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Given that a $3 \times 3$ matrix has only one eigenvalue, what is the dimension of its corresponding eigenspace? It says that the answer is 3. But couldn't we have some matrix

$A = \begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 &\lambda \end{pmatrix}$

in Jordan canonical form. Then $\lambda$ is its only eigenvalue, but there are two Jordan blocks, hence the geometric multiplicity should be 2?

  • 0
    "It says"... Who or what is "it"?2011-10-01
  • 0
    You can only say that the dimension is 3 (full dimension) iff the matrix is diagonalizable (eg , if it's normal).2011-10-01

2 Answers 2