spectral theorem - why does it only apply to a symmetric matrix?

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The real spectral theorem asserts that any symmetric matrix can be decomposed into a composition of rotations, reflections and scaling. Why can't a non-symmetric matrix be represented as such? Are there other types of operations other than rotations, reflections and scaling that explain why non-symmetric matrices are left out of this theorem? I understand that the singular value decomposition says that you can decompose a matrix into 3 other matrices - but the matrices are complex and I don't know if you can interpret a complex matrix as a rotation etc. I apologise if this is a silly question and please let me know if the question requires further clarification.




linear-algebra matrices rotations symmetric-matrices reflection




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  • 1




    What happened to your previous question?
    – Lord Shark the Unknown
    Aug 24 at 9:03






  • 1




    "The real spectral theorem asserts that any non-symmetric matrix can be decomposed into a composition of rotations, reflections and scaling." That doesn't appear to be any spectral theorem that I know.
    – Lord Shark the Unknown
    Aug 24 at 9:05






  • 1




    I think you need to remove some of your "non-" to make this question correct.
    – Arthur
    Aug 24 at 9:05






  • 2




    You could just have undeleted your previous question. Anyway, all the matrices in the singular value decomposition are still real, not complex. (Now I suppose you'll delete your question again...)
    – Rahul
    Aug 24 at 9:09







  • 2




    And, rotation matrices are not symmetric so I don't quite see what you mean. My educated guess is that you want to apply the fact that a symmetric real matrix has an orthonormal system of eigenvectors. But, then you are really conjugating a diagonal matrix (= a composition of scalings) with a rotation rather than just taking arbitrary compositions of all those transformations.
    – Jyrki Lahtonen
    Aug 24 at 9:21















up vote
1
down vote

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The real spectral theorem asserts that any symmetric matrix can be decomposed into a composition of rotations, reflections and scaling. Why can't a non-symmetric matrix be represented as such? Are there other types of operations other than rotations, reflections and scaling that explain why non-symmetric matrices are left out of this theorem? I understand that the singular value decomposition says that you can decompose a matrix into 3 other matrices - but the matrices are complex and I don't know if you can interpret a complex matrix as a rotation etc. I apologise if this is a silly question and please let me know if the question requires further clarification.




linear-algebra matrices rotations symmetric-matrices reflection




share|cite|improve this question


















  • 1




    What happened to your previous question?
    – Lord Shark the Unknown
    Aug 24 at 9:03






  • 1




    "The real spectral theorem asserts that any non-symmetric matrix can be decomposed into a composition of rotations, reflections and scaling." That doesn't appear to be any spectral theorem that I know.
    – Lord Shark the Unknown
    Aug 24 at 9:05






  • 1




    I think you need to remove some of your "non-" to make this question correct.
    – Arthur
    Aug 24 at 9:05






  • 2




    You could just have undeleted your previous question. Anyway, all the matrices in the singular value decomposition are still real, not complex. (Now I suppose you'll delete your question again...)
    – Rahul
    Aug 24 at 9:09







  • 2




    And, rotation matrices are not symmetric so I don't quite see what you mean. My educated guess is that you want to apply the fact that a symmetric real matrix has an orthonormal system of eigenvectors. But, then you are really conjugating a diagonal matrix (= a composition of scalings) with a rotation rather than just taking arbitrary compositions of all those transformations.
    – Jyrki Lahtonen
    Aug 24 at 9:21













up vote
1
down vote

favorite









up vote
1
down vote

favorite











The real spectral theorem asserts that any symmetric matrix can be decomposed into a composition of rotations, reflections and scaling. Why can't a non-symmetric matrix be represented as such? Are there other types of operations other than rotations, reflections and scaling that explain why non-symmetric matrices are left out of this theorem? I understand that the singular value decomposition says that you can decompose a matrix into 3 other matrices - but the matrices are complex and I don't know if you can interpret a complex matrix as a rotation etc. I apologise if this is a silly question and please let me know if the question requires further clarification.




linear-algebra matrices rotations symmetric-matrices reflection




share|cite|improve this question














The real spectral theorem asserts that any symmetric matrix can be decomposed into a composition of rotations, reflections and scaling. Why can't a non-symmetric matrix be represented as such? Are there other types of operations other than rotations, reflections and scaling that explain why non-symmetric matrices are left out of this theorem? I understand that the singular value decomposition says that you can decompose a matrix into 3 other matrices - but the matrices are complex and I don't know if you can interpret a complex matrix as a rotation etc. I apologise if this is a silly question and please let me know if the question requires further clarification.





linear-algebra matrices rotations symmetric-matrices reflection





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 24 at 9:16









Rahul

32.5k362159




32.5k362159










asked Aug 24 at 9:01









Christian

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  • 1




    What happened to your previous question?
    – Lord Shark the Unknown
    Aug 24 at 9:03






  • 1




    "The real spectral theorem asserts that any non-symmetric matrix can be decomposed into a composition of rotations, reflections and scaling." That doesn't appear to be any spectral theorem that I know.
    – Lord Shark the Unknown
    Aug 24 at 9:05






  • 1




    I think you need to remove some of your "non-" to make this question correct.
    – Arthur
    Aug 24 at 9:05






  • 2




    You could just have undeleted your previous question. Anyway, all the matrices in the singular value decomposition are still real, not complex. (Now I suppose you'll delete your question again...)
    – Rahul
    Aug 24 at 9:09







  • 2




    And, rotation matrices are not symmetric so I don't quite see what you mean. My educated guess is that you want to apply the fact that a symmetric real matrix has an orthonormal system of eigenvectors. But, then you are really conjugating a diagonal matrix (= a composition of scalings) with a rotation rather than just taking arbitrary compositions of all those transformations.
    – Jyrki Lahtonen
    Aug 24 at 9:21













  • 1




    What happened to your previous question?
    – Lord Shark the Unknown
    Aug 24 at 9:03






  • 1




    "The real spectral theorem asserts that any non-symmetric matrix can be decomposed into a composition of rotations, reflections and scaling." That doesn't appear to be any spectral theorem that I know.
    – Lord Shark the Unknown
    Aug 24 at 9:05






  • 1




    I think you need to remove some of your "non-" to make this question correct.
    – Arthur
    Aug 24 at 9:05






  • 2




    You could just have undeleted your previous question. Anyway, all the matrices in the singular value decomposition are still real, not complex. (Now I suppose you'll delete your question again...)
    – Rahul
    Aug 24 at 9:09







  • 2




    And, rotation matrices are not symmetric so I don't quite see what you mean. My educated guess is that you want to apply the fact that a symmetric real matrix has an orthonormal system of eigenvectors. But, then you are really conjugating a diagonal matrix (= a composition of scalings) with a rotation rather than just taking arbitrary compositions of all those transformations.
    – Jyrki Lahtonen
    Aug 24 at 9:21








1




1




What happened to your previous question?
– Lord Shark the Unknown
Aug 24 at 9:03




What happened to your previous question?
– Lord Shark the Unknown
Aug 24 at 9:03




1




1




"The real spectral theorem asserts that any non-symmetric matrix can be decomposed into a composition of rotations, reflections and scaling." That doesn't appear to be any spectral theorem that I know.
– Lord Shark the Unknown
Aug 24 at 9:05




"The real spectral theorem asserts that any non-symmetric matrix can be decomposed into a composition of rotations, reflections and scaling." That doesn't appear to be any spectral theorem that I know.
– Lord Shark the Unknown
Aug 24 at 9:05




1




1




I think you need to remove some of your "non-" to make this question correct.
– Arthur
Aug 24 at 9:05




I think you need to remove some of your "non-" to make this question correct.
– Arthur
Aug 24 at 9:05




2




2




You could just have undeleted your previous question. Anyway, all the matrices in the singular value decomposition are still real, not complex. (Now I suppose you'll delete your question again...)
– Rahul
Aug 24 at 9:09





You could just have undeleted your previous question. Anyway, all the matrices in the singular value decomposition are still real, not complex. (Now I suppose you'll delete your question again...)
– Rahul
Aug 24 at 9:09





2




2




And, rotation matrices are not symmetric so I don't quite see what you mean. My educated guess is that you want to apply the fact that a symmetric real matrix has an orthonormal system of eigenvectors. But, then you are really conjugating a diagonal matrix (= a composition of scalings) with a rotation rather than just taking arbitrary compositions of all those transformations.
– Jyrki Lahtonen
Aug 24 at 9:21





And, rotation matrices are not symmetric so I don't quite see what you mean. My educated guess is that you want to apply the fact that a symmetric real matrix has an orthonormal system of eigenvectors. But, then you are really conjugating a diagonal matrix (= a composition of scalings) with a rotation rather than just taking arbitrary compositions of all those transformations.
– Jyrki Lahtonen
Aug 24 at 9:21
















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