A matrix is similar to its transpose without jordan form











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Let $A in M_{n×n}mathbb{(C)}$ be invertible. Then



$(A^∗)^{−1} = (A^{−1})^∗$



Let B be a nonsingular matrix such that $A = B^{−1}B^∗$ . Show that $A^{−1}$ is similar to $A^∗$. ($A^*$ is complex conjugate transpose).




I have already shown the first part by showing that $A^*(A^{-1})^*$ is same as $(A^{-1}A)^*$ which is identity.



Now for second part I showed that $A^{-1}$ is transpose of $A^*$. How to show that transpose of two matrix are similar without using the jordan form.






matrices numerical-linear-algebra matrix-decomposition transpose







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  • Sorry for asking. I did it already. But if someone can help me show the proof of last statement, I will be very thankful.
    – Jimmy
    Nov 18 at 12:35












  • "Now for second part I showed that $A^{−1}$ is transpose of $A^ast$." No, it isn't. E.g.$$B=pmatrix{1&1\ 0&1} Rightarrow A=B^{-1}B^ast=pmatrix{0&-1\ 1&1} Rightarrow A^{-1}=pmatrix{1&1\ -1&0}, A^ast=pmatrix{0&1\ -1&1}ne(A^{-1})^T.$$
    – user1551
    Nov 22 at 14:08

















up vote
0
down vote

favorite













Let $A in M_{n×n}mathbb{(C)}$ be invertible. Then



$(A^∗)^{−1} = (A^{−1})^∗$



Let B be a nonsingular matrix such that $A = B^{−1}B^∗$ . Show that $A^{−1}$ is similar to $A^∗$. ($A^*$ is complex conjugate transpose).




I have already shown the first part by showing that $A^*(A^{-1})^*$ is same as $(A^{-1}A)^*$ which is identity.



Now for second part I showed that $A^{-1}$ is transpose of $A^*$. How to show that transpose of two matrix are similar without using the jordan form.






matrices numerical-linear-algebra matrix-decomposition transpose







share|cite|improve this question






















  • Sorry for asking. I did it already. But if someone can help me show the proof of last statement, I will be very thankful.
    – Jimmy
    Nov 18 at 12:35












  • "Now for second part I showed that $A^{−1}$ is transpose of $A^ast$." No, it isn't. E.g.$$B=pmatrix{1&1\ 0&1} Rightarrow A=B^{-1}B^ast=pmatrix{0&-1\ 1&1} Rightarrow A^{-1}=pmatrix{1&1\ -1&0}, A^ast=pmatrix{0&1\ -1&1}ne(A^{-1})^T.$$
    – user1551
    Nov 22 at 14:08















up vote
0
down vote

favorite









up vote
0
down vote

favorite












Let $A in M_{n×n}mathbb{(C)}$ be invertible. Then



$(A^∗)^{−1} = (A^{−1})^∗$



Let B be a nonsingular matrix such that $A = B^{−1}B^∗$ . Show that $A^{−1}$ is similar to $A^∗$. ($A^*$ is complex conjugate transpose).




I have already shown the first part by showing that $A^*(A^{-1})^*$ is same as $(A^{-1}A)^*$ which is identity.



Now for second part I showed that $A^{-1}$ is transpose of $A^*$. How to show that transpose of two matrix are similar without using the jordan form.






matrices numerical-linear-algebra matrix-decomposition transpose







share|cite|improve this question














Let $A in M_{n×n}mathbb{(C)}$ be invertible. Then



$(A^∗)^{−1} = (A^{−1})^∗$



Let B be a nonsingular matrix such that $A = B^{−1}B^∗$ . Show that $A^{−1}$ is similar to $A^∗$. ($A^*$ is complex conjugate transpose).




I have already shown the first part by showing that $A^*(A^{-1})^*$ is same as $(A^{-1}A)^*$ which is identity.



Now for second part I showed that $A^{-1}$ is transpose of $A^*$. How to show that transpose of two matrix are similar without using the jordan form.






matrices numerical-linear-algebra matrix-decomposition transpose




matrices numerical-linear-algebra matrix-decomposition transpose






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 18 at 12:13









Jimmy

17312




17312












  • Sorry for asking. I did it already. But if someone can help me show the proof of last statement, I will be very thankful.
    – Jimmy
    Nov 18 at 12:35












  • "Now for second part I showed that $A^{−1}$ is transpose of $A^ast$." No, it isn't. E.g.$$B=pmatrix{1&1\ 0&1} Rightarrow A=B^{-1}B^ast=pmatrix{0&-1\ 1&1} Rightarrow A^{-1}=pmatrix{1&1\ -1&0}, A^ast=pmatrix{0&1\ -1&1}ne(A^{-1})^T.$$
    – user1551
    Nov 22 at 14:08




















  • Sorry for asking. I did it already. But if someone can help me show the proof of last statement, I will be very thankful.
    – Jimmy
    Nov 18 at 12:35












  • "Now for second part I showed that $A^{−1}$ is transpose of $A^ast$." No, it isn't. E.g.$$B=pmatrix{1&1\ 0&1} Rightarrow A=B^{-1}B^ast=pmatrix{0&-1\ 1&1} Rightarrow A^{-1}=pmatrix{1&1\ -1&0}, A^ast=pmatrix{0&1\ -1&1}ne(A^{-1})^T.$$
    – user1551
    Nov 22 at 14:08


















Sorry for asking. I did it already. But if someone can help me show the proof of last statement, I will be very thankful.
– Jimmy
Nov 18 at 12:35






Sorry for asking. I did it already. But if someone can help me show the proof of last statement, I will be very thankful.
– Jimmy
Nov 18 at 12:35














"Now for second part I showed that $A^{−1}$ is transpose of $A^ast$." No, it isn't. E.g.$$B=pmatrix{1&1\ 0&1} Rightarrow A=B^{-1}B^ast=pmatrix{0&-1\ 1&1} Rightarrow A^{-1}=pmatrix{1&1\ -1&0}, A^ast=pmatrix{0&1\ -1&1}ne(A^{-1})^T.$$
– user1551
Nov 22 at 14:08






"Now for second part I showed that $A^{−1}$ is transpose of $A^ast$." No, it isn't. E.g.$$B=pmatrix{1&1\ 0&1} Rightarrow A=B^{-1}B^ast=pmatrix{0&-1\ 1&1} Rightarrow A^{-1}=pmatrix{1&1\ -1&0}, A^ast=pmatrix{0&1\ -1&1}ne(A^{-1})^T.$$
– user1551
Nov 22 at 14:08












1 Answer
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Just simply use the definition of $A$ in terms of $B$.



If you have $A=B^{-1}B^*$, then
$$A^{-1}=B^{-*}B, quad A^*=BB^{-*}.$$
So
$$A^{-1}=B^{-*}B=B^{-1}BB^{-*}B=B^{-1}A^*B$$
and hence $A^{-1}$ and $A^*$ are similar.






share|cite|improve this answer























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    1 Answer
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    up vote
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    Just simply use the definition of $A$ in terms of $B$.



    If you have $A=B^{-1}B^*$, then
    $$A^{-1}=B^{-*}B, quad A^*=BB^{-*}.$$
    So
    $$A^{-1}=B^{-*}B=B^{-1}BB^{-*}B=B^{-1}A^*B$$
    and hence $A^{-1}$ and $A^*$ are similar.






    share|cite|improve this answer



























      up vote
      2
      down vote













      Just simply use the definition of $A$ in terms of $B$.



      If you have $A=B^{-1}B^*$, then
      $$A^{-1}=B^{-*}B, quad A^*=BB^{-*}.$$
      So
      $$A^{-1}=B^{-*}B=B^{-1}BB^{-*}B=B^{-1}A^*B$$
      and hence $A^{-1}$ and $A^*$ are similar.






      share|cite|improve this answer

























        up vote
        2
        down vote










        up vote
        2
        down vote









        Just simply use the definition of $A$ in terms of $B$.



        If you have $A=B^{-1}B^*$, then
        $$A^{-1}=B^{-*}B, quad A^*=BB^{-*}.$$
        So
        $$A^{-1}=B^{-*}B=B^{-1}BB^{-*}B=B^{-1}A^*B$$
        and hence $A^{-1}$ and $A^*$ are similar.






        share|cite|improve this answer














        Just simply use the definition of $A$ in terms of $B$.



        If you have $A=B^{-1}B^*$, then
        $$A^{-1}=B^{-*}B, quad A^*=BB^{-*}.$$
        So
        $$A^{-1}=B^{-*}B=B^{-1}BB^{-*}B=B^{-1}A^*B$$
        and hence $A^{-1}$ and $A^*$ are similar.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 22 at 14:28

























        answered Nov 22 at 13:41









        Algebraic Pavel

        16.2k31839




        16.2k31839






























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