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**HallsofIvy** If your characteristic equation gives $\displaystyle \lambda_1$ as a "double" eigenvalue and $\displaystyle \lambda_2$ as a "single" eigenvalue, you **can't** have two independent eigenvalues corresponding to $\displaystyle \lambda_2$. The number of independent eigenvectors corresponding to an eigenvalue (the "geometric multiplicity") cannot be larger than the multiplicity of the eigenvalue (the "algebraic multiplicity").

If $\displaystyle \lambda_1$ is a double eigenvalue ("algebraic multiplicity" two) with only one independent eigenvector ("geometric multiplicty one), then the matrix is not diagonalizable.