# Does 1 distinct eigenvalue guarantee 1 eigenvector?

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?

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.
• @kompasak: If a $2 \times 2$ matrix has two linearly independent eigenvectors, regardless of the corresponding eigenvalue, then it is diagonalizable. – copper.hat Nov 27 '13 at 2:23