Is it possible for a $3 times 3$ matrix to have rank 1 but not be diagonalizable?
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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.
linear-algebra eigenvalues-eigenvectors diagonalization
edited Nov 27 at 3:33
Chinnapparaj R
4,7991825
asked Nov 27 at 3:24
SolidSnackDrive
1808
add a comment |
up vote
2
down vote
favorite
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.
linear-algebra eigenvalues-eigenvectors diagonalization
edited Nov 27 at 3:33
Chinnapparaj R
4,7991825
asked Nov 27 at 3:24
SolidSnackDrive
1808
5
$pmatrix{0&1\0&0}$ is a $2times 2$ matrix of rank $1$ that isn't diagonalisable.
– Lord Shark the Unknown
Nov 27 at 3:27
This might help!
– Chinnapparaj R
Nov 27 at 3:28
1
@LordSharktheUnknown Then it follows that $begin{bmatrix}0 & 0 & 1 \0 & 0 & 0\0 & 0 & 0end{bmatrix}$ is not diagonalizable, correct? (Since there is only one eigenvector)
– SolidSnackDrive
Nov 27 at 4:36
@SolidSnackDrive That's not a square matrix, so cannot be diagonalisable.
– Lord Shark the Unknown
Nov 27 at 4:37
Sorry, please see my edit.
– SolidSnackDrive
Nov 27 at 4:37
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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.
linear-algebra eigenvalues-eigenvectors diagonalization
edited Nov 27 at 3:33
Chinnapparaj R
4,7991825
asked Nov 27 at 3:24
SolidSnackDrive
1808
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.
linear-algebra eigenvalues-eigenvectors diagonalization
linear-algebra eigenvalues-eigenvectors diagonalization
edited Nov 27 at 3:33
Chinnapparaj R
4,7991825
asked Nov 27 at 3:24
SolidSnackDrive
1808
edited Nov 27 at 3:33
Chinnapparaj R
4,7991825
asked Nov 27 at 3:24
SolidSnackDrive
1808
edited Nov 27 at 3:33
Chinnapparaj R
4,7991825
edited Nov 27 at 3:33
Chinnapparaj R
4,7991825
edited Nov 27 at 3:33
Chinnapparaj R
4,7991825
4,7991825
asked Nov 27 at 3:24
SolidSnackDrive
1808
asked Nov 27 at 3:24
SolidSnackDrive
1808
asked Nov 27 at 3:24
SolidSnackDrive
1808
1808
5
$pmatrix{0&1\0&0}$ is a $2times 2$ matrix of rank $1$ that isn't diagonalisable.
– Lord Shark the Unknown
Nov 27 at 3:27
This might help!
– Chinnapparaj R
Nov 27 at 3:28
1
@LordSharktheUnknown Then it follows that $begin{bmatrix}0 & 0 & 1 \0 & 0 & 0\0 & 0 & 0end{bmatrix}$ is not diagonalizable, correct? (Since there is only one eigenvector)
– SolidSnackDrive
Nov 27 at 4:36
@SolidSnackDrive That's not a square matrix, so cannot be diagonalisable.
– Lord Shark the Unknown
Nov 27 at 4:37
Sorry, please see my edit.
– SolidSnackDrive
Nov 27 at 4:37
add a comment |
5
$pmatrix{0&1\0&0}$ is a $2times 2$ matrix of rank $1$ that isn't diagonalisable.
– Lord Shark the Unknown
Nov 27 at 3:27
This might help!
– Chinnapparaj R
Nov 27 at 3:28
1
@LordSharktheUnknown Then it follows that $begin{bmatrix}0 & 0 & 1 \0 & 0 & 0\0 & 0 & 0end{bmatrix}$ is not diagonalizable, correct? (Since there is only one eigenvector)
– SolidSnackDrive
Nov 27 at 4:36
@SolidSnackDrive That's not a square matrix, so cannot be diagonalisable.
– Lord Shark the Unknown
Nov 27 at 4:37
Sorry, please see my edit.
– SolidSnackDrive
Nov 27 at 4:37
5
5
$pmatrix{0&1\0&0}$ is a $2times 2$ matrix of rank $1$ that isn't diagonalisable.
– Lord Shark the Unknown
Nov 27 at 3:27
$pmatrix{0&1\0&0}$ is a $2times 2$ matrix of rank $1$ that isn't diagonalisable.
– Lord Shark the Unknown
Nov 27 at 3:27
This might help!
– Chinnapparaj R
Nov 27 at 3:28
This might help!
– Chinnapparaj R
Nov 27 at 3:28
1
1
@LordSharktheUnknown Then it follows that $begin{bmatrix}0 & 0 & 1 \0 & 0 & 0\0 & 0 & 0end{bmatrix}$ is not diagonalizable, correct? (Since there is only one eigenvector)
– SolidSnackDrive
Nov 27 at 4:36
@LordSharktheUnknown Then it follows that $begin{bmatrix}0 & 0 & 1 \0 & 0 & 0\0 & 0 & 0end{bmatrix}$ is not diagonalizable, correct? (Since there is only one eigenvector)
– SolidSnackDrive
Nov 27 at 4:36
@SolidSnackDrive That's not a square matrix, so cannot be diagonalisable.
– Lord Shark the Unknown
Nov 27 at 4:37
@SolidSnackDrive That's not a square matrix, so cannot be diagonalisable.
– Lord Shark the Unknown
Nov 27 at 4:37
Sorry, please see my edit.
– SolidSnackDrive
Nov 27 at 4:37
Sorry, please see my edit.
– SolidSnackDrive
Nov 27 at 4:37
add a comment |
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5
$pmatrix{0&1\0&0}$ is a $2times 2$ matrix of rank $1$ that isn't diagonalisable.
– Lord Shark the Unknown
Nov 27 at 3:27
This might help!
– Chinnapparaj R
Nov 27 at 3:28
1
@LordSharktheUnknown Then it follows that $begin{bmatrix}0 & 0 & 1 \0 & 0 & 0\0 & 0 & 0end{bmatrix}$ is not diagonalizable, correct? (Since there is only one eigenvector)
– SolidSnackDrive
Nov 27 at 4:36
@SolidSnackDrive That's not a square matrix, so cannot be diagonalisable.
– Lord Shark the Unknown
Nov 27 at 4:37
Sorry, please see my edit.
– SolidSnackDrive
Nov 27 at 4:37