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Chapter 7: Problem 18

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

Short Answer

Expert verified
Based on the step-by-step solution provided, identify the eigenvalues and their corresponding eigenvectors. Eigenvalue 1: λ1 = 1 + i Eigenvector 1: v1 = [x1, x1] Eigenvalue 2: λ2 = 1 - i Eigenvector 2: v2 = [x2, -x2]

Step by step solution

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01

Find the Characteristic Equation of the Given Matrix

To find the characteristic equation of a 2x2 matrix, we will first find the determinant of the matrix formed by subtracting lambda from the main diagonal and then setting it equal to zero. $$\det(A - \lambda I) = \begin{vmatrix}1 - \lambda & i \\ -i & 1 - \lambda\end{vmatrix} = (1-\lambda)^2 - (-i)(i) = 0$$
02

Solve the Characteristic Equation for Lambda

Solve the characteristic equation from Step 1 to find the eigenvalues (lambda). $$(1-\lambda)^2 + 1 = 0$$ $$\lambda^2 - 2\lambda + 2 = 0$$ Now, we can solve the quadratic equation using the quadratic formula, \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2\), and \(c = 2\). $$\lambda = \frac{2 \pm \sqrt{(-2)^2 - 4(2)}}{2} = \frac{2 \pm \sqrt{-4}}{2}$$ So, we have two eigenvalues, \(\lambda_1 = 1 + i\) and \(\lambda_2 = 1 - i\).
03

Find the Eigenvectors for the Eigenvalues

Now that we have the eigenvalues, we will find the eigenvectors by solving the system of linear equations given by \((A - \lambda_i I) \cdot \vec{v_i} = 0\), where \(\lambda_i\) refers to the eigenvalues and \(\vec{v_i}\) refers to the eigenvectors. For \(\lambda_1 = 1 + i\), $$(A - (1 + i)I) \cdot \begin{bmatrix}x_1\\ y_1\end{bmatrix} = \begin{bmatrix}1 - (1 + i) & i \\ -i & 1 - (1 + i)\end{bmatrix} \begin{bmatrix}x_1\\ y_1\end{bmatrix} = 0$$ This reduces to: $$\begin{bmatrix}-i & i \\ -i & -i\end{bmatrix}\begin{bmatrix}x_1\\ y_1\end{bmatrix} = 0$$ We only need to solve one of the rows because the second row is a scalar multiple of the first row. So, solve \(-ix_1 + iy_1 = 0\), which results in \(y_1 = x_1\). Thus, the eigenvector corresponding to the eigenvalue \(1 + i\) is: $$\vec{v_1} = \begin{bmatrix}x_1\\ x_1\end{bmatrix},\, x_1 \ne 0$$ For \(\lambda_2 = 1 - i\), $$(A - (1 - i)I) \cdot \begin{bmatrix}x_2\\ y_2\end{bmatrix} = \begin{bmatrix}1 - (1 - i) & i \\ -i & 1 - (1 - i)\end{bmatrix} \begin{bmatrix}x_2\\ y_2\end{bmatrix} = 0$$ This reduces to: $$\begin{bmatrix}i & i \\ -i & i\end{bmatrix}\begin{bmatrix}x_2\\ y_2\end{bmatrix} = 0$$ We only need to solve one of the rows because the other row is also a scalar multiple of the first row. Solving \(ix_2 + iy_2 = 0\), which results in \(y_2 = -x_2\). Thus, the eigenvector corresponding to the eigenvalue \(1 - i\) is: $$\vec{v_2} = \begin{bmatrix}x_2\\ -x_2\end{bmatrix},\, x_2 \ne 0$$ So, the eigenvalues and their corresponding eigenvectors of the given matrix are \(\lambda_1 = 1 + i\) with \(\vec{v_1} = \begin{bmatrix}x_1\\ x_1\end{bmatrix}\) and \(\lambda_2 = 1 - i\) with \(\vec{v_2} = \begin{bmatrix}x_2\\ -x_2\end{bmatrix}\).

Key Concepts

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

Eigenvalues
Eigenvalues are crucial in understanding transformations represented by matrices. When you apply a matrix to a vector, the resulting vector can change in magnitude and/or direction. However, some special vectors only change in magnitude or just remain unchanged. These are associated with what we call eigenvalues.
In general, an eigenvalue \( \lambda \) gives us a scalar that tells how the magnitude of the vector changes when the matrix is applied. So, finding eigenvalues for a matrix \( A \) typically involves solving the characteristic equation, which leads to complex or real solutions, representing the eigenvalues.
For the matrix given in the exercise, the eigenvalues turned out to be complex numbers, \( 1 + i \) and \( 1 - i \). These numbers show how each vector eigenvalue is scaled during the transformation by the matrix.
Computing eigenvalues is essential for many applications, such as stability analysis and quantum mechanics.
Eigenvectors
Eigenvectors go hand-in-hand with eigenvalues as they describe the direction that is preserved during the transformation of a matrix. If you have an eigenvalue, there's always an accompanying eigenvector (or multiple, depending on the matrix), which points in the direction that remains unchanged, although it can be scaled.
For example, in our matrix, after finding the eigenvalues \( 1 + i \) and \( 1 - i \), we determine their respective eigenvectors that make the equation \((A - \lambda I) \cdot \vec{v} = 0\) true.
In this problem, corresponding to the eigenvalue \( 1 + i \), we have the eigenvector \( \begin{bmatrix} x_1 \ x_1 \end{bmatrix} \) where all the elements are equal. For the eigenvalue \( 1 - i \), the eigenvector is \( \begin{bmatrix} x_2 \ -x_2 \end{bmatrix} \), with one element being the negative of the other.
These eigenvectors show the directions that if matrices transform them, they will not be rotated, just scaled.
Characteristic Equation
The characteristic equation is a key to finding both eigenvalues and eigenvectors. It originates from the expression \( \det(A - \lambda I) = 0 \), where \( A \) is your matrix, \( \lambda \) is the eigenvalue, and \( I \) is the identity matrix.
This equation is derived from the requirement that multiplying a matrix by its eigenvector should yield a result that's a scalar multiple of that vector. Solving this equation involves finding for what values of \( \lambda \) this is true.
In the exercise, the characteristic equation is solved as follows: \((1-\lambda)^2 + 1 = 0\) transforms into \(\lambda^2 - 2\lambda + 2 = 0\), finding our eigenvalues via the quadratic formula.
The characteristic equation is a fascinating tool because it reveals these special scalars (eigenvalues) that describe how transformations affect their vectors.
Complex Numbers
Complex numbers are numbers that include a real part and an imaginary part. In many mathematical situations, including linear algebra and transformations, they play an essential role.
The imaginary unit \( i \) is the square root of \(-1\) and is crucial in forming complex numbers. A complex number is of the form \( a + bi \), where \( a \) and \( b \) are real numbers. In our problem, the eigenvalues \( 1 + i \) and \( 1 - i \) are both complex, showcasing how the matrix affects its vectors not just by scaling them, but sometimes through rotational properties.
Understanding complex numbers allows interpretations of transformations, especially in systems that exhibit oscillatory or wave-like behavior. They let us represent more intricate systems than simple real numbers might allow.

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Most popular questions from this chapter

Consider the system $$ \mathbf{x}^{\prime}=\mathbf{A x}=\left(\begin{array}{rrr}{5} & {-3} & {-2} \\\ {8} & {-5} & {-2} \\ {-4} & {-5} & {-4} \\ {-4} & {3} & {3}\end{array}\right) \mathbf{x} $$ (a) Show that \(r=1\) is a triple eigenvalue of the coefficient matrix \(\mathbf{A},\) and that there are only two linearly independent eigenvectors, which we may take as $$ \xi^{(1)}=\left(\begin{array}{l}{1} \\ {0} \\ {2}\end{array}\right), \quad \xi^{(2)}=\left(\begin{array}{r}{0} \\ {2} \\ {-3}\end{array}\right) $$ Find two linearly independent solutions \(\mathbf{x}^{(1)}(t)\) and \(\mathbf{x}^{(2)}(t)\) of Eq. (i). (b) To find a third solution assume that \(\mathbf{x}=\xi t e^{t}+\mathbf{\eta} e^{\lambda} ;\) thow that \(\xi\) and \(\eta\) must satisfy $$ (\mathbf{A}-\mathbf{1}) \xi=0 $$ \((\mathbf{A}-\mathbf{I}) \mathbf{\eta}=\mathbf{\xi}\) (c) Show that \(\xi=c_{1} \xi^{(1)}+c_{2} \mathbf{\xi}^{(2)},\) where \(c_{1}\) and \(c_{2}\) are arbitrary constants, is the most general solution of Eq. (iii). Show that in order to solve Eq. (iv) it is necessary that \(c_{1}=c_{2}\) (d) It is convenient to choose \(c_{1}=c_{2}=2 .\) For this choice show that $$ \xi=\left(\begin{array}{r}{2} \\ {4} \\ {-2}\end{array}\right), \quad \mathbf{\eta}=\left(\begin{array}{r}{0} \\ {0} \\ {-1}\end{array}\right) $$ where we have dropped the multiples of \(\xi^{(1)}\) and \(\xi^{(2)}\) that appear in \(\eta\). Use the results given in Eqs. (v) to find a third linearly independent solution \(\mathbf{x}^{(3)}\) of Eq. (i). (e) Write down a fundamental matrix \(\Psi(t)\) for the system (i). (f) Form a matrix T with the cigenvector \(\xi^{(1)}\) in the first column and with the eigenvector \(\xi\) and the generalized eigenvector \(\eta\) from Eqs. (v) in the other two columns. Find \(\mathbf{T}^{-1}\) and form the product \(\mathbf{J}=\mathbf{T}^{-1} \mathbf{A} \mathbf{T}\). The matrix \(\mathbf{J}\) is the Jordan form of \(\mathbf{A} .\)

Solve the given system of equations in each of Problems 20 through 23. Assume that \(t>0 .\) $$ r_{1}=-1, \quad \xi^{(1)}=\left(\begin{array}{c}{-1} \\\ {2}\end{array}\right): \quad r_{2}=-2, \quad \xi^{(2)}=\left(\begin{array}{c}{1} \\ {2}\end{array}\right) $$

In each of Problems 24 through 27 the eigenvalues and eigenvectors of a matrix \(\mathrm{A}\) are given. Consider the corresponding system \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\). $$ \begin{array}{l}{\text { (a) Sketch a phase portrait of the system. }} \\\ {\text { (b) Sketch the trajectory passing through the initial point }(2,3) \text { . }} \\ {\text { (c) For the trajectory in part (b) sketch the graphs of } x_{1} \text { versus } t \text { and of } x_{2} \text { versus } t \text { on the }} \\ {\text { same set of axes. }}\end{array} $$ $$ r_{1}=1, \quad \xi^{(1)}=\left(\begin{array}{l}{1} \\ {2}\end{array}\right) ; \quad r_{2}=2, \quad \xi^{(2)}=\left(\begin{array}{r}{1} \\\ {-2}\end{array}\right) $$

In each of Problems 23 and 24 ; (a) Find the eigenvalues of the given system. (b) Choose an initial point (other than the origin) and draw the corresponding trajectory in the \(x_{1} x_{2}\) -plane. Also draw the trajectories in the \(x_{1} x_{1}-\) and \(x_{2} x_{3}-\) planes. (c) For the initial point in part (b) draw the corresponding trajectory in \(x_{1} x_{2} x_{3}\) -space. $$ \mathbf{x}^{\prime}=\left(\begin{array}{ccc}{-\frac{1}{4}} & {1} & {0} \\\ {-1} & {-\frac{1}{4}} & {0} \\ {0} & {0} & {-\frac{1}{4}}\end{array}\right) \mathbf{x} $$

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{-\frac{5}{4}} & {\frac{3}{4}} \\\ {\frac{3}{4}} & {-\frac{5}{4}}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{2 t} \\ {e^{t}}\end{array}\right) $$
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