Diagonalizability, Multiplicities & Defective Matrices
“AM vs GM, modal matrix P, similarity diagonalization A = P D P^{-1}, and computing matrix powers A^k.”
1.Diagonalization & The Multiplicity Theorem
§ 3.1.A matrix A is Diagonalizable if it is similar to a diagonal matrix: A = P D P^{-1}.
Algebraic vs Geometric Multiplicity:
- AM(lambda): How many times (lambda - lambda_0) divides det(A - lambda I).
- GM(lambda): Number of linearly independent eigenvectors for lambda, equal to Nullity(A - lambda I) = n - Rank(A - lambda I).
The Defective Matrix Trap:
If GM(lambda) < AM(lambda) for any eigenvalue, the matrix is defective and lacks a full basis of eigenvectors. It CANNOT be diagonalized.
Power Calculation Application:
Computing A^{100} directly requires 99 matrix multiplications. Under diagonalization:
A^{100} = (P D P^{-1})^{100} = P D^{100} P^{-1}.
2. Worked Practice Problem: Diagonalizability Test & Matrix Power
Check GM = n - rank(A - lambda*I) to confirm sufficient independent eigenvectors.
Problem: Let A = [ [1, 2], [0, 3] ]. Verify if A is diagonalizable. Construct the modal matrix P and diagonal matrix D, and compute A^4.
Step 1: Find eigenvalues of triangular matrix A:
Since A is upper triangular, its eigenvalues are the diagonal entries: lambda1 = 1, lambda2 = 3.
Since both eigenvalues are distinct (lambda1 != lambda2), A is GUARANTEED to be diagonalizable (AM = GM = 1 for each).
Step 2: Find eigenvector for lambda1 = 1: (A - 1*I) v = 0:
[ [0, 2], [0, 2] ] [ v1, v2 ]^T = 0 => 2*v2 = 0 => v2 = 0, v1 = 1.
p1 = [ 1, 0 ]^T.
Step 3: Find eigenvector for lambda2 = 3: (A - 3*I) v = 0:
[ [-2, 2], [0, 0] ] [ v1, v2 ]^T = 0 => -2v1 + 2v2 = 0 => v1 = v2.
p2 = [ 1, 1 ]^T.
Step 4: Construct P, P^{-1}, and D:
P = [ [1, 1], [0, 1] ]; D = [ [1, 0], [0, 3] ].
det(P) = 1. P^{-1} = [ [1, -1], [0, 1] ].
Step 5: Compute A^4 = P D^4 P^{-1}:
D^4 = [ [1^4, 0], [0, 3^4] ] = [ [1, 0], [0, 81] ].
P D^4 = [ [1, 1], [0, 1] ] [ [1, 0], [0, 81] ] = [ [1, 81], [0, 81] ].
A^4 = [ [1, 81], [0, 81] ] * [ [1, -1], [0, 1] ] = [ [1, -1 + 81], [0, 81] ] = [ [1, 80], [0, 81] ].
Conclusion: A is diagonalizable, and A^4 = [ [1, 80], [0, 81] ].
Primary Academic References & Exam Sources
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