Eigenvectors of $n^{th}$ root of identity matrix
2
Let $A^n=I,$ where $A$ is $ntimes n,$ and assume that $A^kneq I,$ for all $1leq k<n.$ Since its characteristic polynomial is $x^n-1$, the distinct $n^{th}$ roots of unity are its eigenvalues,
Thus there are a full set of linearly independent eigenvectors.
- What do they look like?
- If we assume $A$ is orthogonal, what do they look like?
- If we assume $A$ is real, can one say anything more?
linear-algebra eigenvalues-eigenvectors
asked Nov 25 at 4:19
kodlu
3,380716
add a comment |
2
Let $A^n=I,$ where $A$ is $ntimes n,$ and assume that $A^kneq I,$ for all $1leq k<n.$ Since its characteristic polynomial is $x^n-1$, the distinct $n^{th}$ roots of unity are its eigenvalues,
Thus there are a full set of linearly independent eigenvectors.
- What do they look like?
- If we assume $A$ is orthogonal, what do they look like?
- If we assume $A$ is real, can one say anything more?
linear-algebra eigenvalues-eigenvectors
asked Nov 25 at 4:19
kodlu
3,380716
Hint: since the eigenvalues are distinct it follows the eigenvectors are linearly independent and $A$ is diagonalizable. So write $A=USU^*$.
– Alex R.
Nov 25 at 4:24
2
They can look like anything in general, in the sense that they can literally be any basis of the underlying vector space.
– Qiaochu Yuan
Nov 25 at 7:56
The roots are not necessarily distinct. For that you need that $A^{n-1}ne I$.
– I like Serena
Nov 25 at 11:25
Please see edit, thanks for the comment.
– kodlu
Nov 25 at 11:50
add a comment |
2
2
2
Let $A^n=I,$ where $A$ is $ntimes n,$ and assume that $A^kneq I,$ for all $1leq k<n.$ Since its characteristic polynomial is $x^n-1$, the distinct $n^{th}$ roots of unity are its eigenvalues,
Thus there are a full set of linearly independent eigenvectors.
- What do they look like?
- If we assume $A$ is orthogonal, what do they look like?
- If we assume $A$ is real, can one say anything more?
linear-algebra eigenvalues-eigenvectors
asked Nov 25 at 4:19
kodlu
3,380716
Let $A^n=I,$ where $A$ is $ntimes n,$ and assume that $A^kneq I,$ for all $1leq k<n.$ Since its characteristic polynomial is $x^n-1$, the distinct $n^{th}$ roots of unity are its eigenvalues,
Thus there are a full set of linearly independent eigenvectors.
- What do they look like?
- If we assume $A$ is orthogonal, what do they look like?
- If we assume $A$ is real, can one say anything more?
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked Nov 25 at 4:19
kodlu
3,380716
asked Nov 25 at 4:19
kodlu
3,380716
edited Nov 25 at 11:49
asked Nov 25 at 4:19
kodlu
3,380716
asked Nov 25 at 4:19
kodlu
3,380716
asked Nov 25 at 4:19
kodlu
3,380716
3,380716
Hint: since the eigenvalues are distinct it follows the eigenvectors are linearly independent and $A$ is diagonalizable. So write $A=USU^*$.
– Alex R.
Nov 25 at 4:24
2
They can look like anything in general, in the sense that they can literally be any basis of the underlying vector space.
– Qiaochu Yuan
Nov 25 at 7:56
The roots are not necessarily distinct. For that you need that $A^{n-1}ne I$.
– I like Serena
Nov 25 at 11:25
Please see edit, thanks for the comment.
– kodlu
Nov 25 at 11:50
add a comment |
Hint: since the eigenvalues are distinct it follows the eigenvectors are linearly independent and $A$ is diagonalizable. So write $A=USU^*$.
– Alex R.
Nov 25 at 4:24
2
They can look like anything in general, in the sense that they can literally be any basis of the underlying vector space.
– Qiaochu Yuan
Nov 25 at 7:56
The roots are not necessarily distinct. For that you need that $A^{n-1}ne I$.
– I like Serena
Nov 25 at 11:25
Please see edit, thanks for the comment.
– kodlu
Nov 25 at 11:50
Hint: since the eigenvalues are distinct it follows the eigenvectors are linearly independent and $A$ is diagonalizable. So write $A=USU^*$.
– Alex R.
Nov 25 at 4:24
Hint: since the eigenvalues are distinct it follows the eigenvectors are linearly independent and $A$ is diagonalizable. So write $A=USU^*$.
– Alex R.
Nov 25 at 4:24
2
2
They can look like anything in general, in the sense that they can literally be any basis of the underlying vector space.
– Qiaochu Yuan
Nov 25 at 7:56
They can look like anything in general, in the sense that they can literally be any basis of the underlying vector space.
– Qiaochu Yuan
Nov 25 at 7:56
The roots are not necessarily distinct. For that you need that $A^{n-1}ne I$.
– I like Serena
Nov 25 at 11:25
The roots are not necessarily distinct. For that you need that $A^{n-1}ne I$.
– I like Serena
Nov 25 at 11:25
Please see edit, thanks for the comment.
– kodlu
Nov 25 at 11:50
Please see edit, thanks for the comment.
– kodlu
Nov 25 at 11:50
add a comment |
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Hint: since the eigenvalues are distinct it follows the eigenvectors are linearly independent and $A$ is diagonalizable. So write $A=USU^*$.
– Alex R.
Nov 25 at 4:24
2
They can look like anything in general, in the sense that they can literally be any basis of the underlying vector space.
– Qiaochu Yuan
Nov 25 at 7:56
The roots are not necessarily distinct. For that you need that $A^{n-1}ne I$.
– I like Serena
Nov 25 at 11:25
Please see edit, thanks for the comment.
– kodlu
Nov 25 at 11:50