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Chapter 15: Quick Eigenvalue Trick

The Magic Formula

For any 2×2 matrix, eigenvalues can be computed instantly!

\[ \lambda = m \pm \sqrt{m^2 - p} \]

Where:

• m = (a+d)/2 = mean of diagonal (half of trace)

• p = ad - bc = determinant

No quadratic formula needed!

import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns

sns.set_style('whitegrid')
plt.rcParams['figure.figsize'] = (12, 8)

The Three Key Facts

Fact 1: Sum of Eigenvalues = Trace

\[ \lambda_1 + \lambda_2 = a + d \]

So the mean is: m = (λ₁ + λ₂)/2 = (a+d)/2

Fact 2: Product of Eigenvalues = Determinant

\[ \lambda_1 \cdot \lambda_2 = \det(A) = ad - bc \]

Call this: p = λ₁ · λ₂

Fact 3: From Mean and Product to Values

If two numbers have mean m and product p:

\[ \text{The numbers are: } m \pm \sqrt{m^2 - p} \]

def demonstrate_quick_formula():
    """Show the quick eigenvalue formula."""
    A = np.array([[3, 1], [4, 1]])
    
    print("Quick Eigenvalue Formula\n")
    print(f"Matrix A =\n{A}\n")
    
    a, b, c, d = A[0,0], A[0,1], A[1,0], A[1,1]
    
    # Quick formula
    m = (a + d) / 2
    p = a * d - b * c
    disc = m**2 - p
    
    print("Step 1: m = (a+d)/2 = ({:.0f}+{:.0f})/2 = {:.1f}".format(a, d, m))
    print("Step 2: p = ad-bc = {:.0f}·{:.0f} - {:.0f}·{:.0f} = {:.0f}".format(a, d, b, c, p))
    print("Step 3: m²-p = {:.2f}² - {:.0f} = {:.2f}".format(m, p, disc))
    print("        √(m²-p) = {:.3f}\n".format(np.sqrt(disc)))
    
    lambda1 = m + np.sqrt(disc)
    lambda2 = m - np.sqrt(disc)
    
    print("Result: λ = {:.1f} ± {:.3f}".format(m, np.sqrt(disc)))
    print("        λ₁ = {:.3f}".format(lambda1))
    print("        λ₂ = {:.3f}\n".format(lambda2))
    
    eigenvalues_np = np.linalg.eigvals(A)
    print(f"NumPy verification: {eigenvalues_np}")
    print("Match: ✓")

demonstrate_quick_formula()

Mental Math Examples

Example 1: Diagonal Matrix

\[\begin{split} A = \begin{bmatrix} 5 & 0 \\\\ 0 & 3 \end{bmatrix} \end{split}\]

• m = (5+3)/2 = 4

• p = 5·3 = 15

• m²-p = 16-15 = 1

• λ = 4 ± 1 = 5, 3

Example 2: Symmetric

\[\begin{split} A = \begin{bmatrix} 6 & 2 \\\\ 2 & 3 \end{bmatrix} \end{split}\]

• m = 4.5

• p = 14

• m²-p = 6.25

• λ = 4.5 ± 2.5 = 7, 2

Summary

The Formula

\[ \boxed{\lambda = m \pm \sqrt{m^2 - p}} \]

Where:

• m = (a+d)/2 (mean of diagonal)

• p = ad - bc (determinant)

When to Use

✅ Perfect for 2×2 matrices ✅ Mental math ✅ Quick checks

❌ Not for 3×3 or larger

Memorize: “m plus or minus the square root of m squared minus p” 🎵

Next: Abstract vector spaces!