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Chapter 4: Problem 26

Find all (real) values of \(k\) for which \(A\) is diagonalizable. $$A=\left[\begin{array}{ll} k & 1 \\ 1 & 0 \end{array}\right]$$

Short Answer

Expert verified
The matrix \( A \) is diagonalizable for \( k > 2 \) or \( k < -2 \).

Step by step solution

01

Definition of Diagonalizability

A matrix is diagonalizable if it can be expressed in the form \( PDP^{-1} \), where \( D \) is a diagonal matrix, and \( P \) is an invertible matrix. A necessary and sufficient condition for diagonalizability is that the matrix has enough linearly independent eigenvectors to form a basis for the vector space.
02

Compute the Characteristic Polynomial

The characteristic polynomial is found by evaluating \( \det(A - \lambda I) \), where \( I \) is the identity matrix. For matrix \( A \), this is calculated as follows:\[A - \lambda I = \begin{pmatrix} k - \lambda & 1 \ 1 & -\lambda \end{pmatrix}\]The determinant is then:\[\det(A - \lambda I) = (k-\lambda)(-\lambda) - 1 = -\lambda^2 - k\lambda - 1\]
03

Solve the Characteristic Equation

Set the characteristic polynomial equal to zero and solve for \( \lambda \):\[-\lambda^2 - k\lambda - 1 = 0\]Rearranging gives:\[\lambda^2 + k\lambda + 1 = 0\]Use the discriminant \( \Delta = k^2 - 4 \times 1 \times 1 = k^2 - 4 \). A matrix is diagonalizable if it has distinct eigenvalues, which occur when \( \Delta > 0 \).
04

Find Conditions on k

For the quadratic to have distinct real roots, the discriminant must be positive \( \Delta > 0 \). This gives the inequality:\[k^2 - 4 > 0\]Solving this inequality:\[k^2 > 4 \k > 2 \quad \text{or} \quad k < -2\]
05

Verify Conditions

Verify the solution by considering a few values inside and outside the range. For \( k = 3 \):\[\lambda^2 + 3\lambda + 1 = 0 \Rightarrow \Delta = 9 - 4 = 5 > 0\]For \( k = 0 \):\[\lambda^2 + 1 = 0 \Rightarrow \Delta = 0 - 4 = -4 < 0\]Thus, \( k > 2 \) or \( k < -2 \) are valid solutions for diagonalizability.

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Key Concepts

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

Eigenvalues
Eigenvalues are crucial because they tell us about the behavior of linear transformations represented by matrices. In the context of diagonalizability, eigenvalues help determine if a matrix can be expressed in a simpler form, known as a diagonal matrix. An eigenvalue for a square matrix is a scalar \( \lambda \) such that there exists a non-zero vector \( \mathbf{v} \) (the eigenvector) satisfying the equation: \[ A\mathbf{v} = \lambda \mathbf{v}. \] This equation essentially states that applying the matrix \( A \) on \( \mathbf{v} \) is the same as scaling the vector by \( \lambda \).
  • To find eigenvalues, we solve the characteristic equation formed by setting \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix.
  • The roots of this equation are the eigenvalues of the matrix \( A \).
Characteristic Polynomial
The characteristic polynomial is a polynomial expression that is instrumental in identifying the eigenvalues of a matrix. Given a matrix \( A \), the characteristic polynomial is derived from the determinant of the matrix \( A - \lambda I \), where \( \lambda \) is a variable and \( I \) is the identity matrix.
  • For a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the characteristic polynomial is \( \lambda^2 - (a+d)\lambda + (ad - bc) \).
  • The roots of this polynomial equation, when set to zero, provide the eigenvalues of the matrix.
In the exercise example, the characteristic polynomial \( \lambda^2 + k\lambda + 1 \) was derived from the matrix \( A \), allowing us to examine the conditions under which the matrix is diagonalizable.
Discriminant
The discriminant of a quadratic equation, like that of the characteristic polynomial, is key to determining the number and nature of its roots. For a general quadratic equation in the form \( ax^2 + bx + c = 0 \), the discriminant is given by \( \Delta = b^2 - 4ac.\)
  • If \( \Delta > 0 \), there are two distinct real roots, indicating distinct eigenvalues for a matrix.
  • If \( \Delta = 0 \), there is exactly one real root (or a repeated root), leading to a lack of distinct eigenvalues.
  • If \( \Delta < 0 \), the roots are complex 鈥� not suitable for real diagonalizability conditions.
In diagonalizing matrices, a positive discriminant ensures that the matrix has distinct eigenvalues, which is crucial for diagonalizability as it ensures the availability of enough independent eigenvectors.
Linearly Independent Eigenvectors
Linearly independent eigenvectors are necessary for a matrix to be diagonalizable because they enable the formation of a basis for the vector space. If a matrix has \( n \) independent eigenvectors (where \( n \) is the size of the matrix), it can be transformed into a diagonal matrix.
  • When the characteristic polynomial has distinct roots, each eigenvalue leads to different eigenvectors.
  • These eigenvectors can be combined to form a matrix \( P \) in the diagonalization process \( PDP^{-1} \).
The diagonalization relies on whether these eigenvectors are linearly independent. If a matrix lacks this independence (i.e., not enough unique eigenvectors), it cannot be diagonalized, even if it has the correct number of eigenvalues.

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Most popular questions from this chapter

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.) $$\begin{array}{l} x^{\prime}=x+3 y, \quad x(0)=0 \\ y^{\prime}=2 x+2 y, \quad y(0)=5 \end{array}$$

\(P\) is the transition matrix of a regular Markov chain. Find the long range transition matrix \(L\) of \(P\). $$P=\left[\begin{array}{ccc} \frac{1}{2} & \frac{1}{3} & \frac{1}{6} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{3} \\ 0 & \frac{1}{6} & \frac{1}{2} \end{array}\right]$$

Exercise 32 in Section 4.3 demonstrates that every polynomial is (plus or minus) the characteristic polynomial of its own companion matrix. Therefore, the roots of a polynomial p are the eigenvalues of \(C(p)\). Hence, we can use the methods of this section to approximate the roots of any polynomial when exact results are not readily available. Apply the shifted inverse power method to the companion matrix \(C(p)\) of \(p\) to approximate the root of \(p\) closest to \(\alpha\) to three decimal places. $$p(x)=x^{2}-x-3, \alpha=2$$

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. $$A=\left[\begin{array}{rrr} 1 & -1 & 0 \\ 1 & 1 & 0 \\ 1 & -1 & 1 \end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]$$

Prove that the eigenvalues of \(A=\left[\begin{array}{cccc}0 & 1 & 0 & 0 \\ 2 & 5 & 0 & 0 \\ \frac{1}{2} & 0 & 3 & \frac{1}{2} \\ 0 & 0 & \frac{3}{4} & 7\end{array}\right]\) are all real, and locate each of these eigenvalues within a closed interval on the real line.
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