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Lesson 2: The Characteristic Equation

Estimated time: 40-50 minutes

Learning Objectives

The Characteristic Polynomial

The characteristic equation det(A - lambda I) = 0 produces a polynomial in lambda. For an n x n matrix, this is a degree-n polynomial.

Characteristic Polynomial: p(lambda) = det(A - lambda I). This is a polynomial of degree n with leading term (-lambda)^n (or equivalently (-1)^n * lambda^n).

The 2x2 Case

2x2 Formula: For A = [a b; c d], the characteristic polynomial is:

p(lambda) = lambda^2 - (a+d)*lambda + (ad - bc) = lambda^2 - trace(A)*lambda + det(A).

Worked Example 1

A = [6 -1; 2 3]. trace = 9, det = 18 + 2 = 20.

p(lambda) = lambda^2 - 9*lambda + 20 = (lambda - 4)(lambda - 5) = 0.

Eigenvalues: lambda = 4 and lambda = 5.

Worked Example 2: Repeated Eigenvalue

A = [3 1; 0 3]. trace = 6, det = 9.

p(lambda) = lambda^2 - 6*lambda + 9 = (lambda - 3)^2 = 0.

Only eigenvalue: lambda = 3 with algebraic multiplicity 2.

The 3x3 Case

For 3x3 matrices, expand det(A - lambda I) using cofactor expansion. The result is a cubic polynomial.

Worked Example 3: 3x3 Characteristic Polynomial

A = [2 0 0; 0 3 -1; 0 -1 3].

A - lambda I = [2-lambda, 0, 0; 0, 3-lambda, -1; 0, -1, 3-lambda].

Expand along row 1: det = (2-lambda) * det([3-lambda, -1; -1, 3-lambda]).

= (2-lambda)[(3-lambda)^2 - 1] = (2-lambda)(lambda^2 - 6*lambda + 8) = (2-lambda)(lambda-2)(lambda-4).

Eigenvalues: lambda = 2 (algebraic multiplicity 2) and lambda = 4 (algebraic multiplicity 1).

Worked Example 4

A = [1 2 0; 0 3 0; 0 0 5].

Since A is upper triangular, det(A - lambda I) = (1-lambda)(3-lambda)(5-lambda) = 0.

Eigenvalues: lambda = 1, 3, 5 -- just read them from the diagonal!

Algebraic vs. Geometric Multiplicity

Algebraic Multiplicity: The number of times lambda appears as a root of the characteristic polynomial.

Geometric Multiplicity: The dimension of the eigenspace E_lambda = null(A - lambda I). This equals the number of free variables when you row reduce (A - lambda I).

Key Inequality

For every eigenvalue lambda: 1 ≤ geometric multiplicity ≤ algebraic multiplicity.

Worked Example 5: When Multiplicities Differ

A = [3 1; 0 3]. Eigenvalue lambda = 3 has algebraic multiplicity 2.

A - 3I = [0 1; 0 0]. Row reduce: [0 1; 0 0]. One free variable (x1). Geometric multiplicity = 1.

Eigenspace: E_3 = span{(1, 0)}. Only one linearly independent eigenvector despite algebraic multiplicity 2.

Worked Example 6: When Multiplicities Match

A = [3 0; 0 3] = 3I. Eigenvalue lambda = 3 has algebraic multiplicity 2.

A - 3I = [0 0; 0 0]. Two free variables. Geometric multiplicity = 2.

Eigenspace: E_3 = R^2 (every nonzero vector is an eigenvector). Multiplicities match!

Complex Eigenvalues

Not every real matrix has real eigenvalues. The characteristic polynomial may have complex roots.

Worked Example 7

A = [0 -1; 1 0] (90-degree rotation).

p(lambda) = lambda^2 + 1 = 0. Eigenvalues: lambda = i and lambda = -i.

No real eigenvectors exist -- a rotation does not preserve any real direction!

Complex Eigenvalues of Real Matrices: Complex eigenvalues always come in conjugate pairs: if lambda = a + bi is an eigenvalue, so is lambda-bar = a - bi.

Check Your Understanding

1. Find the characteristic polynomial of A = [2 5; 1 4].

Answer: p(lambda) = lambda^2 - 6*lambda + 3. Using the quadratic formula: lambda = (6 +/- sqrt(24))/2 = 3 +/- sqrt(6).

2. A 3x3 matrix has characteristic polynomial (lambda - 1)^2(lambda + 3). What are the eigenvalues and their algebraic multiplicities?

Answer: lambda = 1 with algebraic multiplicity 2, and lambda = -3 with algebraic multiplicity 1.

3. For A = [3 1; 0 3], find the geometric multiplicity of lambda = 3.

Answer: A - 3I = [0 1; 0 0]. One free variable, so geometric multiplicity = 1. This is less than the algebraic multiplicity of 2.

Key Takeaways

Next Lesson

Eigenspaces and Diagonalization.

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