**Computing Eigenvalues and Bases of Eigenspaces**To determine the eigenvalues and bases of eigenspaces for a given matrix $ A $, follow these steps:Given matrix:\[A = \begin{bmatrix}-2 & 1 & 4 \\4 & 1 & -4 \\-1 & 1 & 3 \end{bmatrix}\]1. **Compute the Eigenvalues**: Eigenvalues ($\lambda$) of a matrix $ A $ are found by solving the characteristic equation: \[ \det(A - \lambda I) = 0 \] Here, $ I $ is the identity matrix of the same dimension as $ A $, and $ \det $ denotes the determinant.2. **Construct the Matrix $ A - \lambda I $**: \[ A - \lambda I = \begin{bmatrix} -2-\lambda & 1 & 4 \\ 4 & 1-\lambda & -4 \\ -1 & 1 & 3-\lambda \end{bmatrix} \]3. **Compute the Determinant of $ A - \lambda I $**: \[ \det(A - \lambda I) = \begin{vmatrix} -2-\lambda & 1 & 4 \\ 4 & 1-\lambda & -4 \\ -1 & 1 & 3-\lambda \end{vmatrix} \] Calculate this determinant to find the characteristic polynomial and set it equal to zero to solve for the eigenvalues.4. **Solve for the Eigenvalues**: Find the roots of the characteristic polynomial obtained from the determinant calculation. These roots are the eigenvalues $ \lambda_1, \lambda_2, \lambda_3 $.5. **Find the Eigenspaces**: For each eigenvalue $\lambda_i$, solve the system of linear equations: \[ (A - \lambda_i I)\mathbf{x} = \mathbf{0} \] where $\mathbf{x}$ is the eigenvector corresponding to $\lambda_i$.6. **Form the Basis for Each Eigenspace**: The nontrivial solutions to the above system give the **Problem Statement**Compute the eigenvalues and bases of eigenspaces of the following matrix:\[ A = \begin{bmatrix} -2 & -2 \\ 6 & 5 \end{bmatrix}. \]**Solution**To find the eigenvalues $\lambda$ of the matrix $A$, we need to solve the characteristic equation $\det(A - \lambda I) = 0$, where $I$ is the identity matrix of the same dimension as $A$.1. Construct the matrix $A - \lambda I$:\[ A - \lambda I = \begin{bmatrix} -2 - \lambda & -2 \\ 6 & 5 - \lambda \end{bmatrix}. \]2. Compute the determinant of the matrix $A - \lambda I$:\[ \det(A - \lambda I) = \begin{vmatrix} -2 - \lambda & -2 \\ 6 & 5 - \lambda \end{vmatrix} = (-2 - \lambda)(5 - \lambda) - (-2)(6).\]3. Expand the determinant:\[ (-2 - \lambda)(5 - \lambda) + 12 = \lambda^2 - 3\lambda - 10 + 12 = \lambda^2 - 3\lambda + 2.\]4. Set the determinant equal to zero:\[ \lambda^2 - 3\lambda + 2 = 0.\]5. Solve the quadratic equation for $\lambda$:\[ (\lambda - 1)(\lambda - 2) = 0.\]6. Therefore, the eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 2$.**Finding Eigenspaces**Next, we need to find the eigenvectors corresponding to each eigenvalue to determine the bases of the eigenspaces.1. For $\lambda_1 = 1$:\[ A - \lambda_1 I = \begin{bmatrix} -2 - 1 & -2 \\ 6 & 5 - 1 \end{bmatrix} = \begin{bmatrix} -3 & -2 \\ 6 & 4 \end{bmatrix}. \]

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Compute the eigenvalues and bases of eigenspaces of the following matrix -2 1 4 4 1 -4 -1 1 A = 3

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

**Computing Eigenvalues and Bases of Eigenspaces**

To determine the eigenvalues and bases of eigenspaces for a given matrix $ A $, follow these steps:

Given matrix:

\[
A = \begin{bmatrix}
-2 & 1 & 4 \\
4 & 1 & -4 \\
-1 & 1 & 3
\end{bmatrix}
\]

1. **Compute the Eigenvalues**:

Eigenvalues ($\lambda$) of a matrix $ A $ are found by solving the characteristic equation:

\[
\det(A - \lambda I) = 0
\]

Here, $ I $ is the identity matrix of the same dimension as $ A $, and $ \det $ denotes the determinant.

2. **Construct the Matrix $ A - \lambda I $**:

\[
A - \lambda I = \begin{bmatrix}
-2-\lambda & 1 & 4 \\
4 & 1-\lambda & -4 \\
-1 & 1 & 3-\lambda
\end{bmatrix}
\]

3. **Compute the Determinant of $ A - \lambda I $**:

\[
\det(A - \lambda I) = \begin{vmatrix}
-2-\lambda & 1 & 4 \\
4 & 1-\lambda & -4 \\
-1 & 1 & 3-\lambda
\end{vmatrix}
\]

Calculate this determinant to find the characteristic polynomial and set it equal to zero to solve for the eigenvalues.

4. **Solve for the Eigenvalues**:

Find the roots of the characteristic polynomial obtained from the determinant calculation. These roots are the eigenvalues $ \lambda_1, \lambda_2, \lambda_3 $.

5. **Find the Eigenspaces**:

For each eigenvalue $\lambda_i$, solve the system of linear equations:

\[
(A - \lambda_i I)\mathbf{x} = \mathbf{0}
\]

where $\mathbf{x}$ is the eigenvector corresponding to $\lambda_i$.

6. **Form the Basis for Each Eigenspace**:

The nontrivial solutions to the above system give the

**Problem Statement**

Compute the eigenvalues and bases of eigenspaces of the following matrix:

\[
A = \begin{bmatrix}
-2 & -2 \\
6 & 5
\end{bmatrix}.
\]

**Solution**

To find the eigenvalues $\lambda$ of the matrix $A$, we need to solve the characteristic equation $\det(A - \lambda I) = 0$, where $I$ is the identity matrix of the same dimension as $A$.

1. Construct the matrix $A - \lambda I$:
\[
A - \lambda I = \begin{bmatrix}
-2 - \lambda & -2 \\
6 & 5 - \lambda
\end{bmatrix}.
\]

2. Compute the determinant of the matrix $A - \lambda I$:

\[
\det(A - \lambda I) = \begin{vmatrix}
-2 - \lambda & -2 \\
6 & 5 - \lambda
\end{vmatrix} = (-2 - \lambda)(5 - \lambda) - (-2)(6).
\]

3. Expand the determinant:

\[
(-2 - \lambda)(5 - \lambda) + 12 = \lambda^2 - 3\lambda - 10 + 12 = \lambda^2 - 3\lambda + 2.
\]

4. Set the determinant equal to zero:

\[
\lambda^2 - 3\lambda + 2 = 0.
\]

5. Solve the quadratic equation for $\lambda$:

\[
(\lambda - 1)(\lambda - 2) = 0.
\]

6. Therefore, the eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 2$.

**Finding Eigenspaces**

Next, we need to find the eigenvectors corresponding to each eigenvalue to determine the bases of the eigenspaces.

1. For $\lambda_1 = 1$:

\[
A - \lambda_1 I = \begin{bmatrix}
-2 - 1 & -2 \\
6 & 5 - 1
\end{bmatrix} = \begin{bmatrix}
-3 & -2 \\
6 & 4
\end{bmatrix}.
\]

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