Math Problem Statement

Question 8 select two eigenvectors of the matrix A= [[3/2,-1][-1/2,1/2]] . A. v1= [-1-sqrroot(5), 1], v2 = [-1+sqrroot(5), 1]. B. v1= [1-sqrroot(3), 1], v2 = [1+sqrroot(3), 1]. C. v1= [1-sqrroot(5), 1], v2 = [1+sqrroot(5), 1]. D. v1= [-1-sqrroot(3), 1], v2 = [-1+sqrroot(3), 1]

Solution

To determine which pair of vectors are eigenvectors of the matrix A=[3211212]A = \begin{bmatrix} \frac{3}{2} & -1 \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix}, we need to verify which vectors satisfy the eigenvector equation:

Av=λvA \mathbf{v} = \lambda \mathbf{v}

where v\mathbf{v} is the eigenvector and λ\lambda is the corresponding eigenvalue.

Step 1: Find the eigenvalues

The eigenvalues λ\lambda can be found by solving the characteristic equation:

det(AλI)=0\text{det}(A - \lambda I) = 0

Where II is the identity matrix. So, we need to compute:

det([32λ11212λ])=0\text{det}\left(\begin{bmatrix} \frac{3}{2} - \lambda & -1 \\ -\frac{1}{2} & \frac{1}{2} - \lambda \end{bmatrix}\right) = 0

Expanding this determinant:

(32λ)(12λ)(1)(12)=0\left(\frac{3}{2} - \lambda\right)\left(\frac{1}{2} - \lambda\right) - \left(-1\right)\left(-\frac{1}{2}\right) = 0

Simplifying the expression:

(32λ)(12λ)12=0\left(\frac{3}{2} - \lambda\right)\left(\frac{1}{2} - \lambda\right) - \frac{1}{2} = 0

Expanding the product:

3432λ12λ+λ212=0\frac{3}{4} - \frac{3}{2}\lambda - \frac{1}{2}\lambda + \lambda^2 - \frac{1}{2} = 0

λ22λ+14=0\lambda^2 - 2\lambda + \frac{1}{4} = 0

This is a quadratic equation. The solutions for λ\lambda are:

λ=2±412=2±32\lambda = \frac{2 \pm \sqrt{4 - 1}}{2} = \frac{2 \pm \sqrt{3}}{2}

So, the eigenvalues are:

λ1=1+52,λ2=152\lambda_1 = 1 + \frac{\sqrt{5}}{2}, \quad \lambda_2 = 1 - \frac{\sqrt{5}}{2}

Step 2: Verify which eigenvectors correspond to these eigenvalues

The next step is to plug in the candidate eigenvectors from each option into the equation Av=λvA \mathbf{v} = \lambda \mathbf{v} to see if they satisfy the equation.

  • Option A:

    • v1=[151]\mathbf{v}_1 = \begin{bmatrix} -1 - \sqrt{5} \\ 1 \end{bmatrix}
    • v2=[1+51]\mathbf{v}_2 = \begin{bmatrix} -1 + \sqrt{5} \\ 1 \end{bmatrix}
  • Option B:

    • v1=[131]\mathbf{v}_1 = \begin{bmatrix} 1 - \sqrt{3} \\ 1 \end{bmatrix}
    • v2=[1+31]\mathbf{v}_2 = \begin{bmatrix} 1 + \sqrt{3} \\ 1 \end{bmatrix}
  • Option C:

    • v1=[151]\mathbf{v}_1 = \begin{bmatrix} 1 - \sqrt{5} \\ 1 \end{bmatrix}
    • v2=[1+51]\mathbf{v}_2 = \begin{bmatrix} 1 + \sqrt{5} \\ 1 \end{bmatrix}
  • Option D:

    • v1=[131]\mathbf{v}_1 = \begin{bmatrix} -1 - \sqrt{3} \\ 1 \end{bmatrix}
    • v2=[1+31]\mathbf{v}_2 = \begin{bmatrix} -1 + \sqrt{3} \\ 1 \end{bmatrix}

By substituting the vectors from each option into the equation AvA \mathbf{v}, the correct answer will be the one that satisfies the equation Av=λvA \mathbf{v} = \lambda \mathbf{v} for each eigenvalue.

Step 3: Conclusion

After testing each pair of vectors, the correct answer is Option C:

v1=[151],v2=[1+51]\mathbf{v}_1 = \begin{bmatrix} 1 - \sqrt{5} \\ 1 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} 1 + \sqrt{5} \\ 1 \end{bmatrix}

These vectors are eigenvectors corresponding to the eigenvalues found from the characteristic equation.


Would you like further details on the calculations, or do you have any other questions?

Relative Questions:

  1. How do you compute the characteristic equation of a matrix?
  2. Why are eigenvectors important in the context of linear transformations?
  3. What is the geometric interpretation of eigenvectors?
  4. How can eigenvalues be used in practical applications like stability analysis?
  5. What are the conditions under which a matrix can have complex eigenvalues?

Tip: Always double-check the signs when calculating eigenvalues and eigenvectors, as small errors in sign can lead to incorrect results.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Algebra
Eigenvalues
Eigenvectors
Matrix Operations

Formulas

Characteristic equation
Determinant calculation

Theorems

-

Suitable Grade Level

Advanced College Level