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Chapter 5: Problem 9

Diagonalize the matrices in Exercises \(7-20,\) if possible. The eigenvalues for Exercises \(11-16\) are as follows: \((11) \lambda=1,2,3\) (12) \(\lambda=2,8 ;(13) \lambda=5,1 ;(14) \lambda=5,4 ;(15) \lambda=3,1 ;(16)\) \(\lambda=2,1 .\) For Exercise \(18,\) one eigenvalue is \(\lambda=5\) and one eigenvector is \((-2,1,2) .\) \(\left[\begin{array}{ll}{3} & {-1} \\ {1} & {5}\end{array}\right]\)

Short Answer

Expert verified
The matrix cannot be diagonalized because it has only one linearly independent eigenvector.

Step by step solution

01

Identify the Matrix

The given matrix is:\[A = \begin{bmatrix} 3 & -1 \ 1 & 5 \end{bmatrix}\]We need to determine whether this matrix can be diagonalized.
02

Find the Characteristic Polynomial

To find the eigenvalues, we first need the characteristic polynomial. This is given by solving the determinant \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix.
03

Compute the Determinant

Compute \( A - \lambda I \):\[A - \lambda I = \begin{bmatrix} 3 & -1 \ 1 & 5 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \ 0 & \lambda \end{bmatrix} = \begin{bmatrix} 3 - \lambda & -1 \ 1 & 5 - \lambda \end{bmatrix}\]Compute the determinant:\[\det(A - \lambda I) = (3 - \lambda)(5 - \lambda) - (-1)(1)\]\[= (3 - \lambda)(5 - \lambda) + 1 = \lambda^2 - 8\lambda + 16\]
04

Find Eigenvalues

Set the characteristic polynomial equal to zero and solve for \( \lambda \):\[\lambda^2 - 8\lambda + 16 = 0\]Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):\[\lambda = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} = \frac{8 \pm \sqrt{64 - 64}}{2} = 4\]There is only one eigenvalue, \( \lambda = 4 \), with multiplicity 2.
05

Check for Diagonalizability

A matrix is diagonalizable if the number of linearly independent eigenvectors is equal to the size of the matrix. For \( A \), being a 2x2 matrix, we need two linearly independent eigenvectors. However, with only one eigenvalue with algebraic multiplicity 2, we need to check if there are two linearly independent eigenvectors.
06

Solve for Eigenvectors

Solve \((A - 4I)\mathbf{x} = \mathbf{0}\):\[(A - 4I) = \begin{bmatrix} 3 - 4 & -1 \ 1 & 5 - 4 \end{bmatrix} = \begin{bmatrix} -1 & -1 \ 1 & 1 \end{bmatrix}\]Row-reduce:\[\begin{bmatrix} -1 & -1 \ 1 & 1 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 \ 0 & 0 \end{bmatrix}\]The solution to \((A - 4I)\mathbf{x} = \mathbf{0}\) gives the eigenvector \( \mathbf{x} = k \begin{bmatrix} 1 \ -1 \end{bmatrix} \). Only one linearly independent eigenvector exists.

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Key Concepts

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

Eigenvalues
When we dive into the realm of linear algebra, one of the key concepts we encounter is that of eigenvalues. For a given square matrix, eigenvalues are scalar values that provide significant insight into the matrix's behavior. To find these, we determine when the equation \( (A - \lambda I)\mathbf{x} = \mathbf{0} \) holds true, where \( A \) is the matrix, \( \lambda \) represents the eigenvalue, \( I \) is the identity matrix, and \( \mathbf{x} \) is the eigenvector.

The critical step involves solving the characteristic equation, \( \det(A - \lambda I) = 0 \). This equation allows us to find possible values of \( \lambda \). For the matrix \( A = \begin{bmatrix} 3 & -1 \ 1 & 5 \end{bmatrix} \), the eigenvalues were found to be \( \lambda = 4 \) with an algebraic multiplicity of 2. This means that although the equation suggests two potential solutions, they both turn out to be the same value.
  • Algebraic multiplicity refers to the number of times a particular eigenvalue appears.
  • Finding eigenvalues is crucial for understanding the matrix's scaling effect.
Eigenvectors
Once eigenvalues are found, we turn our attention to eigenvectors. These vectors maintain their direction under the transformation associated with a matrix. To determine these vectors, we solve \( (A - \lambda I)\mathbf{x} = \mathbf{0} \), where \( \lambda \) is a known eigenvalue.

For the matrix \( A = \begin{bmatrix} 3 & -1 \ 1 & 5 \end{bmatrix} \), with the eigenvalue \( \lambda = 4 \), we simplify the matrix subtraction to find eigenvectors. Simplifying and reducing leads to the equation forming a dependent system, yielding the eigenvector \( \mathbf{x} = k \begin{bmatrix} 1 \ -1 \end{bmatrix} \), where \( k \) is any scalar constant.
  • Eigenvectors correspond to distinct directional characteristics of the matrix.
  • Even if an eigenvalue repeats, the associated eigenvectors may not be fully independent.
Characteristic Polynomial
At the heart of finding eigenvalues is the characteristic polynomial. This polynomial is derived from the expression \( \det(A - \lambda I) \), where the determinant is set to zero to define the polynomial. This polynomial encapsulates the essence of the matrix's dependent behavior.

For our matrix \( A = \begin{bmatrix} 3 & -1 \ 1 & 5 \end{bmatrix} \), the characteristic polynomial is computed as \( \lambda^2 - 8\lambda + 16 \). By solving this quadratic equation, we obtain \( \lambda = 4 \) for both roots. Hence, the coefficients and powers of \( \lambda \) in the polynomial directly inform us about the matrix's eigenvalues.
  • The order of the polynomial (given here by the degree 2) suggests the number of eigenvalues.
  • Solving the polynomial can reveal complex eigenvalues for certain matrices.
Diagonalizable Matrix
A matrix is termed diagonalizable if it can be rewritten in a diagonal form by utilizing its eigenvalues and eigenvectors. Specifically, a matrix \( A \) is diagonalizable if there exists a matrix \( P \) such that \( P^{-1}AP = D \), where \( D \) is a diagonal matrix containing the eigenvalues of \( A \). Being diagonalizable simplifies computing matrix powers and exponentials.

For the matrix \( A = \begin{bmatrix} 3 & -1 \ 1 & 5 \end{bmatrix} \), to achieve diagonalization, we require sufficient linearly independent eigenvectors. Unfortunately, with a single eigenvalue, \( \lambda = 4 \), and only one linearly independent eigenvector, our matrix does not meet the criteria for diagonalizability.
  • A diagonalizable matrix simplifies many linear algebra operations.
  • The prerequisite is having a complete set of linearly independent eigenvectors.
  • Checking the number of independent eigenvectors against the matrix size is crucial.

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Most popular questions from this chapter

In Exercises \(9-18\) , construct the general solution of \(\mathbf{x}^{\prime}=A \mathbf{x}\) involving complex eigenfunctions and then obtain the general real solution. Describe the shapes of typical trajectories. $$ A=\left[\begin{array}{ll}{-2} & {1} \\ {-8} & {2}\end{array}\right] $$

In Exercises \(13-16,\) define \(T : \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) by \(T(\mathbf{x})=A \mathbf{x}\) . Find a basis \(\mathcal{B}\) for \(\mathbb{R}^{2}\) with the property that \([T]_{\mathcal{B}}\) is diagonal. $$ A=\left[\begin{array}{rr}{2} & {-6} \\ {-1} & {3}\end{array}\right] $$

A herd of American buffalo (bison) can be modeled by a stage matrix similar to that for the spotted owls. The females can be divided into calves (up to 1 year old), yearlings (1 to 2 years), and adults. Suppose an average of 42 female calves are born each year per 100 adult females. (Only adults produce offspring. Each year, about 60\(\%\) of the calves survive, 75\(\%\) of the yearlings survive, and 95\(\%\) of the adults survive. For \(k \geq 0,\) let \(\mathbf{x}_{k}=\left(c_{k}, y_{k}, a_{k}\right),\) where the entries in \(\mathbf{x}_{k}\) are the numbers of females in each life stage at year \(k .\) a. Construct the stage-matrix \(A\) for the buffalo herd, such that \(\mathbf{x}_{k+1}=A \mathbf{x}_{k}\) for \(k \geq 0\) b. [M] Show that the buffalo herd is growing, determine the expected growth rate after many years, and give the expected numbers of calves and yearlings present per 100 adults.

Define \(T : \mathbb{P}_{3} \rightarrow \mathbb{R}^{4}\) by \(T(\mathbf{p})=\left[\begin{array}{c}{\mathbf{p}(-3)} \\ {\mathbf{p}(-1)} \\\ {\mathbf{p}(1)} \\ {\mathbf{p}(3)}\end{array}\right]\) a. Show that \(T\) is a linear transformation. b. Find the matrix for \(T\) relative to the basis \(\left\\{1, t, t^{2}, t^{3}\right\\}\) for \(\mathbb{P}_{3}\) and the standard basis for \(\mathbb{R}^{4} .\)

Show that if \(A\) is both diagonalizable and invertible, then so is \(A^{-1} .\)
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