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Exploring Eigenvalues and Eigenvectors

20/12/23·3 min read

Introduction

Eigenvalues and eigenvectors are key concepts in linear algebra with numerous applications in areas such as physics, engineering, computer science, and data analysis. In this post, we'll explore their definitions, properties, and how to calculate them.

Definitions

For a square matrix ARn×nA \in \mathbb{R}^{n \times n}, an eigenvector v\mathbf{v} and an eigenvalue λ\lambda satisfy the equation:

Av=λv,A \mathbf{v} = \lambda \mathbf{v},

where v0\mathbf{v} \neq \mathbf{0} is the eigenvector, and λ\lambda is the eigenvalue, which is a scalar.

Calculating Eigenvalues

To find the eigenvalues, we solve the characteristic equation:

det(AλI)=0,\det(A - \lambda I) = 0,

where II is the identity matrix of the same dimension as AA. The determinant yields a polynomial in λ\lambda, known as the characteristic polynomial.

Example

Consider the matrix:

A=[4123].A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}.

The characteristic equation is:

det([4123]λ[1001])=0.\det\left(\begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right) = 0.

Simplify:

det[4λ123λ]=0.\det\begin{bmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{bmatrix} = 0.

Expanding the determinant:

(4λ)(3λ)(21)=0,(4 - \lambda)(3 - \lambda) - (2 \cdot 1) = 0,

which simplifies to:

λ27λ+10=0.\lambda^2 - 7\lambda + 10 = 0.

Factoring gives:

(λ5)(λ2)=0.(\lambda - 5)(\lambda - 2) = 0.

Thus, the eigenvalues are λ=5\lambda = 5 and λ=2\lambda = 2.

Finding Eigenvectors

For each eigenvalue, substitute λ\lambda back into AλIA - \lambda I and solve the equation (AλI)v=0\left(A - \lambda I\right)\mathbf{v} = \mathbf{0}.

Example (Continued)

For λ=5\lambda = 5:

A5I=[1122].A - 5I = \begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix}.

Solve:

[1122][xy]=[00].\begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.

This yields x=yx = y. Thus, the eigenvector corresponding to λ=5\lambda = 5 is:

v1=[11].\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}.

Similarly, for λ=2\lambda = 2:

A2I=[2121].A - 2I = \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}.

Solve:

[2121][xy]=[00].\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.

This yields x=yx = -y. Thus, the eigenvector corresponding to λ=2\lambda = 2 is:

v2=[11].\mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}.

Applications

Eigenvalues and eigenvectors have wide-ranging applications, including:

  1. Stability Analysis: Used to study the stability of systems in engineering and physics.
  2. Principal Component Analysis (PCA): In data science, PCA uses eigenvectors to reduce the dimensionality of data.
  3. Quantum Mechanics: Operators in quantum mechanics have eigenvalues corresponding to measurable quantities.

Conclusion

Understanding eigenvalues and eigenvectors is fundamental for solving problems in various scientific and engineering domains. By mastering their calculation and interpretation, you can unlock powerful tools for analysis and problem-solving.