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I am trying to figure out when 2x2 matrices are not diagonalizable. Right now, my conditions are:

  • the matrix has only 1 distinct eigenvalue

  • the matrix yields only 1 linearly independent eigenvector

But when I know that there are 2 eigenvalues, can I safely assume that each eigenvalue will have at least 1 linearly independent eigenvector?

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By definition, an eigenvalue always has an eigenvector. And it's easy to prove that eigenvectors for distinct eigenvalues are linearly independent. If a $2 \times 2$ matrix has only one eigenvalue, there may be either $1$ or $2$ linearly independent eigenvectors.

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  • $\begingroup$ Also, if a 2x2 matrix yields one eigenvalue, but two linearly independent eigenvectors, is the matrix still diagonalizable? $\endgroup$ – kompasak Nov 27 '13 at 2:19
  • $\begingroup$ @kompasak Distinct tells you nothing. You want linearly independent or not. Eigenspaces contain many different eigenvectors, what matters is the dimension. $\endgroup$ – Pedro Tamaroff Nov 27 '13 at 2:21
  • $\begingroup$ @kompasak: If a $2 \times 2$ matrix has two linearly independent eigenvectors, regardless of the corresponding eigenvalue, then it is diagonalizable. $\endgroup$ – copper.hat Nov 27 '13 at 2:23
  • $\begingroup$ Okay, thank you everyone, I understand it now. $\endgroup$ – kompasak Nov 27 '13 at 2:26

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