How To Diagonalize A Matrix 3x3

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Treneri

May 15, 2025 · 5 min read

How To Diagonalize A Matrix 3x3
How To Diagonalize A Matrix 3x3

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    How to Diagonalize a 3x3 Matrix: A Comprehensive Guide

    Diagonalizing a matrix is a fundamental concept in linear algebra with far-reaching applications in various fields, including physics, computer science, and engineering. This comprehensive guide will walk you through the process of diagonalizing a 3x3 matrix, explaining the underlying theory and providing step-by-step examples. We'll cover eigenvalue calculations, eigenvector determination, and the construction of the diagonal matrix. By the end, you'll have a solid understanding of this crucial linear algebra technique.

    Understanding Diagonalization

    Before diving into the mechanics, let's clarify what diagonalization entails. A square matrix A is diagonalizable if it can be expressed in the form:

    A = PDP⁻¹

    where:

    • A is the original square matrix (in our case, 3x3).
    • P is an invertible matrix whose columns are the eigenvectors of A.
    • D is a diagonal matrix whose diagonal entries are the eigenvalues of A.
    • P⁻¹ is the inverse of matrix P.

    This decomposition simplifies many matrix operations, such as calculating powers of a matrix or solving systems of differential equations. The key to diagonalization lies in finding the eigenvalues and eigenvectors of the matrix.

    Finding Eigenvalues: The Characteristic Equation

    The eigenvalues of a matrix A are the scalar values λ (lambda) that satisfy the equation:

    Av = λv

    where:

    • A is the matrix.
    • v is the eigenvector (a non-zero vector).
    • λ is the eigenvalue.

    This equation can be rewritten as:

    (A - λI)v = 0

    where I is the identity matrix. For a non-trivial solution (v ≠ 0), the determinant of (A - λI) must be zero:

    det(A - λI) = 0

    This equation is known as the characteristic equation. Solving this equation for λ gives us the eigenvalues of the matrix. For a 3x3 matrix, the characteristic equation will be a cubic polynomial equation.

    Step-by-Step Example: Finding Eigenvalues

    Let's consider a sample 3x3 matrix:

    A =  [ 2  1  1 ]
         [ 1  2  1 ]
         [ 1  1  2 ]
    
    1. Form (A - λI):
    A - λI =  [ 2-λ  1   1 ]
              [ 1  2-λ  1 ]
              [ 1   1  2-λ ]
    
    1. Calculate the determinant: This involves expanding the determinant along a row or column. The resulting characteristic equation will be a cubic polynomial in λ. For this example, the determinant calculation yields:

      det(A - λI) = (2-λ)((2-λ)² - 1) - ( (2-λ) - 1) + (1 - (2-λ)) = (2-λ)(λ²-4λ+3) + λ -1 + λ -1 = -λ³ + 6λ² - 9λ + 4

    2. Solve the cubic equation: Solving the cubic equation det(A - λI) = 0 can be challenging. You can use numerical methods (like the Newton-Raphson method) or utilize software like MATLAB or Python's NumPy library to find the roots. In this case, the roots (eigenvalues) are:

      λ₁ = 1, λ₂ = 1, λ₃ = 4

    Finding Eigenvectors

    Once you have the eigenvalues, the next step is to find the corresponding eigenvectors. For each eigenvalue λᵢ, we solve the equation:

    (A - λᵢI)vᵢ = 0

    This system of linear equations will have infinitely many solutions, all scalar multiples of a single eigenvector. We typically find one eigenvector for each eigenvalue by row-reducing the augmented matrix (A - λᵢI | 0) and expressing the solution in terms of free variables.

    Step-by-Step Example: Finding Eigenvectors

    Using our example matrix and eigenvalues:

    1. For λ₁ = 1:

      (A - λ₁I) = [ 1 1 1 ] [ 1 1 1 ] [ 1 1 1 ]

      Row-reducing gives:

      [ 1 1 1 ] [ 0 0 0 ] [ 0 0 0 ]

      This leads to the equation x₁ + x₂ + x₃ = 0. We can choose two free variables (e.g., x₂ = s, x₃ = t) and express x₁ as x₁ = -s - t. Therefore, the eigenvector v₁ is given by:

      v₁ = [-s - t, s, t] = s[-1, 1, 0] + t[-1, 0, 1]

      We can choose two linearly independent eigenvectors for this repeated eigenvalue: [-1, 1, 0] and [-1, 0, 1].

    2. For λ₃ = 4:

      (A - λ₃I) = [ -2 1 1 ] [ 1 -2 1 ] [ 1 1 -2 ]

      Row-reducing this matrix will yield a system of equations with only one free variable, leading to an eigenvector:

      v₃ = [1, 1, 1]

    Constructing the Diagonal Matrix and P Matrix

    Now that we have the eigenvalues and eigenvectors, we can construct the diagonal matrix D and the matrix P.

    • D: The diagonal matrix D has the eigenvalues on its diagonal:
    D =  [ 1  0  0 ]
         [ 0  1  0 ]
         [ 0  0  4 ]
    
    • P: The matrix P has the eigenvectors as its columns:
    P =  [ -1 -1  1 ]
         [  1  0  1 ]
         [  0  1  1 ]
    

    Note that the order of eigenvectors in P must correspond to the order of eigenvalues in D.

    Calculating the Inverse of P (P⁻¹)

    To complete the diagonalization, we need the inverse of matrix P, denoted as P⁻¹. There are several methods for calculating the inverse of a matrix, including using the adjugate matrix or employing Gaussian elimination. For this example, the inverse of P is:

    P⁻¹ = [ -1/3  -1/3  2/3 ]
           [ -1/3   2/3 -1/3 ]
           [  1/3   1/3  1/3 ]
    

    Verification: A = PDP⁻¹

    Finally, to verify the diagonalization, multiply the matrices PDP⁻¹:

    PDP⁻¹ = [ -1 -1  1 ] [ 1  0  0 ] [ -1/3  -1/3  2/3 ] = [ 2  1  1 ]
             [  1  0  1 ] [ 0  1  0 ] [ -1/3   2/3 -1/3 ] = [ 1  2  1 ]
             [  0  1  1 ] [ 0  0  4 ] [  1/3   1/3  1/3 ] = [ 1  1  2 ]
    

    This confirms that we have successfully diagonalized the matrix A.

    Handling Cases with Repeated Eigenvalues

    When a matrix has repeated eigenvalues, the diagonalizability depends on the number of linearly independent eigenvectors associated with the repeated eigenvalue. If the number of linearly independent eigenvectors is equal to the algebraic multiplicity of the repeated eigenvalue, the matrix is diagonalizable. If not, it is not diagonalizable. In our example, we had a repeated eigenvalue (λ = 1) with two linearly independent eigenvectors, thus ensuring diagonalizability. Matrices with repeated eigenvalues that are not diagonalizable are called defective matrices.

    Conclusion

    Diagonalizing a 3x3 matrix is a powerful tool in linear algebra, enabling simplified computations and offering insights into the matrix's properties. This process involves finding the eigenvalues by solving the characteristic equation, determining the corresponding eigenvectors, and constructing the diagonal matrix and the invertible matrix P. While the process might seem involved, understanding the underlying concepts and following a systematic approach will allow you to confidently diagonalize matrices. Remember to always check your work by verifying A = PDP⁻¹! The use of computational tools can significantly aid in solving the cubic equation and handling matrix inversions, especially for larger and more complex matrices. Mastering this technique opens doors to a deeper understanding of linear algebra and its applications across numerous disciplines.

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