Connection between selfadjoint and normal matices












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$begingroup$



Let $A, B in mathbb{C}^{ntimes n}$ be selfadjoint, such that $[A,B] := AB − BA = 0$ . Show that $C := A + iB$ is normal matrix.




Could someone give me a hint on this problem ? I think that as selfadjoint matrices are also normal, I could first prove that $A$ and $B$ are normal, so therefore $A$ + $iB$ gives us a normal matrix. However I am not sure if that is the right way to prove that.






linear-algebra matrices self-adjoint-operators







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  • 3




    $begingroup$
    Hint: Just multiply $CC^*$ and $C^*C$ out, and compare. (No, the normality of two matrices does not automatically imply the normality of their sum.)
    $endgroup$
    – darij grinberg
    Jan 11 at 16:01












  • $begingroup$
    (A+iB)(A* + i B*) = (A*+iB*)(A+i B) But does that prove that $C$ is normal?
    $endgroup$
    – Kai
    Jan 11 at 16:07






  • 1




    $begingroup$
    You are incorrectly computing $C^ast$; keep in mind that $i$ becomes $overline{i} = -i$.
    $endgroup$
    – darij grinberg
    Jan 11 at 16:09






  • 2




    $begingroup$
    Also, recall that $C$ is normal if and only if $CC^ast = C^ast C$. This is the very definition of "normal".
    $endgroup$
    – darij grinberg
    Jan 11 at 16:09
















0












$begingroup$



Let $A, B in mathbb{C}^{ntimes n}$ be selfadjoint, such that $[A,B] := AB − BA = 0$ . Show that $C := A + iB$ is normal matrix.




Could someone give me a hint on this problem ? I think that as selfadjoint matrices are also normal, I could first prove that $A$ and $B$ are normal, so therefore $A$ + $iB$ gives us a normal matrix. However I am not sure if that is the right way to prove that.






linear-algebra matrices self-adjoint-operators







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Hint: Just multiply $CC^*$ and $C^*C$ out, and compare. (No, the normality of two matrices does not automatically imply the normality of their sum.)
    $endgroup$
    – darij grinberg
    Jan 11 at 16:01












  • $begingroup$
    (A+iB)(A* + i B*) = (A*+iB*)(A+i B) But does that prove that $C$ is normal?
    $endgroup$
    – Kai
    Jan 11 at 16:07






  • 1




    $begingroup$
    You are incorrectly computing $C^ast$; keep in mind that $i$ becomes $overline{i} = -i$.
    $endgroup$
    – darij grinberg
    Jan 11 at 16:09






  • 2




    $begingroup$
    Also, recall that $C$ is normal if and only if $CC^ast = C^ast C$. This is the very definition of "normal".
    $endgroup$
    – darij grinberg
    Jan 11 at 16:09














0












0








0





$begingroup$



Let $A, B in mathbb{C}^{ntimes n}$ be selfadjoint, such that $[A,B] := AB − BA = 0$ . Show that $C := A + iB$ is normal matrix.




Could someone give me a hint on this problem ? I think that as selfadjoint matrices are also normal, I could first prove that $A$ and $B$ are normal, so therefore $A$ + $iB$ gives us a normal matrix. However I am not sure if that is the right way to prove that.






linear-algebra matrices self-adjoint-operators







share|cite|improve this question











$endgroup$





Let $A, B in mathbb{C}^{ntimes n}$ be selfadjoint, such that $[A,B] := AB − BA = 0$ . Show that $C := A + iB$ is normal matrix.




Could someone give me a hint on this problem ? I think that as selfadjoint matrices are also normal, I could first prove that $A$ and $B$ are normal, so therefore $A$ + $iB$ gives us a normal matrix. However I am not sure if that is the right way to prove that.






linear-algebra matrices self-adjoint-operators




linear-algebra matrices self-adjoint-operators






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share|cite|improve this question













share|cite|improve this question




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edited Jan 11 at 18:15









Viktor Glombik

782527




782527










asked Jan 11 at 15:45









KaiKai

446




446








  • 3




    $begingroup$
    Hint: Just multiply $CC^*$ and $C^*C$ out, and compare. (No, the normality of two matrices does not automatically imply the normality of their sum.)
    $endgroup$
    – darij grinberg
    Jan 11 at 16:01












  • $begingroup$
    (A+iB)(A* + i B*) = (A*+iB*)(A+i B) But does that prove that $C$ is normal?
    $endgroup$
    – Kai
    Jan 11 at 16:07






  • 1




    $begingroup$
    You are incorrectly computing $C^ast$; keep in mind that $i$ becomes $overline{i} = -i$.
    $endgroup$
    – darij grinberg
    Jan 11 at 16:09






  • 2




    $begingroup$
    Also, recall that $C$ is normal if and only if $CC^ast = C^ast C$. This is the very definition of "normal".
    $endgroup$
    – darij grinberg
    Jan 11 at 16:09














  • 3




    $begingroup$
    Hint: Just multiply $CC^*$ and $C^*C$ out, and compare. (No, the normality of two matrices does not automatically imply the normality of their sum.)
    $endgroup$
    – darij grinberg
    Jan 11 at 16:01












  • $begingroup$
    (A+iB)(A* + i B*) = (A*+iB*)(A+i B) But does that prove that $C$ is normal?
    $endgroup$
    – Kai
    Jan 11 at 16:07






  • 1




    $begingroup$
    You are incorrectly computing $C^ast$; keep in mind that $i$ becomes $overline{i} = -i$.
    $endgroup$
    – darij grinberg
    Jan 11 at 16:09






  • 2




    $begingroup$
    Also, recall that $C$ is normal if and only if $CC^ast = C^ast C$. This is the very definition of "normal".
    $endgroup$
    – darij grinberg
    Jan 11 at 16:09








3




3




$begingroup$
Hint: Just multiply $CC^*$ and $C^*C$ out, and compare. (No, the normality of two matrices does not automatically imply the normality of their sum.)
$endgroup$
– darij grinberg
Jan 11 at 16:01






$begingroup$
Hint: Just multiply $CC^*$ and $C^*C$ out, and compare. (No, the normality of two matrices does not automatically imply the normality of their sum.)
$endgroup$
– darij grinberg
Jan 11 at 16:01














$begingroup$
(A+iB)(A* + i B*) = (A*+iB*)(A+i B) But does that prove that $C$ is normal?
$endgroup$
– Kai
Jan 11 at 16:07




$begingroup$
(A+iB)(A* + i B*) = (A*+iB*)(A+i B) But does that prove that $C$ is normal?
$endgroup$
– Kai
Jan 11 at 16:07




1




1




$begingroup$
You are incorrectly computing $C^ast$; keep in mind that $i$ becomes $overline{i} = -i$.
$endgroup$
– darij grinberg
Jan 11 at 16:09




$begingroup$
You are incorrectly computing $C^ast$; keep in mind that $i$ becomes $overline{i} = -i$.
$endgroup$
– darij grinberg
Jan 11 at 16:09




2




2




$begingroup$
Also, recall that $C$ is normal if and only if $CC^ast = C^ast C$. This is the very definition of "normal".
$endgroup$
– darij grinberg
Jan 11 at 16:09




$begingroup$
Also, recall that $C$ is normal if and only if $CC^ast = C^ast C$. This is the very definition of "normal".
$endgroup$
– darij grinberg
Jan 11 at 16:09










1 Answer
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2












$begingroup$

Your approach won't work, because the sum of two normal matrices is, in general, not normal.



Instead, simply check the definition of normality:



begin{align*}[C,C^*] &= CC^*-C^*C \&= (A+iB)(A+iB)^*-(A+iB)^*(A+iB)
\&=(A+iB)(A^*+(iB)^*)-(A^*+(iB)^*)(A+iB)
\&= AA^*+iBA^*+A(iB)^* + (iB)(iB)^* - A^*A-(iB)^*A-A^*iB-(iB)^*(iB)
\&= (AA^*-A^*A) + (A^*iB-iBA^*)+(Abar{i}B^*-bar{i}B^*A)+(iBbar{i}B^*-bar{i}B^*iB)
\&= [A,A^*] + i[A^*,B] + bar{i}[A,B^*] + ibar{i}[B,B^*]
\&= [A,A] + i[A,B] + bar{i}[A,B] + [B,B]mbox{ since $A$ and $B$ are self-adjoint}
\&= 0mbox{ since [A,B] = 0, and $[X,X] = 0$ for all $X$}end{align*}






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    $begingroup$

    Your approach won't work, because the sum of two normal matrices is, in general, not normal.



    Instead, simply check the definition of normality:



    begin{align*}[C,C^*] &= CC^*-C^*C \&= (A+iB)(A+iB)^*-(A+iB)^*(A+iB)
    \&=(A+iB)(A^*+(iB)^*)-(A^*+(iB)^*)(A+iB)
    \&= AA^*+iBA^*+A(iB)^* + (iB)(iB)^* - A^*A-(iB)^*A-A^*iB-(iB)^*(iB)
    \&= (AA^*-A^*A) + (A^*iB-iBA^*)+(Abar{i}B^*-bar{i}B^*A)+(iBbar{i}B^*-bar{i}B^*iB)
    \&= [A,A^*] + i[A^*,B] + bar{i}[A,B^*] + ibar{i}[B,B^*]
    \&= [A,A] + i[A,B] + bar{i}[A,B] + [B,B]mbox{ since $A$ and $B$ are self-adjoint}
    \&= 0mbox{ since [A,B] = 0, and $[X,X] = 0$ for all $X$}end{align*}






    share|cite|improve this answer











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      2












      $begingroup$

      Your approach won't work, because the sum of two normal matrices is, in general, not normal.



      Instead, simply check the definition of normality:



      begin{align*}[C,C^*] &= CC^*-C^*C \&= (A+iB)(A+iB)^*-(A+iB)^*(A+iB)
      \&=(A+iB)(A^*+(iB)^*)-(A^*+(iB)^*)(A+iB)
      \&= AA^*+iBA^*+A(iB)^* + (iB)(iB)^* - A^*A-(iB)^*A-A^*iB-(iB)^*(iB)
      \&= (AA^*-A^*A) + (A^*iB-iBA^*)+(Abar{i}B^*-bar{i}B^*A)+(iBbar{i}B^*-bar{i}B^*iB)
      \&= [A,A^*] + i[A^*,B] + bar{i}[A,B^*] + ibar{i}[B,B^*]
      \&= [A,A] + i[A,B] + bar{i}[A,B] + [B,B]mbox{ since $A$ and $B$ are self-adjoint}
      \&= 0mbox{ since [A,B] = 0, and $[X,X] = 0$ for all $X$}end{align*}






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Your approach won't work, because the sum of two normal matrices is, in general, not normal.



        Instead, simply check the definition of normality:



        begin{align*}[C,C^*] &= CC^*-C^*C \&= (A+iB)(A+iB)^*-(A+iB)^*(A+iB)
        \&=(A+iB)(A^*+(iB)^*)-(A^*+(iB)^*)(A+iB)
        \&= AA^*+iBA^*+A(iB)^* + (iB)(iB)^* - A^*A-(iB)^*A-A^*iB-(iB)^*(iB)
        \&= (AA^*-A^*A) + (A^*iB-iBA^*)+(Abar{i}B^*-bar{i}B^*A)+(iBbar{i}B^*-bar{i}B^*iB)
        \&= [A,A^*] + i[A^*,B] + bar{i}[A,B^*] + ibar{i}[B,B^*]
        \&= [A,A] + i[A,B] + bar{i}[A,B] + [B,B]mbox{ since $A$ and $B$ are self-adjoint}
        \&= 0mbox{ since [A,B] = 0, and $[X,X] = 0$ for all $X$}end{align*}






        share|cite|improve this answer











        $endgroup$



        Your approach won't work, because the sum of two normal matrices is, in general, not normal.



        Instead, simply check the definition of normality:



        begin{align*}[C,C^*] &= CC^*-C^*C \&= (A+iB)(A+iB)^*-(A+iB)^*(A+iB)
        \&=(A+iB)(A^*+(iB)^*)-(A^*+(iB)^*)(A+iB)
        \&= AA^*+iBA^*+A(iB)^* + (iB)(iB)^* - A^*A-(iB)^*A-A^*iB-(iB)^*(iB)
        \&= (AA^*-A^*A) + (A^*iB-iBA^*)+(Abar{i}B^*-bar{i}B^*A)+(iBbar{i}B^*-bar{i}B^*iB)
        \&= [A,A^*] + i[A^*,B] + bar{i}[A,B^*] + ibar{i}[B,B^*]
        \&= [A,A] + i[A,B] + bar{i}[A,B] + [B,B]mbox{ since $A$ and $B$ are self-adjoint}
        \&= 0mbox{ since [A,B] = 0, and $[X,X] = 0$ for all $X$}end{align*}







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 11 at 17:35









        Viktor Glombik

        782527




        782527










        answered Jan 11 at 16:08









        user3482749user3482749

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