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

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.

• $\pmatrix{0&1\\0&0}$ is a $2\times 2$ matrix of rank $1$ that isn't diagonalisable. – Lord Shark the Unknown Nov 27 '18 at 3:27
• This might help! – Chinnapparaj R Nov 27 '18 at 3:28
• @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) – SolidSnackDrive Nov 27 '18 at 4:36
• @SolidSnackDrive That's not a square matrix, so cannot be diagonalisable. – Lord Shark the Unknown Nov 27 '18 at 4:37
• Sorry, please see my edit. – SolidSnackDrive Nov 27 '18 at 4:37