Does a square matrix always commute with its transpose? [duplicate]












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  • Commutativity of matrix and its transpose

    2 answers




Does a square matrix always commute with its transpose? which means If M is a square matrix, then $M^TM=MM^T$? I just couldn't give a counterexample.






linear-algebra







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 Jan 2 at 23:30


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    See en.wikipedia.org/wiki/Normal_matrix
    $endgroup$
    – Kavi Rama Murthy
    Jan 2 at 23:28










  • $begingroup$
    HINT: Test with some non-diagonalizable matrix.
    $endgroup$
    – Giuseppe Negro
    Jan 2 at 23:28
















1












$begingroup$



This question already has an answer here:




  • Commutativity of matrix and its transpose

    2 answers




Does a square matrix always commute with its transpose? which means If M is a square matrix, then $M^TM=MM^T$? I just couldn't give a counterexample.






linear-algebra







share|cite|improve this question











$endgroup$



marked as duplicate by amWhy linear-algebra
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 Jan 2 at 23:30


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    See en.wikipedia.org/wiki/Normal_matrix
    $endgroup$
    – Kavi Rama Murthy
    Jan 2 at 23:28










  • $begingroup$
    HINT: Test with some non-diagonalizable matrix.
    $endgroup$
    – Giuseppe Negro
    Jan 2 at 23:28














1












1








1





$begingroup$



This question already has an answer here:




  • Commutativity of matrix and its transpose

    2 answers




Does a square matrix always commute with its transpose? which means If M is a square matrix, then $M^TM=MM^T$? I just couldn't give a counterexample.






linear-algebra







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Commutativity of matrix and its transpose

    2 answers




Does a square matrix always commute with its transpose? which means If M is a square matrix, then $M^TM=MM^T$? I just couldn't give a counterexample.





This question already has an answer here:




  • Commutativity of matrix and its transpose

    2 answers







linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 21:46









Rohit Pandey

1,4171023




1,4171023










asked Jan 2 at 23:24









Yibei HeYibei He

3139




3139




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 Jan 2 at 23:30


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 Jan 2 at 23:30


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    See en.wikipedia.org/wiki/Normal_matrix
    $endgroup$
    – Kavi Rama Murthy
    Jan 2 at 23:28










  • $begingroup$
    HINT: Test with some non-diagonalizable matrix.
    $endgroup$
    – Giuseppe Negro
    Jan 2 at 23:28


















  • $begingroup$
    See en.wikipedia.org/wiki/Normal_matrix
    $endgroup$
    – Kavi Rama Murthy
    Jan 2 at 23:28










  • $begingroup$
    HINT: Test with some non-diagonalizable matrix.
    $endgroup$
    – Giuseppe Negro
    Jan 2 at 23:28
















$begingroup$
See en.wikipedia.org/wiki/Normal_matrix
$endgroup$
– Kavi Rama Murthy
Jan 2 at 23:28




$begingroup$
See en.wikipedia.org/wiki/Normal_matrix
$endgroup$
– Kavi Rama Murthy
Jan 2 at 23:28












$begingroup$
HINT: Test with some non-diagonalizable matrix.
$endgroup$
– Giuseppe Negro
Jan 2 at 23:28




$begingroup$
HINT: Test with some non-diagonalizable matrix.
$endgroup$
– Giuseppe Negro
Jan 2 at 23:28










3 Answers
3






active

oldest

votes


















4












$begingroup$

Counterexample:
$$
M = pmatrix{0&1\0&0}
$$
More generally, an upper triangular matrix will commute with its transpose if and only if it is diagonal. If $M$ commutes with its transpose, it is called a "normal" matrix.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Here is a simple counter example: Let
    $
    M = left( begin{matrix} 0 & 1 \ 0 & 0 end{matrix} right),
    $
    then $M M^T = left( begin{matrix} 1 & 0 \ 0 & 0 end{matrix} right)$
    and $M^T M = left( begin{matrix} 0 & 0 \ 0 & 1 end{matrix} right)$.



    A real matrix $M$ which satisfies $M^T M = M M^T$ is called normal.
    A general version of the Spectral Theorem states that $M$ is normal if and only
    if there is a diagonal matrix $D$ and a unitary matrix $P$ such that
    $M = P D P^{-1}$. (Here $D$ and $P$ may be complex.)






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      This version of the spectral theorem requires that we allow $P$ and $D$ to be complex, and that $P$ is unitary as a complex matrix.
      $endgroup$
      – Omnomnomnom
      Jan 3 at 0:02










    • $begingroup$
      Corrected, thanks!
      $endgroup$
      – 1213
      Jan 3 at 14:28



















    0












    $begingroup$

    No. Here is a counter example (python code):



    import numpy as np
    
m = np.array([[1,2,3],[1.1,0.1,0.2],[1,1,100]])
    
np.dot(m,m.T)[0,0]
    
np.dot(m.T,m)[0,0]


    The first entry is 14 and the second is 3.21/






    share|cite|improve this answer









    $endgroup$




















      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Counterexample:
      $$
      M = pmatrix{0&1\0&0}
      $$
      More generally, an upper triangular matrix will commute with its transpose if and only if it is diagonal. If $M$ commutes with its transpose, it is called a "normal" matrix.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Counterexample:
        $$
        M = pmatrix{0&1\0&0}
        $$
        More generally, an upper triangular matrix will commute with its transpose if and only if it is diagonal. If $M$ commutes with its transpose, it is called a "normal" matrix.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Counterexample:
          $$
          M = pmatrix{0&1\0&0}
          $$
          More generally, an upper triangular matrix will commute with its transpose if and only if it is diagonal. If $M$ commutes with its transpose, it is called a "normal" matrix.






          share|cite|improve this answer









          $endgroup$



          Counterexample:
          $$
          M = pmatrix{0&1\0&0}
          $$
          More generally, an upper triangular matrix will commute with its transpose if and only if it is diagonal. If $M$ commutes with its transpose, it is called a "normal" matrix.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 23:29









          OmnomnomnomOmnomnomnom

          128k791185




          128k791185























              2












              $begingroup$

              Here is a simple counter example: Let
              $
              M = left( begin{matrix} 0 & 1 \ 0 & 0 end{matrix} right),
              $
              then $M M^T = left( begin{matrix} 1 & 0 \ 0 & 0 end{matrix} right)$
              and $M^T M = left( begin{matrix} 0 & 0 \ 0 & 1 end{matrix} right)$.



              A real matrix $M$ which satisfies $M^T M = M M^T$ is called normal.
              A general version of the Spectral Theorem states that $M$ is normal if and only
              if there is a diagonal matrix $D$ and a unitary matrix $P$ such that
              $M = P D P^{-1}$. (Here $D$ and $P$ may be complex.)






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                This version of the spectral theorem requires that we allow $P$ and $D$ to be complex, and that $P$ is unitary as a complex matrix.
                $endgroup$
                – Omnomnomnom
                Jan 3 at 0:02










              • $begingroup$
                Corrected, thanks!
                $endgroup$
                – 1213
                Jan 3 at 14:28
















              2












              $begingroup$

              Here is a simple counter example: Let
              $
              M = left( begin{matrix} 0 & 1 \ 0 & 0 end{matrix} right),
              $
              then $M M^T = left( begin{matrix} 1 & 0 \ 0 & 0 end{matrix} right)$
              and $M^T M = left( begin{matrix} 0 & 0 \ 0 & 1 end{matrix} right)$.



              A real matrix $M$ which satisfies $M^T M = M M^T$ is called normal.
              A general version of the Spectral Theorem states that $M$ is normal if and only
              if there is a diagonal matrix $D$ and a unitary matrix $P$ such that
              $M = P D P^{-1}$. (Here $D$ and $P$ may be complex.)






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                This version of the spectral theorem requires that we allow $P$ and $D$ to be complex, and that $P$ is unitary as a complex matrix.
                $endgroup$
                – Omnomnomnom
                Jan 3 at 0:02










              • $begingroup$
                Corrected, thanks!
                $endgroup$
                – 1213
                Jan 3 at 14:28














              2












              2








              2





              $begingroup$

              Here is a simple counter example: Let
              $
              M = left( begin{matrix} 0 & 1 \ 0 & 0 end{matrix} right),
              $
              then $M M^T = left( begin{matrix} 1 & 0 \ 0 & 0 end{matrix} right)$
              and $M^T M = left( begin{matrix} 0 & 0 \ 0 & 1 end{matrix} right)$.



              A real matrix $M$ which satisfies $M^T M = M M^T$ is called normal.
              A general version of the Spectral Theorem states that $M$ is normal if and only
              if there is a diagonal matrix $D$ and a unitary matrix $P$ such that
              $M = P D P^{-1}$. (Here $D$ and $P$ may be complex.)






              share|cite|improve this answer











              $endgroup$



              Here is a simple counter example: Let
              $
              M = left( begin{matrix} 0 & 1 \ 0 & 0 end{matrix} right),
              $
              then $M M^T = left( begin{matrix} 1 & 0 \ 0 & 0 end{matrix} right)$
              and $M^T M = left( begin{matrix} 0 & 0 \ 0 & 1 end{matrix} right)$.



              A real matrix $M$ which satisfies $M^T M = M M^T$ is called normal.
              A general version of the Spectral Theorem states that $M$ is normal if and only
              if there is a diagonal matrix $D$ and a unitary matrix $P$ such that
              $M = P D P^{-1}$. (Here $D$ and $P$ may be complex.)







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 3 at 14:27

























              answered Jan 2 at 23:36









              12131213

              1967




              1967












              • $begingroup$
                This version of the spectral theorem requires that we allow $P$ and $D$ to be complex, and that $P$ is unitary as a complex matrix.
                $endgroup$
                – Omnomnomnom
                Jan 3 at 0:02










              • $begingroup$
                Corrected, thanks!
                $endgroup$
                – 1213
                Jan 3 at 14:28


















              • $begingroup$
                This version of the spectral theorem requires that we allow $P$ and $D$ to be complex, and that $P$ is unitary as a complex matrix.
                $endgroup$
                – Omnomnomnom
                Jan 3 at 0:02










              • $begingroup$
                Corrected, thanks!
                $endgroup$
                – 1213
                Jan 3 at 14:28
















              $begingroup$
              This version of the spectral theorem requires that we allow $P$ and $D$ to be complex, and that $P$ is unitary as a complex matrix.
              $endgroup$
              – Omnomnomnom
              Jan 3 at 0:02




              $begingroup$
              This version of the spectral theorem requires that we allow $P$ and $D$ to be complex, and that $P$ is unitary as a complex matrix.
              $endgroup$
              – Omnomnomnom
              Jan 3 at 0:02












              $begingroup$
              Corrected, thanks!
              $endgroup$
              – 1213
              Jan 3 at 14:28




              $begingroup$
              Corrected, thanks!
              $endgroup$
              – 1213
              Jan 3 at 14:28











              0












              $begingroup$

              No. Here is a counter example (python code):



              import numpy as np
              
m = np.array([[1,2,3],[1.1,0.1,0.2],[1,1,100]])
              
np.dot(m,m.T)[0,0]
              
np.dot(m.T,m)[0,0]


              The first entry is 14 and the second is 3.21/






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                No. Here is a counter example (python code):



                import numpy as np
                
m = np.array([[1,2,3],[1.1,0.1,0.2],[1,1,100]])
                
np.dot(m,m.T)[0,0]
                
np.dot(m.T,m)[0,0]


                The first entry is 14 and the second is 3.21/






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  No. Here is a counter example (python code):



                  import numpy as np
                  
m = np.array([[1,2,3],[1.1,0.1,0.2],[1,1,100]])
                  
np.dot(m,m.T)[0,0]
                  
np.dot(m.T,m)[0,0]


                  The first entry is 14 and the second is 3.21/






                  share|cite|improve this answer









                  $endgroup$



                  No. Here is a counter example (python code):



                  import numpy as np
                  
m = np.array([[1,2,3],[1.1,0.1,0.2],[1,1,100]])
                  
np.dot(m,m.T)[0,0]
                  
np.dot(m.T,m)[0,0]


                  The first entry is 14 and the second is 3.21/







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 2 at 23:29









                  Rohit PandeyRohit Pandey

                  1,4171023




                  1,4171023















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