--- title: "Eigenvalues, eigenvectors, modal matrix [2,0,1;0,2,−1;…]" subject: "Mathematics III" chapter: 1 key: "semester_3/mathematics_iii:eigenvalues-eigenvectors-modal-matrix-2-0" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/eigenvalues-eigenvectors-modal-matrix-2-0 site: BCT Engineering --- # Eigenvalues, eigenvectors, modal matrix [2,0,1;0,2,−1;…] Mathematics III · Chapter 1: Determinants and Matrices - Asked in: 2076 Chaitra, 5 marks - Video: [3 x 3 eigenvalues and eigenvectors](https://www.youtube.com/watch?v=qa9fI6qvUQY) — Prime Newtons - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/eigenvalues-eigenvectors-modal-matrix-2-0 - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1 ## Question Find the eigenvalues and eigenvectors of the matrix [2 0 1; 0 2 −1; 0 0 2] and also find its modal matrix. ## Short answer Eigenvalue is 2 (triple); independent eigenvectors are (1,0,0)ᵀ and (0,1,0)ᵀ. The matrix is defective, so a modal matrix P with a generalised eigenvector (0,0,1)ᵀ gives Jordan form J = [[2,0,1],[0,2,0],[0,0,2]]. ## Answer A = [2 0 1; 0 2 −1; 0 0 2] #### Eigenvalues A is upper triangular, so |A − λI| = (2 − λ)³ = 0 and **λ = 2, 2, 2**. #### Eigenvectors ```text A − 2I = [0 0 1] [0 0 −1] [0 0 0] ``` Rank = 1, so there are 3 − 1 = 2 independent eigenvectors. The equation is z = 0, with x, y free: ```text X₁ = (1, 0, 0)ᵀ , X₂ = (0, 1, 0)ᵀ ``` (Any k₁X₁ + k₂X₂ ≠ 0 is an eigenvector.) #### Modal matrix A 3 × 3 matrix needs three independent eigenvectors to be diagonalized. Here the algebraic multiplicity of λ = 2 is 3 but the geometric multiplicity is 2, so A is **defective**: no modal matrix P exists with P⁻¹AP diagonal. To complete a modal matrix we add a **generalised eigenvector** X₃ with (A − 2I)X₃ = u, where u is an eigenvector. Take X₃ = (0, 0, 1)ᵀ. Then ```text (A − 2I)X₃ = (1, −1, 0)ᵀ = u ``` u = (1, −1, 0)ᵀ is a genuine eigenvector, since it is the combination X₁ − X₂ of the two eigenvectors found above. Take the columns u, X₁ = (1, 0, 0)ᵀ and X₃. The modal matrix is ```text P = [ 1 1 0] [−1 0 0] |P| = 1 ≠ 0 [ 0 0 1] ``` For this P (Au = 2u, AX₁ = 2X₁, AX₃ = 2X₃ + u): ```text P⁻¹AP = J = [2 0 1] [0 2 0] [0 0 2] ``` which is a Jordan form, the nearest to diagonal possible. **Answer: eigenvalue 2 (three times) with eigenvectors (1, 0, 0)ᵀ and (0, 1, 0)ᵀ; the matrix is not diagonalizable, and the modal matrix (with a generalised eigenvector) is P above, giving the Jordan form J.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is λ = 2, 2, 2? A. The trace is 2 B. A is upper triangular, so eigenvalues are its diagonal entries C. A is symmetric D. |A| = 2 Answer: B. |A - λI| = (2 - λ)^3 for a triangular matrix with diagonal 2, 2, 2. (Key point: Eigenvalues of triangular matrix) 2. A - 2I has rank 1. How many independent eigenvectors does λ = 2 have? A. 3 B. 1 C. 0 D. 2 Answer: D. The number is 3 - rank = 2. (Key point: Number of eigenvectors) 3. Why is A not diagonalizable? A. Its eigenvalues are repeated, which always prevents diagonalization B. Its determinant is zero C. Geometric multiplicity (2) is less than algebraic multiplicity (3) D. It has complex eigenvalues Answer: C. A 3x3 matrix needs 3 independent eigenvectors; here there are only 2, so A is defective. (Key point: Diagonalizability) 4. How is the third column of the modal matrix found? A. Solve (A - 2I)X3 = X1 for an eigenvector X1 B. Take any vector independent of the other two C. Solve (A - 2I)X3 = 0 again D. Square an eigenvector Answer: A. A generalised eigenvector satisfies (A - 2I)X3 = X1, giving Jordan form. 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