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Chapter 4: Problem 29

Find all of the eigenvalues of the matrix A over the complex numbers \(\mathbb{C}\). Give bases for each of the corresponding eigenspaces. $$A=\left[\begin{array}{ll} 1 & i \\ i & 1 \end{array}\right]$$

Short Answer

Expert verified
Eigenvalues: \(1+i, 1-i\); Bases: \(\{[1, 1]\}\) and \(\{[i, -i]\}\).

Step by step solution

01

Find the Characteristic Polynomial of Matrix

First, calculate the characteristic polynomial of the matrix \(A\). The characteristic polynomial is given by \(det(A - \lambda I)\), where \(I\) is the identity matrix and \(\lambda\) is the eigenvalue.For the given matrix \(A\):\[A - \lambda I = \begin{bmatrix} 1-\lambda & i \ i & 1-\lambda \end{bmatrix}\]Compute the determinant:\[\det(A - \lambda I) = (1-\lambda)(1-\lambda) - (i)(i) = (1-\lambda)^2 + 1\] The characteristic polynomial is then:\[\lambda^2 - 2\lambda + 2\]
02

Solve the Characteristic Polynomial

To find the eigenvalues, solve the characteristic polynomial equation \(\lambda^2 - 2\lambda + 2 = 0\).This is a quadratic equation, and the roots can be found using the quadratic formula:\[\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = 1\), \(b = -2\), and \(c = 2\).Calculate:\[\lambda = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2}\]\(\sqrt{-4} = 2i\), thus:\[\lambda = \frac{2 \pm 2i}{2} = 1 \pm i\]The eigenvalues are \(\lambda_1 = 1+i\) and \(\lambda_2 = 1-i\).
03

Find the Eigenvectors for Each Eigenvalue

For \(\lambda_1 = 1+i\), solve \((A - \lambda_1 I)\mathbf{v} = 0\):\[\begin{bmatrix} 1 - (1+i) & i \ i & 1 - (1+i) \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = 0\]\[\begin{bmatrix} -i & i \ i & -i \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = 0\]This gives the equations:\[-ix + iy = 0\] \[ix - iy = 0\] Both reduce to \(ix = iy\) implying \(x = y\).Therefore, the eigenvector corresponding to \(\lambda_1\) is \(\begin{bmatrix} 1 \ 1 \end{bmatrix}\).For \(\lambda_2 = 1-i\), solve \((A - \lambda_2 I) \mathbf{v} = 0\): \[\begin{bmatrix} 1 - (1-i) & i \ i & 1 - (1-i) \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = 0\]\[\begin{bmatrix} i & i \ i & i \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = 0\]This results in:\[ix + iy = 0\] \[ix + iy = 0\]Again, implying \(x = -y\).Thus, the eigenvector corresponding to \(\lambda_2\) is \(\begin{bmatrix} i \ -i \end{bmatrix}\).
04

Provide Bases for Eigenspaces

The bases for each eigenspace are the sets of eigenvectors.- Eigenspace for \(\lambda = 1+i\) has a basis \(\left\{ \begin{bmatrix} 1 \ 1 \end{bmatrix} \right\}\).- Eigenspace for \(\lambda = 1-i\) has a basis \(\left\{ \begin{bmatrix} i \ -i \end{bmatrix} \right\}\)

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Key Concepts

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

Eigenspaces
An eigenspace is a set of all eigenvectors associated with a particular eigenvalue, along with the zero vector. To understand eigenspaces better, think of them as directions where certain transformations involving matrices simply "stretch" the vector or flip its direction. They don't change its direction in the way a normal transformation may do.
This is why finding a basis for an eigenspace is important. A basis provides a set of vectors that span the entire eigenspace.

For matrix \(A\) given in the exercise, the eigenspaces are derived from its eigenvectors:
  • The eigenvector \(\begin{bmatrix} 1 \ 1 \end{bmatrix}\) forms a basis for the eigenspace associated with eigenvalue \(1+i\).
  • The eigenvector \(\begin{bmatrix} i \ -i \end{bmatrix}\) forms a basis for the eigenspace associated with eigenvalue \(1-i\).
Understanding these can help students realize that even though complex numbers are at play, the concept of direction remains unchanged.
Complex Numbers
Complex numbers are numbers that have two parts: a real part and an imaginary part. They are expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, defined as \(\sqrt{-1}\).
Complex numbers come in handy especially in solving polynomial equations where real numbers alone wouldn鈥檛 be enough to provide solutions.

In our exercise, the eigenvalues are complex: \(1+i\) and \(1-i\).
  • The real part is \(1\).
  • The imaginary part is \(i\) or \(-i\), indicating rotation and scaling in two-dimensional space.
Working with complex numbers may feel intimidating at first, but they are quite natural in higher mathematics, providing valuable insights into symmetries and solving equations that have no real solutions.
Characteristic Polynomial
The characteristic polynomial is a fundamental concept in linear algebra, which is used to find the eigenvalues of a matrix. It is obtained by taking the determinant of the matrix \(A - \lambda I\), where \(A\) is the matrix in question, \(\lambda\) is an eigenvalue, and \(I\) is the identity matrix.
This polynomial provides a bridge between matrices and algebraic equations, allowing one to solve for eigenvalues as roots of the polynomial.

Let's break down the polynomial from our problem:
  • Given matrix \(A\), the characteristic polynomial \(\lambda^2 - 2\lambda + 2\) was derived by calculating \(det(A - \lambda I)\).
  • Solving this polynomial reveals the eigenvalues \(1+i\) and \(1-i\).
This mechanism demonstrates how matrices can be dissected into simpler structures by treating them like algebraic entities.
Eigenvectors
Eigenvectors are special vectors that, when a linear transformation is applied, do not change direction, though they may be scaled by some factor called the eigenvalue.
Eigenvectors provide critical insights into the geometric structure of transformations, and they are crucial for simplifying matrix operations in higher dimensions.

In our exercise:
  • For eigenvalue \(1+i\), the eigenvector \(\begin{bmatrix} 1 \ 1 \end{bmatrix}\) was derived, indicating a line of stretching or contraction parallel to itself.
  • For eigenvalue \(1-i\), the eigenvector \(\begin{bmatrix} i \ -i \end{bmatrix}\) was determined, again showcasing a consistent direction albeit in a complex plane.
Eigenvectors often appear with associated eigenvalues because together they provide a complete description of the action of a matrix on its spaces. Understanding them allows for deeper comprehension of linear transformations and their effects.

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Most popular questions from this chapter

(a) Give an example to show that if \(A\) can be partitioned as \\[A=\left[\begin{array}{ll}P & Q \\\R & S\end{array}\right]\\] where \(P, Q, R,\) and \(S\) are all square, then it is not necessarily true that \\[\operatorname{det} A=(\operatorname{det} P)(\operatorname{det} S)-(\operatorname{det} Q)(\operatorname{det} R)\\] (b) Assume that \(A\) is partitioned as in part (a) and that \(P\) is invertible. Let \\[B=\left[\begin{array}{cc}P^{-1} & O \\\\-R P^{-1} & I\end{array}\right]\\] Compute det \((B A)\) using Exercise 69 and use the result to show that \\[\operatorname{det} A=\operatorname{det} P \operatorname{det}\left(S-R P^{-1} Q\right)\\] [The matrix \(S-R P^{-1} Q\) is called the Schur complement of \(P\) in \(A\), after Issai Schur \((1875-1941)\) who was born in Belarus but spent most of his life in Germany. He is known mainly for his fundamental work on the representation theory of groups, but he also worked in number theory, analysis, and other areas. (c) Assume that \(A\) is partitioned as in part (a), that \(P\) is invertible, and that \(P R=R P\). Prove that \\[\operatorname{det} A=\operatorname{det}(P S-R Q)\\]

(a) Show that the companion matrix \(C(p)\) of \(p(x)=\) \(x^{2}+a x+b\) has characteristic polynomial \(\lambda^{2}+a \lambda+b\) (b) Show that if \(\lambda\) is an eigenvalue of the companion matrix \(C(p)\) in part (a), then \(\left[\begin{array}{l}\lambda \\ 1\end{array}\right]\) is an eigenvector of \(C(p)\) corresponding to \(\lambda\)

Find the Perron root and the corresponding Perron eigenvector of \(A\). $$A=\left[\begin{array}{ll} 1 & 3 \\ 2 & 0 \end{array}\right]$$

The power method does not converge to the dominant eigenvalue and eigenvector. Verify this, using the given initial vector \(\mathbf{x}_{0} .\) Compute the exact eigenvalues and eigenvectors and explain what is happening. $$A=\left[\begin{array}{rrr} 1 & -1 & 0 \\ 1 & 1 & 0 \\ 1 & -1 & 1 \end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]$$

Find the general solution to the given system of differential equations. Then find the specific solution that satisfies the initial conditions. (Consider all functions to be functions of t.) $$\begin{array}{l} y_{1}^{\prime}=y_{1}-y_{2}, \quad y_{1}(0)=1 \\ y_{2}^{\prime}=y_{1}+y_{2}, \quad y_{2}(0)=1 \end{array}$$
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