factor \(\lambda\), without changing their direction.
Eigenvalues express this factor, while eigenvectors represent the direction. This includes understanding:
  • The significance of linear transformations and how they can be decomposed into simpler actions using eigenvectors.
  • The connection between eigenvalues and matrix stability, important in numerous applications, including systems dynamics and machine learning.
Mastering the concepts of linear algebra opens doors to analyzing more complex matrices and transformations across various fields.

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Most popular questions from this chapter

A mass \(m\) on a spring with constant \(k\) satisfies the differential equation (see Section 3.8 ) \(m u^{\prime \prime}+k u=0\) where \(u(t)\) is the displacement at time \(t\) of the mass from its equilibrium position. (a) Let \(x_{1}=u\) and \(x_{2}=u^{\prime}\); show that the resulting system is \(\mathbf{x}^{\prime}=\left(\begin{array}{rr}{0} & {1} \\ {-k / m} & {0}\end{array}\right) \mathbf{x}\) (b) Find the eigenvalues of the matrix for the system in part (a). (c) Sketch several trajectories of the system. Choose one of your trajectories and sketch the corresponding graphs of \(x_{1}\) versus \(t\) and of \(x_{2}\) versus \(t\), Sketch both graphs on one set of axes. (d) What is the relation between the eigenvalues of the coefficient matrix and the natural frequency of the spring-mass system?

Verify that the given vector is the general solution of the corresponding homogeneous system, and then solve the non-homogeneous system. Assume that \(t>0 .\) $$ t \mathbf{x}^{\prime}=\left(\begin{array}{cc}{2} & {-1} \\ {3} & {-2}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{1-t^{2}} \\ {2 t}\end{array}\right), \quad \mathbf{x}^{(t)}=c_{1}\left(\begin{array}{c}{1} \\\ {1}\end{array}\right) t+c_{2}\left(\begin{array}{c}{1} \\\ {3}\end{array}\right) t^{-1} $$

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{-4} & {2} \\ {2} & {-1}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{t^{-1}} \\ {2 t^{-1}+4}\end{array}\right), \quad t>0 $$

In this problem we indicate how to show that \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\), as given by Eqs. (9), are linearly independent. Let \(r_{1}=\lambda+i \mu\) and \(\bar{r}_{1}=\lambda-i \mu\) be a pair of conjugate eigenvalues of the coefficient matrix \(\mathbf{A}\) of \(\mathrm{Fq}(1)\); let \(\xi^{(1)}=\mathbf{a}+i \mathbf{b}\) and \(\bar{\xi}^{(1)}=\mathbf{a}-i \mathbf{b}\) be the corresponding eigenvectors. Recall that it was stated in Section 7.3 that if \(r_{1} \neq \bar{r}_{1},\) then \(\boldsymbol{\xi}^{(1)}\) and \(\bar{\xi}^{(1)}\) are linearly independent. (a) First we show that a and b are linearly independent. Consider the equation \(c_{1} \mathrm{a}+\) \(c_{2} \mathrm{b}=0 .\) Express a and \(\mathrm{b}\) in terms of \(\xi^{(1)}\) and \(\bar{\xi}^{(1)},\) and then show that \(\left(c_{1}-i c_{2}\right) \xi^{(1)}+\) \(\left(c_{1}+i c_{2}\right) \bar{\xi}^{(1)}=0\) (b) Show that \(c_{1}-i c_{2}=0\) and \(c_{1}+i c_{2}=0\) and then that \(c_{1}=0\) and \(c_{2}=0 .\) Consequently, a and b are linearly independent. (c) To show that \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) are linearly independent consider the equation \(c_{1} \mathbf{u}\left(t_{0}\right)+\) \(c_{2} \mathbf{v}\left(t_{0}\right)=\mathbf{0}\), where \(t_{0}\) is an arbitrary point. Rewrite this equation in terms of a and \(\mathbf{b}\), and then proceed as in part (b) to show that \(c_{1}=0\) and \(c_{2}=0 .\) Hence \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) are linearly independent at the arbitrary point \(t_{0}\). Therefore they are linearly independent at every point and on every interval.

Consider a \(2 \times 2\) system \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\). If we assume that \(r_{1} \neq r_{2}\), the general solution is \(\mathbf{x}=c_{1} \xi^{(1)} e^{t_{1}^{\prime}}+c_{2} \xi^{(2)} e^{\prime 2},\) provided that \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent In this problem we establish the linear independence of \(\xi^{(1)}\) and \(\xi^{(2)}\) by assuming that they are linearly dependent, and then showing that this leads to a contradiction. $$ \begin{array}{l}{\text { (a) Note that } \xi \text { (i) satisfies the matrix equation }\left(\mathbf{A}-r_{1} \mathbf{I}\right) \xi^{(1)}=\mathbf{0} ; \text { similarly, note that }} \\ {\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(2)}=\mathbf{0}} \\ {\text { (b) Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(1)}=\left(r_{1}-r_{2}\right) \mathbf{\xi}^{(1)}} \\\ {\text { (c) Suppose that } \xi^{(1)} \text { and } \xi^{(2)} \text { are linearly dependent. Then } c_{1} \xi^{(1)}+c_{2} \xi^{(2)}=\mathbf{0} \text { and at least }}\end{array} $$ $$ \begin{array}{l}{\text { one of } c_{1} \text { and } c_{2} \text { is not zero; suppose that } c_{1} \neq 0 . \text { Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=\mathbf{0}} \\ {\text { and also show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=c_{1}\left(r_{1}-r_{2}\right) \boldsymbol{\xi}^{(1)} \text { . Hence } c_{1}=0, \text { which is }} \\\ {\text { a contradiction. Therefore } \xi^{(1)} \text { and } \boldsymbol{\xi}^{(2)} \text { are linearly independent. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Modify the argument of part (c) in case } c_{1} \text { is zero but } c_{2} \text { is not. }} \\ {\text { (e) Carry out a similar argument for the case in which the order } n \text { is equal to } 3 \text { ; note that }} \\ {\text { the procedure can be extended to cover an arbitrary value of } n .}\end{array} $$
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