what are linearly independent eigenvectors example

Linearly independent eigenvectors are a set of eigenvectors corresponding to different eigenvalues or the same eigenvalue of a matrix, where no eigenvector in the set can be expressed as a linear combination of the others. This property is crucial in many areas of linear algebra and its applications, such as diagonalization, solving systems of differential equations, and more.

### Example:

Consider the matrix \( A \):

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

To find the eigenvectors of \( A \), we first find the eigenvalues by solving the characteristic equation:

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

where \( I \) is the identity matrix and \( \lambda \) represents the eigenvalues.

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

The determinant of this matrix is:

\[ \text{det}(A - \lambda I) = (4 - \lambda)(3 - \lambda) - (2)(1) \]

\[ = 12 - 4\lambda - 3\lambda + \lambda^2 - 2 \]

\[ = \lambda^2 - 7\lambda + 10 \]

\[ = (\lambda - 5)(\lambda - 2) \]

So, the eigenvalues are:

\[ \lambda_1 = 5, \quad \lambda_2 = 2 \]

Next, we find the eigenvectors for each eigenvalue.

#### For \( \lambda_1 = 5 \):

Solve \( (A - 5I) \mathbf{v} = 0 \):

\[ \begin{pmatrix} 4 - 5 & 1 \\ 2 & 3 - 5 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \]

This simplifies to the system of equations:

\[ -v_1 + v_2 = 0 \]
\[ 2v_1 - 2v_2 = 0 \]

From the first equation, \( v_1 = v_2 \). An eigenvector for \( \lambda_1 = 5 \) is:

\[ \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \]

#### For \( \lambda_2 = 2 \):

Solve \( (A - 2I) \mathbf{v} = 0 \):

\[ \begin{pmatrix} 4 - 2 & 1 \\ 2 & 3 - 2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \]

This simplifies to the system of equations:

\[ 2v_1 + v_2 = 0 \]
\[ 2v_1 + v_2 = 0 \]

From this equation, \( v_2 = -2v_1 \). An eigenvector for \( \lambda_2 = 2 \) is:

\[ \mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix} \]

### Summary:

The matrix \( A \) has two eigenvalues \( \lambda_1 = 5 \) and \( \lambda_2 = 2 \), with corresponding eigenvectors:

\[ \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \quad \text{and} \quad \mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix} \]

These eigenvectors are linearly independent, as they cannot be expressed as scalar multiples of each other. Therefore, they form a basis for the eigenspace associated with their respective eigenvalues.

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### Relative Questions:

1. How do you determine if a set of vectors is linearly independent?
2. Can a matrix have more eigenvectors than its size?
3. What is the significance of eigenvectors in solving differential equations?
4. How do you diagonalize a matrix using its eigenvalues and eigenvectors?
5. Can eigenvectors corresponding to the same eigenvalue be linearly dependent?
6. What are some real-world applications of eigenvalues and eigenvectors?
7. How are eigenvalues and eigenvectors related to the determinant of a matrix?
8. What happens if a matrix does not have enough linearly independent eigenvectors to form a basis?

### Tip:

When working with eigenvalues and eigenvectors, always double-check your calculations, especially when solving the characteristic polynomial, to ensure accuracy in finding the correct eigenvalues and corresponding eigenvectors.

Explore the concept of linearly independent eigenvectors using an example with detailed calculations and explanations. Learn how to find eigenvectors for a matrix, including solving the characteristic equation and verifying independence. Suitable for college-level students and enthusiasts of linear algebra.

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