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Is it possible for a $3 \times 3$ matrix to have rank $1$ but not be diagonalizable?

If the matrix only has the top left entry, then obviously it is already diagonal. But what about other two entries? I know that a condition for the matrix to be diagonalizable is for the matrix to have $3$ linearly independent eigenvectors, but I am unsure how to prove whether that will always be the case for any rank $1$, $3 \times 3$ matrix.

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    $\begingroup$ $\pmatrix{0&1\\0&0}$ is a $2\times 2$ matrix of rank $1$ that isn't diagonalisable. $\endgroup$ – Lord Shark the Unknown Nov 27 '18 at 3:27
  • $\begingroup$ This might help! $\endgroup$ – Chinnapparaj R Nov 27 '18 at 3:28
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    $\begingroup$ @LordSharktheUnknown Then it follows that $\begin{bmatrix}0 & 0 & 1 \\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}$ is not diagonalizable, correct? (Since there is only one eigenvector) $\endgroup$ – SolidSnackDrive Nov 27 '18 at 4:36
  • $\begingroup$ @SolidSnackDrive That's not a square matrix, so cannot be diagonalisable. $\endgroup$ – Lord Shark the Unknown Nov 27 '18 at 4:37
  • $\begingroup$ Sorry, please see my edit. $\endgroup$ – SolidSnackDrive Nov 27 '18 at 4:37

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