Project ForgeEngineering Mathematics
Linear Algebra
GATE 2007
Medium
The number of linearly independent eigenvectors of [[2,1],[0,2]] is
- A0
- 1
- C2
- Dinfinite
Explanation
The repeated eigenvalue is λ=2. Solving (A−2I)v=0 gives [[0,1],[0,0]]v=0, forcing v2=0 with v1 free — only ONE linearly independent eigenvector direction (1,0), even though the eigenvalue's algebraic multiplicity is 2 (a defective/non-diagonalizable matrix).
Source: GATE ME 2007 (2007)
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