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.







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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." – Rhys Steele, Adrian Keister, Mostafa Ayaz, Isaac Browne
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    What if the operator is multiplication by the imaginary unit? $i^* = -i$.
    – mr_e_man
    Jul 30 at 8:18














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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.







share|cite|improve this question











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." – Rhys Steele, Adrian Keister, Mostafa Ayaz, Isaac Browne
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    What if the operator is multiplication by the imaginary unit? $i^* = -i$.
    – mr_e_man
    Jul 30 at 8:18












up vote
1
down vote

favorite









up vote
1
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.







share|cite|improve this question











If a linear operator $X$ on a Hilbert space satisfies $X^*=-X$, then is $X$ equal to $0$? $X^*$ is the adjoint.









share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 30 at 8:07









user92646

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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." – Rhys Steele, Adrian Keister, Mostafa Ayaz, Isaac Browne
If this question can be reworded to fit the rules in the help center, please edit the question.




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." – Rhys Steele, Adrian Keister, Mostafa Ayaz, Isaac Browne
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 2




    What if the operator is multiplication by the imaginary unit? $i^* = -i$.
    – mr_e_man
    Jul 30 at 8:18












  • 2




    What if the operator is multiplication by the imaginary unit? $i^* = -i$.
    – mr_e_man
    Jul 30 at 8:18







2




2




What if the operator is multiplication by the imaginary unit? $i^* = -i$.
– mr_e_man
Jul 30 at 8:18




What if the operator is multiplication by the imaginary unit? $i^* = -i$.
– mr_e_man
Jul 30 at 8:18










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

    votes








    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 $mathbb C^3$)






    share|cite|improve this answer



























      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 $mathbb C^3$)






      share|cite|improve this answer

























        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 $mathbb C^3$)






        share|cite|improve this answer















        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 $mathbb C^3$)







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 30 at 8:47


























        answered Jul 30 at 8:13









        Fred

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        37k1237












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