If an operator $X$ on a Hilbert space satisfies $X^*=-X$, then is $X$ equal to $0$? [closed]
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If a linear operator $X$ on a Hilbert space satisfies $X^*=-X$, then is $X$ equal to $0$? $X^*$ is the adjoint.
operator-theory hilbert-spaces operator-algebras
closed as off-topic by user92646, Rhys Steele, Adrian Keister, Mostafa Ayaz, Isaac Browne Jul 30 at 20:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â Rhys Steele, Adrian Keister, Mostafa Ayaz, Isaac Browne
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If a linear operator $X$ on a Hilbert space satisfies $X^*=-X$, then is $X$ equal to $0$? $X^*$ is the adjoint.
operator-theory hilbert-spaces operator-algebras
closed as off-topic by user92646, Rhys Steele, Adrian Keister, Mostafa Ayaz, Isaac Browne Jul 30 at 20:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â Rhys Steele, Adrian Keister, Mostafa Ayaz, Isaac Browne
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What if the operator is multiplication by the imaginary unit? $i^* = -i$.
â mr_e_man
Jul 30 at 8:18
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up vote
1
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up vote
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down vote
favorite
If a linear operator $X$ on a Hilbert space satisfies $X^*=-X$, then is $X$ equal to $0$? $X^*$ is the adjoint.
operator-theory hilbert-spaces operator-algebras
If a linear operator $X$ on a Hilbert space satisfies $X^*=-X$, then is $X$ equal to $0$? $X^*$ is the adjoint.
operator-theory hilbert-spaces operator-algebras
asked Jul 30 at 8:07
user92646
573310
573310
closed as off-topic by user92646, Rhys Steele, Adrian Keister, Mostafa Ayaz, Isaac Browne Jul 30 at 20:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â Rhys Steele, Adrian Keister, Mostafa Ayaz, Isaac Browne
closed as off-topic by user92646, Rhys Steele, Adrian Keister, Mostafa Ayaz, Isaac Browne Jul 30 at 20:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â Rhys Steele, Adrian Keister, Mostafa Ayaz, Isaac Browne
2
What if the operator is multiplication by the imaginary unit? $i^* = -i$.
â mr_e_man
Jul 30 at 8:18
add a comment |Â
2
What if the operator is multiplication by the imaginary unit? $i^* = -i$.
â mr_e_man
Jul 30 at 8:18
2
2
What if the operator is multiplication by the imaginary unit? $i^* = -i$.
â mr_e_man
Jul 30 at 8:18
What if the operator is multiplication by the imaginary unit? $i^* = -i$.
â mr_e_man
Jul 30 at 8:18
add a comment |Â
1 Answer
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No. Such operators called skew-symmetric. For example the matrix
$X=beginbmatrix0&2&-1\-2&0&-4\1&4&0endbmatrix$
gives a skew-symmetric operator on $mathbb R^3$ (or $mathbb C^3$)
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
No. Such operators called skew-symmetric. For example the matrix
$X=beginbmatrix0&2&-1\-2&0&-4\1&4&0endbmatrix$
gives a skew-symmetric operator on $mathbb R^3$ (or $mathbb C^3$)
add a comment |Â
up vote
4
down vote
No. Such operators called skew-symmetric. For example the matrix
$X=beginbmatrix0&2&-1\-2&0&-4\1&4&0endbmatrix$
gives a skew-symmetric operator on $mathbb R^3$ (or $mathbb C^3$)
add a comment |Â
up vote
4
down vote
up vote
4
down vote
No. Such operators called skew-symmetric. For example the matrix
$X=beginbmatrix0&2&-1\-2&0&-4\1&4&0endbmatrix$
gives a skew-symmetric operator on $mathbb R^3$ (or $mathbb C^3$)
No. Such operators called skew-symmetric. For example the matrix
$X=beginbmatrix0&2&-1\-2&0&-4\1&4&0endbmatrix$
gives a skew-symmetric operator on $mathbb R^3$ (or $mathbb C^3$)
edited Jul 30 at 8:47
answered Jul 30 at 8:13
Fred
37k1237
37k1237
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2
What if the operator is multiplication by the imaginary unit? $i^* = -i$.
â mr_e_man
Jul 30 at 8:18