Eigenvalues, Eigenvectors & Trace-Determinant Relations
“Av = lambda v, characteristic polynomial det(A - lambda I) = 0, trace sum, and determinant product rules.”
1.The Eigensystem: Physical Geometry & Characteristic Equations
§ 3.1.When a matrix A acts on a vector v, it generally rotates and scales v. An Eigenvector is a special direction that undergoes ONLY pure scaling, without rotation: A v = lambda v.
Characteristic Equation:
(A - lambda I) v = 0 has non-trivial solutions iff det(A - lambda I) = 0.
This yields an n-th degree polynomial p(lambda) = (-1)^n lambda^n + c_{n-1} lambda^{n-1} + ... + det(A).
Essential Trace & Determinant Relations:
For any n x n matrix:
- Trace(A) = Sum_{i=1}^n lambda_i (sum of diagonal entries).
- det(A) = Prod_{i=1}^n lambda_i.
These two relations allow fast verification of calculated eigenvalues in GATE examinations.
2. Worked Practice Problem: Eigenvalues & Eigenspaces of 3x3 Matrix
Notice the block-triangular structure to rapidly factor the characteristic polynomial.
Problem: Find all eigenvalues and basis vectors for the eigenspaces of matrix A = [ [2, 1, 1], [1, 2, 1], [0, 0, 1] ].
Step 1: Set up characteristic equation det(A - lambda * I) = 0:
det([ [2 - lambda, 1, 1], [1, 2 - lambda, 1], [0, 0, 1 - lambda] ]) = 0
Expand along row 3:
(1 - lambda) * [ (2 - lambda)^2 - (1)(1) ] = 0
(1 - lambda) [ (lambda^2 - 4lambda + 4) - 1 ] = 0
(1 - lambda) (lambda^2 - 4lambda + 3) = 0
(1 - lambda) (lambda - 1) (lambda - 3) = 0
-(lambda - 1)^2 * (lambda - 3) = 0.
Step 2: Identify eigenvalues and verify with trace and determinant:
lambda1 = 1 (multiplicity 2), lambda2 = 3 (multiplicity 1).
Trace check: 1 + 1 + 3 = 5; Diagonal sum: 2 + 2 + 1 = 5. (Matches).
det check: 1 1 3 = 3; det(A) = 1 * (4 - 1) = 3. (Matches).
Step 3: Eigenspace for lambda = 3: (A - 3*I) v = 0:
[ [-1, 1, 1], [1, -1, 1], [0, 0, -2] ] [ v1, v2, v3 ]^T = 0
Row 3 => -2*v3 = 0 => v3 = 0.
Row 1 => -v1 + v2 = 0 => v1 = v2.
Eigenvector for lambda = 3: v = [ 1, 1, 0 ]^T.
Step 4: Eigenspace for lambda = 1: (A - 1*I) v = 0:
[ [1, 1, 1], [1, 1, 1], [0, 0, 0] ] [ v1, v2, v3 ]^T = 0
Single independent equation: v1 + v2 + v3 = 0.
Two free variables (v2 = s, v3 = t) => v1 = -s - t.
Basis eigenvectors: v_a = [ -1, 1, 0 ]^T, and v_b = [ -1, 0, 1 ]^T.
Conclusion: Eigenvalues are {1, 1, 3} with eigenspace bases { [-1, 1, 0]^T, [-1, 0, 1]^T } for lambda=1 and { [1, 1, 0]^T } for lambda=3.
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