Are the only eigenvalues of an operator $T^k$ the $lambda_n^k$?
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If I have an operator T which has eigenvalues $(lambda_n)_{n in mathbb{N}}$, I understand that $lambda_n^k$ are eigenvalues for $T^k$. But how to show that these are the only eigenvalues? Is this even true?
functional-analysis eigenvalues-eigenvectors
asked Jan 6 at 23:43
roi_saumonroi_saumon
43728
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add a comment |
$begingroup$
If I have an operator T which has eigenvalues $(lambda_n)_{n in mathbb{N}}$, I understand that $lambda_n^k$ are eigenvalues for $T^k$. But how to show that these are the only eigenvalues? Is this even true?
functional-analysis eigenvalues-eigenvectors
asked Jan 6 at 23:43
roi_saumonroi_saumon
43728
$endgroup$
add a comment |
$begingroup$
If I have an operator T which has eigenvalues $(lambda_n)_{n in mathbb{N}}$, I understand that $lambda_n^k$ are eigenvalues for $T^k$. But how to show that these are the only eigenvalues? Is this even true?
functional-analysis eigenvalues-eigenvectors
asked Jan 6 at 23:43
roi_saumonroi_saumon
43728
$endgroup$
If I have an operator T which has eigenvalues $(lambda_n)_{n in mathbb{N}}$, I understand that $lambda_n^k$ are eigenvalues for $T^k$. But how to show that these are the only eigenvalues? Is this even true?
functional-analysis eigenvalues-eigenvectors
functional-analysis eigenvalues-eigenvectors
asked Jan 6 at 23:43
roi_saumonroi_saumon
43728
asked Jan 6 at 23:43
roi_saumonroi_saumon
43728
asked Jan 6 at 23:43
roi_saumonroi_saumon
43728
asked Jan 6 at 23:43
roi_saumonroi_saumon
43728
asked Jan 6 at 23:43
roi_saumonroi_saumon
43728
43728
add a comment |
add a comment |
1 Answer
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Over a field which is not algebraically closed, this is false. Consider
$$A=begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}in M_2(mathbb R).$$
Then $A^2=-I$, which has only the eigenvalues $-1$, but $A$ has no eigenvalues.
But over an algebraically closed field, this is true for operators on finite-dimensional spaces by the Jordan decomposition: squaring the matrix applies to each Jordan block separately.
answered Jan 7 at 0:08
Ashwin TrisalAshwin Trisal
1,2441516
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$begingroup$
Oh, thanks, and what about the infinite dimensional case on $mathbb{C}$?
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– roi_saumon
Jan 7 at 12:21
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Over a field which is not algebraically closed, this is false. Consider
$$A=begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}in M_2(mathbb R).$$
Then $A^2=-I$, which has only the eigenvalues $-1$, but $A$ has no eigenvalues.
But over an algebraically closed field, this is true for operators on finite-dimensional spaces by the Jordan decomposition: squaring the matrix applies to each Jordan block separately.
answered Jan 7 at 0:08
Ashwin TrisalAshwin Trisal
1,2441516
$endgroup$
$begingroup$
Oh, thanks, and what about the infinite dimensional case on $mathbb{C}$?
$endgroup$
– roi_saumon
Jan 7 at 12:21
add a comment |
$begingroup$
Over a field which is not algebraically closed, this is false. Consider
$$A=begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}in M_2(mathbb R).$$
Then $A^2=-I$, which has only the eigenvalues $-1$, but $A$ has no eigenvalues.
But over an algebraically closed field, this is true for operators on finite-dimensional spaces by the Jordan decomposition: squaring the matrix applies to each Jordan block separately.
answered Jan 7 at 0:08
Ashwin TrisalAshwin Trisal
1,2441516
$endgroup$
$begingroup$
Oh, thanks, and what about the infinite dimensional case on $mathbb{C}$?
$endgroup$
– roi_saumon
Jan 7 at 12:21
add a comment |
$begingroup$
Over a field which is not algebraically closed, this is false. Consider
$$A=begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}in M_2(mathbb R).$$
Then $A^2=-I$, which has only the eigenvalues $-1$, but $A$ has no eigenvalues.
But over an algebraically closed field, this is true for operators on finite-dimensional spaces by the Jordan decomposition: squaring the matrix applies to each Jordan block separately.
answered Jan 7 at 0:08
Ashwin TrisalAshwin Trisal
1,2441516
$endgroup$
Over a field which is not algebraically closed, this is false. Consider
$$A=begin{bmatrix} 0 & -1 \ 1 & 0end{bmatrix}in M_2(mathbb R).$$
Then $A^2=-I$, which has only the eigenvalues $-1$, but $A$ has no eigenvalues.
But over an algebraically closed field, this is true for operators on finite-dimensional spaces by the Jordan decomposition: squaring the matrix applies to each Jordan block separately.
answered Jan 7 at 0:08
Ashwin TrisalAshwin Trisal
1,2441516
answered Jan 7 at 0:08
Ashwin TrisalAshwin Trisal
1,2441516
answered Jan 7 at 0:08
Ashwin TrisalAshwin Trisal
1,2441516
answered Jan 7 at 0:08
Ashwin TrisalAshwin Trisal
1,2441516
1,2441516
$begingroup$
Oh, thanks, and what about the infinite dimensional case on $mathbb{C}$?
$endgroup$
– roi_saumon
Jan 7 at 12:21
add a comment |
$begingroup$
Oh, thanks, and what about the infinite dimensional case on $mathbb{C}$?
$endgroup$
– roi_saumon
Jan 7 at 12:21
$begingroup$
Oh, thanks, and what about the infinite dimensional case on $mathbb{C}$?
$endgroup$
– roi_saumon
Jan 7 at 12:21
$begingroup$
Oh, thanks, and what about the infinite dimensional case on $mathbb{C}$?
$endgroup$
– roi_saumon
Jan 7 at 12:21
add a comment |
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