Eigenvectors
A non-zero vector \( \mathbf{v} \) is called an eigenvector of a square matrix \( A \) if there exists a scalar \( \lambda \in \mathbb{R} \) (or \( \mathbb{C} \)) such that: $$ A \mathbf{v} = \lambda \mathbf{v} $$ Here, \( \lambda \) is known as the eigenvalue corresponding to the eigenvector \( \mathbf{v} \).
This equation indicates that the effect of \( A \) on \( \mathbf{v} \) is equivalent to scaling the vector by \( \lambda \).
In simpler terms, the vector’s direction remains unchanged after the transformation; it can only be stretched, compressed, or flipped.
How to Find Eigenvectors
To determine the eigenvectors of a square matrix \( A \) of order \( n \), solve the following equation:
$$ (A - \lambda I) \mathbf{v} = 0 $$
Where \( I \) is the \( n \times n \) identity matrix.
The solutions to this equation provide the eigenvalues \( \lambda \), which can be obtained by solving the characteristic polynomial:
$$ \det(A - \lambda I) = 0 $$
For each eigenvalue \( \lambda \), compute the null space of \( A - \lambda I \), represented as:
$$ \ker(A - \lambda I) $$
This null space provides a basis for the eigenvectors corresponding to \( \lambda \).
A Practical Example
Let’s consider the matrix:
$$ A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} $$
The characteristic equation is:
$$ \det(A - \lambda I) = \det \begin{bmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{bmatrix} = (2-\lambda)^2 - 1 = 0 $$
The eigenvalues are \( \lambda_1 = 3 \) and \( \lambda_2 = 1 \).
Next, calculate the eigenvectors for each eigenvalue.
1] For \( \lambda_1 = 3 \)
Substitute \( \lambda = 3 \) into the matrix:
$$ (A - \lambda I) \mathbf{v} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} - \lambda \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$
$$ (A - \lambda I) \mathbf{v} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} - 3 \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$
$$ (A - \lambda I) \mathbf{v} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} - \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} $$
$$ (A - \lambda I) \mathbf{v} = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} $$
Now solve the equation:
$$ (A - \lambda I) \mathbf{v} = 0 $$
$$ \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = 0 $$
The resulting system of equations is:
$$ \begin{cases} -x + y = 0 \\ x - y = 0 \end{cases} $$
Solving gives \( x = y \):
$$ \mathbf{v}_1 = \begin{bmatrix} x \\ x \end{bmatrix} $$
For example, an eigenvector corresponding to \( \lambda_1 = 3 \) is:
$$ \mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$
2] For \( \lambda_2 = 1 \)
Substitute \( \lambda = 1 \) into the matrix:
$$ (A - \lambda I) \mathbf{v} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} - \lambda \cdot \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$
$$ (A - \lambda I) \mathbf{v} = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} $$
Solve the equation:
$$ (A - \lambda I) \mathbf{v} = 0 $$
$$ \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = 0 $$
The corresponding system of equations simplifies to:
$$ \begin{cases} x + y = 0 \\ x + y = 0 \end{cases} $$
Thus, \( x = -y \):
$$ \mathbf{v}_2 = \begin{bmatrix} x \\ -x \end{bmatrix} $$
For example, an eigenvector corresponding to \( \lambda_2 = 1 \) is:
$$ \mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} $$
3] Conclusion
To summarize, the eigenvectors of the matrix are:
- $$ \mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$
- $$ \mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} $$
And so on.
