Math Problem Statement

1 3 3 1 4 3 -1 3 4 Find eigen values and vectors

Solution

To find the eigenvalues and eigenvectors of the matrix AA, follow these steps:

Given matrix:

A=(133143134)A = \begin{pmatrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ -1 & 3 & 4 \end{pmatrix}

Step 1: Find the Eigenvalues

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

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

Where II is the identity matrix and λ\lambda is the eigenvalue.

  1. Subtract λ\lambda from the diagonal elements of AA:
AλI=(1λ3314λ3134λ)A - \lambda I = \begin{pmatrix} 1-\lambda & 3 & 3 \\ 1 & 4-\lambda & 3 \\ -1 & 3 & 4-\lambda \end{pmatrix}
  1. Calculate the determinant:
det(AλI)=det(1λ3314λ3134λ)\text{det}(A - \lambda I) = \text{det} \begin{pmatrix} 1-\lambda & 3 & 3 \\ 1 & 4-\lambda & 3 \\ -1 & 3 & 4-\lambda \end{pmatrix}

Expanding this determinant:

(1λ)((4λ)(4λ)3×3)3(1(4λ)3(1))+3(1×31×(1))(1-\lambda) \left( (4-\lambda)(4-\lambda) - 3 \times 3 \right) - 3 \left( 1(4-\lambda) - 3(-1) \right) + 3 \left( 1 \times 3 - 1 \times (-1) \right)

Simplify and solve for λ\lambda.

Step 2: Find the Eigenvectors

For each eigenvalue λi\lambda_i, solve:

(AλiI)v=0(A - \lambda_i I) \mathbf{v} = 0

Where v\mathbf{v} is the eigenvector associated with λi\lambda_i.

Calculation:

Let me calculate the eigenvalues and eigenvectors.### Eigenvalues: The eigenvalues of the matrix AA are approximately:

λ1=1+2.95×108i,λ2=12.95×108i,λ3=7\lambda_1 = 1 + 2.95 \times 10^{-8}i, \quad \lambda_2 = 1 - 2.95 \times 10^{-8}i, \quad \lambda_3 = 7

Here, λ1\lambda_1 and λ2\lambda_2 are complex conjugates with very small imaginary parts, so they are almost real.

Eigenvectors:

The corresponding eigenvectors are:

For λ1=1+2.95×108i\lambda_1 = 1 + 2.95 \times 10^{-8}i:

v1(1.57×10162.09×108i0.70710.7071)\mathbf{v}_1 \approx \begin{pmatrix} -1.57 \times 10^{-16} - 2.09 \times 10^{-8}i \\ -0.7071 \\ 0.7071 \end{pmatrix}

For λ2=12.95×108i\lambda_2 = 1 - 2.95 \times 10^{-8}i:

v2(1.57×1016+2.09×108i0.70710.7071)\mathbf{v}_2 \approx \begin{pmatrix} -1.57 \times 10^{-16} + 2.09 \times 10^{-8}i \\ -0.7071 \\ 0.7071 \end{pmatrix}

For λ3=7\lambda_3 = 7:

v3(0.57210.66740.4767)\mathbf{v}_3 \approx \begin{pmatrix} 0.5721 \\ 0.6674 \\ 0.4767 \end{pmatrix}

These eigenvectors are normalized (each has a magnitude of 1).

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

Related Questions:

  1. What is the significance of eigenvalues in a matrix?
  2. How are eigenvectors used in applications such as Principal Component Analysis (PCA)?
  3. Can you explain how the characteristic polynomial is derived from a matrix?
  4. What are some real-world applications of eigenvalues and eigenvectors?
  5. How do complex eigenvalues affect the stability of a system?
  6. What is the geometric interpretation of eigenvectors and eigenvalues?
  7. How can you verify if a vector is an eigenvector of a matrix?
  8. What happens if a matrix has repeated eigenvalues?

Tip: Always ensure eigenvectors are normalized when using them in applications, as their direction is often more important than their magnitude.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Eigenvalues
Eigenvectors
Characteristic Equation
Determinant

Formulas

Characteristic equation
Eigenvalue calculation
Eigenvector calculation

Theorems

Spectral theorem

Suitable Grade Level

College Level