Is $r.b*=b*A$ true?












0












$begingroup$


Let $a, b in Bbb C^n$ and $Ain Bbb C^{n times n}$. If $b^*cdot a=1$ and $r=b^*cdot Acdot a$, is it true that: $rcdot b^*=b^*A$?










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$endgroup$












  • $begingroup$
    Does $*$ refer to the conjugate-transpose?
    $endgroup$
    – Omnomnomnom
    Jan 2 at 18:28










  • $begingroup$
    Yes, ∗ refer to the conjugate-transpose
    $endgroup$
    – Şenay Erdinç
    Jan 2 at 18:37












  • $begingroup$
    My question is may be too simple. I'm not very familiar with matrix-vector expressions
    $endgroup$
    – Şenay Erdinç
    Jan 2 at 18:41
















0












$begingroup$


Let $a, b in Bbb C^n$ and $Ain Bbb C^{n times n}$. If $b^*cdot a=1$ and $r=b^*cdot Acdot a$, is it true that: $rcdot b^*=b^*A$?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Does $*$ refer to the conjugate-transpose?
    $endgroup$
    – Omnomnomnom
    Jan 2 at 18:28










  • $begingroup$
    Yes, ∗ refer to the conjugate-transpose
    $endgroup$
    – Şenay Erdinç
    Jan 2 at 18:37












  • $begingroup$
    My question is may be too simple. I'm not very familiar with matrix-vector expressions
    $endgroup$
    – Şenay Erdinç
    Jan 2 at 18:41














0












0








0





$begingroup$


Let $a, b in Bbb C^n$ and $Ain Bbb C^{n times n}$. If $b^*cdot a=1$ and $r=b^*cdot Acdot a$, is it true that: $rcdot b^*=b^*A$?










share|cite|improve this question











$endgroup$




Let $a, b in Bbb C^n$ and $Ain Bbb C^{n times n}$. If $b^*cdot a=1$ and $r=b^*cdot Acdot a$, is it true that: $rcdot b^*=b^*A$?







vectors matrix-equations






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edited Jan 2 at 18:27









Omnomnomnom

127k790178




127k790178










asked Jan 2 at 18:09









Şenay ErdinçŞenay Erdinç

11




11












  • $begingroup$
    Does $*$ refer to the conjugate-transpose?
    $endgroup$
    – Omnomnomnom
    Jan 2 at 18:28










  • $begingroup$
    Yes, ∗ refer to the conjugate-transpose
    $endgroup$
    – Şenay Erdinç
    Jan 2 at 18:37












  • $begingroup$
    My question is may be too simple. I'm not very familiar with matrix-vector expressions
    $endgroup$
    – Şenay Erdinç
    Jan 2 at 18:41


















  • $begingroup$
    Does $*$ refer to the conjugate-transpose?
    $endgroup$
    – Omnomnomnom
    Jan 2 at 18:28










  • $begingroup$
    Yes, ∗ refer to the conjugate-transpose
    $endgroup$
    – Şenay Erdinç
    Jan 2 at 18:37












  • $begingroup$
    My question is may be too simple. I'm not very familiar with matrix-vector expressions
    $endgroup$
    – Şenay Erdinç
    Jan 2 at 18:41
















$begingroup$
Does $*$ refer to the conjugate-transpose?
$endgroup$
– Omnomnomnom
Jan 2 at 18:28




$begingroup$
Does $*$ refer to the conjugate-transpose?
$endgroup$
– Omnomnomnom
Jan 2 at 18:28












$begingroup$
Yes, ∗ refer to the conjugate-transpose
$endgroup$
– Şenay Erdinç
Jan 2 at 18:37






$begingroup$
Yes, ∗ refer to the conjugate-transpose
$endgroup$
– Şenay Erdinç
Jan 2 at 18:37














$begingroup$
My question is may be too simple. I'm not very familiar with matrix-vector expressions
$endgroup$
– Şenay Erdinç
Jan 2 at 18:41




$begingroup$
My question is may be too simple. I'm not very familiar with matrix-vector expressions
$endgroup$
– Şenay Erdinç
Jan 2 at 18:41










2 Answers
2






active

oldest

votes


















0












$begingroup$

The answer to your question is no. For an example where this doesn't work, take
$$
a = pmatrix{1\0}, quad b = pmatrix{1\0}, quad A = pmatrix{0&1\1&0}
$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Taking the conjugate transpose, we get
    $$(rb^*)^* = b(b^*Aa)^* = ba^*A^*b$$
    Which we need to be equal to
    $$(b^*A)^* = A^*b$$
    Since matrix multiplication is associative, we can explicitly write
    $$ ba^*A^*b = ((ba^*)A^*)b$$
    Then, we have the desired equality for all $b$ if and only if $$ba^* = I$$ for all $b$, which is not possible. Thus, the equality doesn't hold for all $b$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Тhanks a lot for quick answers.
      $endgroup$
      – Şenay Erdinç
      Jan 2 at 19:22











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    The answer to your question is no. For an example where this doesn't work, take
    $$
    a = pmatrix{1\0}, quad b = pmatrix{1\0}, quad A = pmatrix{0&1\1&0}
    $$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The answer to your question is no. For an example where this doesn't work, take
      $$
      a = pmatrix{1\0}, quad b = pmatrix{1\0}, quad A = pmatrix{0&1\1&0}
      $$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The answer to your question is no. For an example where this doesn't work, take
        $$
        a = pmatrix{1\0}, quad b = pmatrix{1\0}, quad A = pmatrix{0&1\1&0}
        $$






        share|cite|improve this answer









        $endgroup$



        The answer to your question is no. For an example where this doesn't work, take
        $$
        a = pmatrix{1\0}, quad b = pmatrix{1\0}, quad A = pmatrix{0&1\1&0}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 19:06









        OmnomnomnomOmnomnomnom

        127k790178




        127k790178























            0












            $begingroup$

            Taking the conjugate transpose, we get
            $$(rb^*)^* = b(b^*Aa)^* = ba^*A^*b$$
            Which we need to be equal to
            $$(b^*A)^* = A^*b$$
            Since matrix multiplication is associative, we can explicitly write
            $$ ba^*A^*b = ((ba^*)A^*)b$$
            Then, we have the desired equality for all $b$ if and only if $$ba^* = I$$ for all $b$, which is not possible. Thus, the equality doesn't hold for all $b$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Тhanks a lot for quick answers.
              $endgroup$
              – Şenay Erdinç
              Jan 2 at 19:22
















            0












            $begingroup$

            Taking the conjugate transpose, we get
            $$(rb^*)^* = b(b^*Aa)^* = ba^*A^*b$$
            Which we need to be equal to
            $$(b^*A)^* = A^*b$$
            Since matrix multiplication is associative, we can explicitly write
            $$ ba^*A^*b = ((ba^*)A^*)b$$
            Then, we have the desired equality for all $b$ if and only if $$ba^* = I$$ for all $b$, which is not possible. Thus, the equality doesn't hold for all $b$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Тhanks a lot for quick answers.
              $endgroup$
              – Şenay Erdinç
              Jan 2 at 19:22














            0












            0








            0





            $begingroup$

            Taking the conjugate transpose, we get
            $$(rb^*)^* = b(b^*Aa)^* = ba^*A^*b$$
            Which we need to be equal to
            $$(b^*A)^* = A^*b$$
            Since matrix multiplication is associative, we can explicitly write
            $$ ba^*A^*b = ((ba^*)A^*)b$$
            Then, we have the desired equality for all $b$ if and only if $$ba^* = I$$ for all $b$, which is not possible. Thus, the equality doesn't hold for all $b$.






            share|cite|improve this answer









            $endgroup$



            Taking the conjugate transpose, we get
            $$(rb^*)^* = b(b^*Aa)^* = ba^*A^*b$$
            Which we need to be equal to
            $$(b^*A)^* = A^*b$$
            Since matrix multiplication is associative, we can explicitly write
            $$ ba^*A^*b = ((ba^*)A^*)b$$
            Then, we have the desired equality for all $b$ if and only if $$ba^* = I$$ for all $b$, which is not possible. Thus, the equality doesn't hold for all $b$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 2 at 19:09









            AlexanderJ93AlexanderJ93

            6,158823




            6,158823












            • $begingroup$
              Тhanks a lot for quick answers.
              $endgroup$
              – Şenay Erdinç
              Jan 2 at 19:22


















            • $begingroup$
              Тhanks a lot for quick answers.
              $endgroup$
              – Şenay Erdinç
              Jan 2 at 19:22
















            $begingroup$
            Тhanks a lot for quick answers.
            $endgroup$
            – Şenay Erdinç
            Jan 2 at 19:22




            $begingroup$
            Тhanks a lot for quick answers.
            $endgroup$
            – Şenay Erdinç
            Jan 2 at 19:22


















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