Sufficient condition for Unitary equivalence












2















Let $A = (a_{ij} )$ and $B = (b_{ij} )$ be similar, and
$sum ^n
_{i,j=1} left|a_{ij}right|^2
=
sum ^n
_{i,j=1} left|b_{ij}right|^2 $.
Then $A$ and $B$ are unitary equivalent.




I have to prove or disprove this statement. I have tried proving this statement in various ways and come to conclusion this statement must be false. But I haven't been able to find any example. The hint given is if $A$ and $B$ are unitary equivalent then $(A+A^*)$ and $(B+B^*)$ are also.



So first can someone tell that if this statement is true or not? Then if it is true then how to prove it otherwise how to find example to disprove it?






matrices vector-spaces eigenvalues-eigenvectors matrix-decomposition







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  • You need to clarify the meaning of unitary equivalence between two square matrices $A$ and $B$. Some authors mean that $B=UAV$ for some unitary matrices $U$ and $V$, but some others simply mean that $A$ and $B$ are unitarily similar.
    – user1551
    Nov 27 at 5:12
















2















Let $A = (a_{ij} )$ and $B = (b_{ij} )$ be similar, and
$sum ^n
_{i,j=1} left|a_{ij}right|^2
=
sum ^n
_{i,j=1} left|b_{ij}right|^2 $.
Then $A$ and $B$ are unitary equivalent.




I have to prove or disprove this statement. I have tried proving this statement in various ways and come to conclusion this statement must be false. But I haven't been able to find any example. The hint given is if $A$ and $B$ are unitary equivalent then $(A+A^*)$ and $(B+B^*)$ are also.



So first can someone tell that if this statement is true or not? Then if it is true then how to prove it otherwise how to find example to disprove it?






matrices vector-spaces eigenvalues-eigenvectors matrix-decomposition







share|cite|improve this question
























  • You need to clarify the meaning of unitary equivalence between two square matrices $A$ and $B$. Some authors mean that $B=UAV$ for some unitary matrices $U$ and $V$, but some others simply mean that $A$ and $B$ are unitarily similar.
    – user1551
    Nov 27 at 5:12














2












2








2








Let $A = (a_{ij} )$ and $B = (b_{ij} )$ be similar, and
$sum ^n
_{i,j=1} left|a_{ij}right|^2
=
sum ^n
_{i,j=1} left|b_{ij}right|^2 $.
Then $A$ and $B$ are unitary equivalent.




I have to prove or disprove this statement. I have tried proving this statement in various ways and come to conclusion this statement must be false. But I haven't been able to find any example. The hint given is if $A$ and $B$ are unitary equivalent then $(A+A^*)$ and $(B+B^*)$ are also.



So first can someone tell that if this statement is true or not? Then if it is true then how to prove it otherwise how to find example to disprove it?






matrices vector-spaces eigenvalues-eigenvectors matrix-decomposition







share|cite|improve this question
















Let $A = (a_{ij} )$ and $B = (b_{ij} )$ be similar, and
$sum ^n
_{i,j=1} left|a_{ij}right|^2
=
sum ^n
_{i,j=1} left|b_{ij}right|^2 $.
Then $A$ and $B$ are unitary equivalent.




I have to prove or disprove this statement. I have tried proving this statement in various ways and come to conclusion this statement must be false. But I haven't been able to find any example. The hint given is if $A$ and $B$ are unitary equivalent then $(A+A^*)$ and $(B+B^*)$ are also.



So first can someone tell that if this statement is true or not? Then if it is true then how to prove it otherwise how to find example to disprove it?






matrices vector-spaces eigenvalues-eigenvectors matrix-decomposition




matrices vector-spaces eigenvalues-eigenvectors matrix-decomposition






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edited Nov 26 at 22:33









Davide Giraudo

125k16150259




125k16150259










asked Nov 18 at 15:07









Jimmy

17212




17212












  • You need to clarify the meaning of unitary equivalence between two square matrices $A$ and $B$. Some authors mean that $B=UAV$ for some unitary matrices $U$ and $V$, but some others simply mean that $A$ and $B$ are unitarily similar.
    – user1551
    Nov 27 at 5:12


















  • You need to clarify the meaning of unitary equivalence between two square matrices $A$ and $B$. Some authors mean that $B=UAV$ for some unitary matrices $U$ and $V$, but some others simply mean that $A$ and $B$ are unitarily similar.
    – user1551
    Nov 27 at 5:12
















You need to clarify the meaning of unitary equivalence between two square matrices $A$ and $B$. Some authors mean that $B=UAV$ for some unitary matrices $U$ and $V$, but some others simply mean that $A$ and $B$ are unitarily similar.
– user1551
Nov 27 at 5:12




You need to clarify the meaning of unitary equivalence between two square matrices $A$ and $B$. Some authors mean that $B=UAV$ for some unitary matrices $U$ and $V$, but some others simply mean that $A$ and $B$ are unitarily similar.
– user1551
Nov 27 at 5:12










2 Answers
2






active

oldest

votes


















3














The answer is no.
Take $A=begin{pmatrix}60& 90& 88\-35& 80& -82\21& 19& -70end{pmatrix}$ and $B=A^T$. ($A$ is random).



Recall that when $A,B$ are real matrices, they are unitarily similar iff they are orthogonally similar.



$A,B$ are clearly similar over $mathbb{C}$ and satisfy the relation above $tr(AA^T)=tr(A^TA)$. Yet, the only solution of the system in $X$



$AX=XB,A^TX=XB^T$ is $0$. Therefore, $A,B$ are not orthogonally similar.






share|cite|improve this answer





























    1














    This already fails over $mathbb C$.



    I suppose that unitary equivalence is what @user1551 means in their comment. As unitary similarity $Rightarrow$ unitary equivalence, demonstrating a counterexample for unitary equivalence will suffice.



    Consider $A=begin{bmatrix}1&1&0\0&0&0\0&0&-1end{bmatrix}$ and $B=begin{bmatrix}1&frac{1}{sqrt{2}}&0\0&0&frac{1}{sqrt{2}}\0&0&-1end{bmatrix}$. They are similar matrices with identical Frobenius norms. However, as
    $$A=begin{bmatrix}1\0\0end{bmatrix}begin{bmatrix}1&1&0end{bmatrix}
    +begin{bmatrix}0\0\1end{bmatrix}begin{bmatrix}0&0&-1end{bmatrix},
    $$
    $$B=begin{bmatrix}1\0\0end{bmatrix}begin{bmatrix}1&frac{1}{sqrt{2}}&0end{bmatrix}
    +begin{bmatrix}0\frac{1}{sqrt{2}}\-1end{bmatrix}begin{bmatrix}0&0&1end{bmatrix},
    $$
    the singular values of $A$ are $sqrt{2},1,0$ and the singular values of $B$ are $sqrt{frac{3}{2}},sqrt{frac{3}{2}},0$. Since $A$ and $B$ have different singular values, they are not unitarily equivalent and not unitarily similar to each other.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      3














      The answer is no.
      Take $A=begin{pmatrix}60& 90& 88\-35& 80& -82\21& 19& -70end{pmatrix}$ and $B=A^T$. ($A$ is random).



      Recall that when $A,B$ are real matrices, they are unitarily similar iff they are orthogonally similar.



      $A,B$ are clearly similar over $mathbb{C}$ and satisfy the relation above $tr(AA^T)=tr(A^TA)$. Yet, the only solution of the system in $X$



      $AX=XB,A^TX=XB^T$ is $0$. Therefore, $A,B$ are not orthogonally similar.






      share|cite|improve this answer


























        3














        The answer is no.
        Take $A=begin{pmatrix}60& 90& 88\-35& 80& -82\21& 19& -70end{pmatrix}$ and $B=A^T$. ($A$ is random).



        Recall that when $A,B$ are real matrices, they are unitarily similar iff they are orthogonally similar.



        $A,B$ are clearly similar over $mathbb{C}$ and satisfy the relation above $tr(AA^T)=tr(A^TA)$. Yet, the only solution of the system in $X$



        $AX=XB,A^TX=XB^T$ is $0$. Therefore, $A,B$ are not orthogonally similar.






        share|cite|improve this answer
























          3












          3








          3






          The answer is no.
          Take $A=begin{pmatrix}60& 90& 88\-35& 80& -82\21& 19& -70end{pmatrix}$ and $B=A^T$. ($A$ is random).



          Recall that when $A,B$ are real matrices, they are unitarily similar iff they are orthogonally similar.



          $A,B$ are clearly similar over $mathbb{C}$ and satisfy the relation above $tr(AA^T)=tr(A^TA)$. Yet, the only solution of the system in $X$



          $AX=XB,A^TX=XB^T$ is $0$. Therefore, $A,B$ are not orthogonally similar.






          share|cite|improve this answer












          The answer is no.
          Take $A=begin{pmatrix}60& 90& 88\-35& 80& -82\21& 19& -70end{pmatrix}$ and $B=A^T$. ($A$ is random).



          Recall that when $A,B$ are real matrices, they are unitarily similar iff they are orthogonally similar.



          $A,B$ are clearly similar over $mathbb{C}$ and satisfy the relation above $tr(AA^T)=tr(A^TA)$. Yet, the only solution of the system in $X$



          $AX=XB,A^TX=XB^T$ is $0$. Therefore, $A,B$ are not orthogonally similar.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 at 0:03









          loup blanc

          22.5k21750




          22.5k21750























              1














              This already fails over $mathbb C$.



              I suppose that unitary equivalence is what @user1551 means in their comment. As unitary similarity $Rightarrow$ unitary equivalence, demonstrating a counterexample for unitary equivalence will suffice.



              Consider $A=begin{bmatrix}1&1&0\0&0&0\0&0&-1end{bmatrix}$ and $B=begin{bmatrix}1&frac{1}{sqrt{2}}&0\0&0&frac{1}{sqrt{2}}\0&0&-1end{bmatrix}$. They are similar matrices with identical Frobenius norms. However, as
              $$A=begin{bmatrix}1\0\0end{bmatrix}begin{bmatrix}1&1&0end{bmatrix}
              +begin{bmatrix}0\0\1end{bmatrix}begin{bmatrix}0&0&-1end{bmatrix},
              $$
              $$B=begin{bmatrix}1\0\0end{bmatrix}begin{bmatrix}1&frac{1}{sqrt{2}}&0end{bmatrix}
              +begin{bmatrix}0\frac{1}{sqrt{2}}\-1end{bmatrix}begin{bmatrix}0&0&1end{bmatrix},
              $$
              the singular values of $A$ are $sqrt{2},1,0$ and the singular values of $B$ are $sqrt{frac{3}{2}},sqrt{frac{3}{2}},0$. Since $A$ and $B$ have different singular values, they are not unitarily equivalent and not unitarily similar to each other.






              share|cite|improve this answer


























                1














                This already fails over $mathbb C$.



                I suppose that unitary equivalence is what @user1551 means in their comment. As unitary similarity $Rightarrow$ unitary equivalence, demonstrating a counterexample for unitary equivalence will suffice.



                Consider $A=begin{bmatrix}1&1&0\0&0&0\0&0&-1end{bmatrix}$ and $B=begin{bmatrix}1&frac{1}{sqrt{2}}&0\0&0&frac{1}{sqrt{2}}\0&0&-1end{bmatrix}$. They are similar matrices with identical Frobenius norms. However, as
                $$A=begin{bmatrix}1\0\0end{bmatrix}begin{bmatrix}1&1&0end{bmatrix}
                +begin{bmatrix}0\0\1end{bmatrix}begin{bmatrix}0&0&-1end{bmatrix},
                $$
                $$B=begin{bmatrix}1\0\0end{bmatrix}begin{bmatrix}1&frac{1}{sqrt{2}}&0end{bmatrix}
                +begin{bmatrix}0\frac{1}{sqrt{2}}\-1end{bmatrix}begin{bmatrix}0&0&1end{bmatrix},
                $$
                the singular values of $A$ are $sqrt{2},1,0$ and the singular values of $B$ are $sqrt{frac{3}{2}},sqrt{frac{3}{2}},0$. Since $A$ and $B$ have different singular values, they are not unitarily equivalent and not unitarily similar to each other.






                share|cite|improve this answer
























                  1












                  1








                  1






                  This already fails over $mathbb C$.



                  I suppose that unitary equivalence is what @user1551 means in their comment. As unitary similarity $Rightarrow$ unitary equivalence, demonstrating a counterexample for unitary equivalence will suffice.



                  Consider $A=begin{bmatrix}1&1&0\0&0&0\0&0&-1end{bmatrix}$ and $B=begin{bmatrix}1&frac{1}{sqrt{2}}&0\0&0&frac{1}{sqrt{2}}\0&0&-1end{bmatrix}$. They are similar matrices with identical Frobenius norms. However, as
                  $$A=begin{bmatrix}1\0\0end{bmatrix}begin{bmatrix}1&1&0end{bmatrix}
                  +begin{bmatrix}0\0\1end{bmatrix}begin{bmatrix}0&0&-1end{bmatrix},
                  $$
                  $$B=begin{bmatrix}1\0\0end{bmatrix}begin{bmatrix}1&frac{1}{sqrt{2}}&0end{bmatrix}
                  +begin{bmatrix}0\frac{1}{sqrt{2}}\-1end{bmatrix}begin{bmatrix}0&0&1end{bmatrix},
                  $$
                  the singular values of $A$ are $sqrt{2},1,0$ and the singular values of $B$ are $sqrt{frac{3}{2}},sqrt{frac{3}{2}},0$. Since $A$ and $B$ have different singular values, they are not unitarily equivalent and not unitarily similar to each other.






                  share|cite|improve this answer












                  This already fails over $mathbb C$.



                  I suppose that unitary equivalence is what @user1551 means in their comment. As unitary similarity $Rightarrow$ unitary equivalence, demonstrating a counterexample for unitary equivalence will suffice.



                  Consider $A=begin{bmatrix}1&1&0\0&0&0\0&0&-1end{bmatrix}$ and $B=begin{bmatrix}1&frac{1}{sqrt{2}}&0\0&0&frac{1}{sqrt{2}}\0&0&-1end{bmatrix}$. They are similar matrices with identical Frobenius norms. However, as
                  $$A=begin{bmatrix}1\0\0end{bmatrix}begin{bmatrix}1&1&0end{bmatrix}
                  +begin{bmatrix}0\0\1end{bmatrix}begin{bmatrix}0&0&-1end{bmatrix},
                  $$
                  $$B=begin{bmatrix}1\0\0end{bmatrix}begin{bmatrix}1&frac{1}{sqrt{2}}&0end{bmatrix}
                  +begin{bmatrix}0\frac{1}{sqrt{2}}\-1end{bmatrix}begin{bmatrix}0&0&1end{bmatrix},
                  $$
                  the singular values of $A$ are $sqrt{2},1,0$ and the singular values of $B$ are $sqrt{frac{3}{2}},sqrt{frac{3}{2}},0$. Since $A$ and $B$ have different singular values, they are not unitarily equivalent and not unitarily similar to each other.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 27 at 6:39









                  William McGonagall

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