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.

Would you like more details or have any questions?

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.