Chapter 11: Chapter 11

Problem 276

Show that the matrix $\mathrm{A}$ is not diagonalizable where $$ \mathrm{A}=\mid \begin{array}{cc} -3 & 2 \\ \mid-2 & 1 \mid \end{array} $$

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Matrix A is given as: $$ \mathrm{A}=\begin{pmatrix} -3 & 2 \\ -2 & 1 \end{pmatrix} $$ We find the characteristic equation: $$ \det(\mathrm{A} - λI) = λ^2 + 2λ + 1 = (λ + 1)^2 $$ Solving the characteristic equation, we find that there is only one eigenvalue, λ = -1. By calculating the eigenspace for λ = -1: $$ \begin{pmatrix} -2 & 2 \\ -2 & 2 \end{pmatrix} $$ We find that there is only one linearly independent eigenvector: $$ \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$ Since matrix A is a 2x2 matrix and there is only one linearly independent eigenvector instead of the necessary two, the matrix A is not diagonalizable.

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Step 1: Finding the characteristic equation

Calculate the determinant of the matrix (A - λI), where λ is an eigenvalue, and I is the identity matrix. $$ \det(\mathrm{A} - λI) = \det\begin{pmatrix} -3 - λ & 2 \\ -2 & 1 - λ \end{pmatrix} $$ Next, find the determinant of this matrix: $$ (-3 - λ)(1 - λ) - (-2)(2) = λ^2 + 2λ + 1 $$ This is our characteristic equation.

Step 2: Solve for the eigenvalues

Solve the characteristic equation for λ to find the eigenvalues of matrix A. $$ λ^2 + 2λ + 1 = (λ + 1)^2 = 0 $$ This equation has only one solution: λ = -1. Since there is only one eigenvalue, the matrix A will not have enough linearly independent eigenvectors to be diagonalizable. We will now find the eigenspace corresponding to the eigenvalue λ = -1 to confirm this.

Step 3: Finding the eigenspace for λ = -1

Substitute the eigenvalue λ = -1 back into (A - λI) and solve for the eigenvectors. $$ \begin{pmatrix} -3 - (-1) & 2 \\ -2 & 1 - (-1) \end{pmatrix} = \begin{pmatrix} -2 & 2 \\ -2 & 2 \end{pmatrix} $$ Now, solve the linear system formed by this matrix by applying Gaussian elimination or another method. Row 1: $$ -2X_1 + 2X_2 = 0 $$ Row 2: $$ -2X_1 + 2X_2 = 0 $$ This linear system has only one equation after simplification. When we solve for $X_1$, we get: $$ X_1 = X_2 $$ Which means there is only one linearly independent eigenvector for λ = -1: $$ \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$ Since the matrix A is a 2x2 matrix and there is only one linearly independent eigenvector instead of the necessary two, the matrix A is not diagonalizable.

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Chapter 11: Chapter 11

Show that the matrix \(\mathrm{A}\) is not diagonalizable where $$ \mathrm{A}=\mid \begin{array}{cc} -3 & 2 \\ \mid-2 & 1 \mid \end{array} $$

Short Answer

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Step 1: Finding the characteristic equation

Step 2: Solve for the eigenvalues

Step 3: Finding the eigenspace for λ = -1

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More chapters from the book ‘Linear Algebra’

Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 12

Chapter 13

Chapter 14

Chapter 15

Chapter 16

Chapter 17

Chapter 18

Chapter 19

Chapter 20

Chapter 21

Chapter 22

Chapter 23

Chapter 24

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