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

Asked Nov 27 '18 at 03:24

Active Nov 27 '18 at 03:33

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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.

edited Nov 27 '18 at 03:33

Chinnapparaj R

11,589

asked Nov 27 '18 at 03:24

SolidSnackDrive

5

$\pmatrix{0&1\0&0}$ is a $2\times 2$ matrix of rank $1$ that isn't diagonalisable. – Angina Seng Nov 27 '18 at 03:27
This might help! – Chinnapparaj R Nov 27 '18 at 03:28
1

@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 04:36
@SolidSnackDrive That's not a square matrix, so cannot be diagonalisable. – Angina Seng Nov 27 '18 at 04:37
Sorry, please see my edit. – SolidSnackDrive Nov 27 '18 at 04:37

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

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