About orthogonal matrix











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Let $Ain M_3(mathbb{R})$ be an orthogonal matrix with $det(A)=1$. Then to show
$$(Tr(A)-1)^2+sum_{i<j}(a_{ij}-a_{ji})^2=4$$ where $Tr$ denotes trace.



Suppose the matrix $A=left(begin{matrix} a_{11} & a_{12} & a_{13}\
a_{21} & a_{22} &a_{23}\
a_{31} & a_{32} & a_{33}
end{matrix}right)$
in $M_3(mathbb{R})$ be orthogonal. Then
$Tr(AA^T)=sum_{i,j}a_{ij}^2=3$. If we expand the left hand side of the equation in question, we have
$$LHS=4+2(a_{11}a_{22}+a_{11}a_{33}+a_{22}a_{33})-2(a_{12}a_{21}+a_{13}a_{31}+a_{23}a_{32})-2(a_{11}+a_{22}+a_{33})$$
$$=4+2(a_{11}a_{22}-a_{12}a_{21})+2(a_{11}a_{33}-a_{13}a_{31})+2(a_{22}a_{33}-a_{23}a_{32})-2(a_{11}+a_{22}+a_{33})$$
$$=4+minor(a_{33})+minor(a_{22})+minor(a_{11})-2(a_{11}+a_{22}+a_{33})$$



Is there any theorem/result that can be used to say that the remaining part of the last expression excluding 4 is zero? Or is there any way we can manipulate $a_{ij}$ so that it becomes zero?
I tried hard but i see no way out. Any help is highly appreciated.










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    up vote
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    Let $Ain M_3(mathbb{R})$ be an orthogonal matrix with $det(A)=1$. Then to show
    $$(Tr(A)-1)^2+sum_{i<j}(a_{ij}-a_{ji})^2=4$$ where $Tr$ denotes trace.



    Suppose the matrix $A=left(begin{matrix} a_{11} & a_{12} & a_{13}\
    a_{21} & a_{22} &a_{23}\
    a_{31} & a_{32} & a_{33}
    end{matrix}right)$
    in $M_3(mathbb{R})$ be orthogonal. Then
    $Tr(AA^T)=sum_{i,j}a_{ij}^2=3$. If we expand the left hand side of the equation in question, we have
    $$LHS=4+2(a_{11}a_{22}+a_{11}a_{33}+a_{22}a_{33})-2(a_{12}a_{21}+a_{13}a_{31}+a_{23}a_{32})-2(a_{11}+a_{22}+a_{33})$$
    $$=4+2(a_{11}a_{22}-a_{12}a_{21})+2(a_{11}a_{33}-a_{13}a_{31})+2(a_{22}a_{33}-a_{23}a_{32})-2(a_{11}+a_{22}+a_{33})$$
    $$=4+minor(a_{33})+minor(a_{22})+minor(a_{11})-2(a_{11}+a_{22}+a_{33})$$



    Is there any theorem/result that can be used to say that the remaining part of the last expression excluding 4 is zero? Or is there any way we can manipulate $a_{ij}$ so that it becomes zero?
    I tried hard but i see no way out. Any help is highly appreciated.










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      Let $Ain M_3(mathbb{R})$ be an orthogonal matrix with $det(A)=1$. Then to show
      $$(Tr(A)-1)^2+sum_{i<j}(a_{ij}-a_{ji})^2=4$$ where $Tr$ denotes trace.



      Suppose the matrix $A=left(begin{matrix} a_{11} & a_{12} & a_{13}\
      a_{21} & a_{22} &a_{23}\
      a_{31} & a_{32} & a_{33}
      end{matrix}right)$
      in $M_3(mathbb{R})$ be orthogonal. Then
      $Tr(AA^T)=sum_{i,j}a_{ij}^2=3$. If we expand the left hand side of the equation in question, we have
      $$LHS=4+2(a_{11}a_{22}+a_{11}a_{33}+a_{22}a_{33})-2(a_{12}a_{21}+a_{13}a_{31}+a_{23}a_{32})-2(a_{11}+a_{22}+a_{33})$$
      $$=4+2(a_{11}a_{22}-a_{12}a_{21})+2(a_{11}a_{33}-a_{13}a_{31})+2(a_{22}a_{33}-a_{23}a_{32})-2(a_{11}+a_{22}+a_{33})$$
      $$=4+minor(a_{33})+minor(a_{22})+minor(a_{11})-2(a_{11}+a_{22}+a_{33})$$



      Is there any theorem/result that can be used to say that the remaining part of the last expression excluding 4 is zero? Or is there any way we can manipulate $a_{ij}$ so that it becomes zero?
      I tried hard but i see no way out. Any help is highly appreciated.










      share|cite|improve this question















      Let $Ain M_3(mathbb{R})$ be an orthogonal matrix with $det(A)=1$. Then to show
      $$(Tr(A)-1)^2+sum_{i<j}(a_{ij}-a_{ji})^2=4$$ where $Tr$ denotes trace.



      Suppose the matrix $A=left(begin{matrix} a_{11} & a_{12} & a_{13}\
      a_{21} & a_{22} &a_{23}\
      a_{31} & a_{32} & a_{33}
      end{matrix}right)$
      in $M_3(mathbb{R})$ be orthogonal. Then
      $Tr(AA^T)=sum_{i,j}a_{ij}^2=3$. If we expand the left hand side of the equation in question, we have
      $$LHS=4+2(a_{11}a_{22}+a_{11}a_{33}+a_{22}a_{33})-2(a_{12}a_{21}+a_{13}a_{31}+a_{23}a_{32})-2(a_{11}+a_{22}+a_{33})$$
      $$=4+2(a_{11}a_{22}-a_{12}a_{21})+2(a_{11}a_{33}-a_{13}a_{31})+2(a_{22}a_{33}-a_{23}a_{32})-2(a_{11}+a_{22}+a_{33})$$
      $$=4+minor(a_{33})+minor(a_{22})+minor(a_{11})-2(a_{11}+a_{22}+a_{33})$$



      Is there any theorem/result that can be used to say that the remaining part of the last expression excluding 4 is zero? Or is there any way we can manipulate $a_{ij}$ so that it becomes zero?
      I tried hard but i see no way out. Any help is highly appreciated.







      matrices orthogonality






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          I think the LHS should be
          $$LHS=4+2(minor(a_{33})+minor(a_{22})+minor(a_{11}))−2(a_{11}+a_{22}+a_{33})$$
          and you can simplify this expression with the help of the characteristic plynomial of $A$. You can write
          $$Det(tI-a)=t^3-c_2t^2+c_1t-c_0$$
          where $I$ is the identity matrix and $c_i$ is the sum of all principal minor of $A$ of size $3-i$. In particular, $c_0=Det(A)=1$, $c_1=minor(a_{33})+minor(a_{22})+minor(a_{11})$ using your notations and $c_2=Tr(A)$. Finally, $A$ is a matrix of $M_3(mathbb{R})$ so it has at least one real eigenvalue, $A$ is orthogonal so this eigenvalue is either $1$ or $-1$ and $det(A)=1$ so $A$ has at least an eigenvalue equal to $1$. This means that $Det(I-A)=0$ hence $1-c_2+c_1-c_0=0$ and then
          $$minor(a_{33})+minor(a_{22})+minor(a_{11})=a_{11}+a_{22}+a_{33}$$
          and fially $LHS=4$.






          share|cite|improve this answer























          • sorry, my bad! I missed the integer 2
            – Dastan
            yesterday










          • Can you tell me where can I find the proof of the formula of determinant you have used here?Moreover, I din't really get what you meant by saying "A is a matrix of $M_3(mathbb{R})$ so it has real eigen values," because a real matrix may not have real eigen values.
            – Dastan
            yesterday












          • Is there any way to solve the problem without having resorting to the theory of eigen values?
            – Dastan
            yesterday










          • Anyway, thanks a lot!!
            – Dastan
            yesterday










          • You can find more information on the coefficients of characteristic polynomials with some references here: math.stackexchange.com/questions/28650/…. In your case, you can always directly expand $Det(tI-A)$ without using this formula. Otherwise, i meant that any matrix of $M_3(mathbb{R})$ have at least one real eigenvalue. That's because the eigenvalues of an $ntimes n$ matrix are the roots of the characteristic polynomial with degree $n$ and when $n$ is odd then this polynomial has at least one real eigenvalue.
            – rezoons
            23 hours ago











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          I think the LHS should be
          $$LHS=4+2(minor(a_{33})+minor(a_{22})+minor(a_{11}))−2(a_{11}+a_{22}+a_{33})$$
          and you can simplify this expression with the help of the characteristic plynomial of $A$. You can write
          $$Det(tI-a)=t^3-c_2t^2+c_1t-c_0$$
          where $I$ is the identity matrix and $c_i$ is the sum of all principal minor of $A$ of size $3-i$. In particular, $c_0=Det(A)=1$, $c_1=minor(a_{33})+minor(a_{22})+minor(a_{11})$ using your notations and $c_2=Tr(A)$. Finally, $A$ is a matrix of $M_3(mathbb{R})$ so it has at least one real eigenvalue, $A$ is orthogonal so this eigenvalue is either $1$ or $-1$ and $det(A)=1$ so $A$ has at least an eigenvalue equal to $1$. This means that $Det(I-A)=0$ hence $1-c_2+c_1-c_0=0$ and then
          $$minor(a_{33})+minor(a_{22})+minor(a_{11})=a_{11}+a_{22}+a_{33}$$
          and fially $LHS=4$.






          share|cite|improve this answer























          • sorry, my bad! I missed the integer 2
            – Dastan
            yesterday










          • Can you tell me where can I find the proof of the formula of determinant you have used here?Moreover, I din't really get what you meant by saying "A is a matrix of $M_3(mathbb{R})$ so it has real eigen values," because a real matrix may not have real eigen values.
            – Dastan
            yesterday












          • Is there any way to solve the problem without having resorting to the theory of eigen values?
            – Dastan
            yesterday










          • Anyway, thanks a lot!!
            – Dastan
            yesterday










          • You can find more information on the coefficients of characteristic polynomials with some references here: math.stackexchange.com/questions/28650/…. In your case, you can always directly expand $Det(tI-A)$ without using this formula. Otherwise, i meant that any matrix of $M_3(mathbb{R})$ have at least one real eigenvalue. That's because the eigenvalues of an $ntimes n$ matrix are the roots of the characteristic polynomial with degree $n$ and when $n$ is odd then this polynomial has at least one real eigenvalue.
            – rezoons
            23 hours ago















          up vote
          0
          down vote



          accepted










          I think the LHS should be
          $$LHS=4+2(minor(a_{33})+minor(a_{22})+minor(a_{11}))−2(a_{11}+a_{22}+a_{33})$$
          and you can simplify this expression with the help of the characteristic plynomial of $A$. You can write
          $$Det(tI-a)=t^3-c_2t^2+c_1t-c_0$$
          where $I$ is the identity matrix and $c_i$ is the sum of all principal minor of $A$ of size $3-i$. In particular, $c_0=Det(A)=1$, $c_1=minor(a_{33})+minor(a_{22})+minor(a_{11})$ using your notations and $c_2=Tr(A)$. Finally, $A$ is a matrix of $M_3(mathbb{R})$ so it has at least one real eigenvalue, $A$ is orthogonal so this eigenvalue is either $1$ or $-1$ and $det(A)=1$ so $A$ has at least an eigenvalue equal to $1$. This means that $Det(I-A)=0$ hence $1-c_2+c_1-c_0=0$ and then
          $$minor(a_{33})+minor(a_{22})+minor(a_{11})=a_{11}+a_{22}+a_{33}$$
          and fially $LHS=4$.






          share|cite|improve this answer























          • sorry, my bad! I missed the integer 2
            – Dastan
            yesterday










          • Can you tell me where can I find the proof of the formula of determinant you have used here?Moreover, I din't really get what you meant by saying "A is a matrix of $M_3(mathbb{R})$ so it has real eigen values," because a real matrix may not have real eigen values.
            – Dastan
            yesterday












          • Is there any way to solve the problem without having resorting to the theory of eigen values?
            – Dastan
            yesterday










          • Anyway, thanks a lot!!
            – Dastan
            yesterday










          • You can find more information on the coefficients of characteristic polynomials with some references here: math.stackexchange.com/questions/28650/…. In your case, you can always directly expand $Det(tI-A)$ without using this formula. Otherwise, i meant that any matrix of $M_3(mathbb{R})$ have at least one real eigenvalue. That's because the eigenvalues of an $ntimes n$ matrix are the roots of the characteristic polynomial with degree $n$ and when $n$ is odd then this polynomial has at least one real eigenvalue.
            – rezoons
            23 hours ago













          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          I think the LHS should be
          $$LHS=4+2(minor(a_{33})+minor(a_{22})+minor(a_{11}))−2(a_{11}+a_{22}+a_{33})$$
          and you can simplify this expression with the help of the characteristic plynomial of $A$. You can write
          $$Det(tI-a)=t^3-c_2t^2+c_1t-c_0$$
          where $I$ is the identity matrix and $c_i$ is the sum of all principal minor of $A$ of size $3-i$. In particular, $c_0=Det(A)=1$, $c_1=minor(a_{33})+minor(a_{22})+minor(a_{11})$ using your notations and $c_2=Tr(A)$. Finally, $A$ is a matrix of $M_3(mathbb{R})$ so it has at least one real eigenvalue, $A$ is orthogonal so this eigenvalue is either $1$ or $-1$ and $det(A)=1$ so $A$ has at least an eigenvalue equal to $1$. This means that $Det(I-A)=0$ hence $1-c_2+c_1-c_0=0$ and then
          $$minor(a_{33})+minor(a_{22})+minor(a_{11})=a_{11}+a_{22}+a_{33}$$
          and fially $LHS=4$.






          share|cite|improve this answer














          I think the LHS should be
          $$LHS=4+2(minor(a_{33})+minor(a_{22})+minor(a_{11}))−2(a_{11}+a_{22}+a_{33})$$
          and you can simplify this expression with the help of the characteristic plynomial of $A$. You can write
          $$Det(tI-a)=t^3-c_2t^2+c_1t-c_0$$
          where $I$ is the identity matrix and $c_i$ is the sum of all principal minor of $A$ of size $3-i$. In particular, $c_0=Det(A)=1$, $c_1=minor(a_{33})+minor(a_{22})+minor(a_{11})$ using your notations and $c_2=Tr(A)$. Finally, $A$ is a matrix of $M_3(mathbb{R})$ so it has at least one real eigenvalue, $A$ is orthogonal so this eigenvalue is either $1$ or $-1$ and $det(A)=1$ so $A$ has at least an eigenvalue equal to $1$. This means that $Det(I-A)=0$ hence $1-c_2+c_1-c_0=0$ and then
          $$minor(a_{33})+minor(a_{22})+minor(a_{11})=a_{11}+a_{22}+a_{33}$$
          and fially $LHS=4$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered yesterday









          rezoons

          414




          414












          • sorry, my bad! I missed the integer 2
            – Dastan
            yesterday










          • Can you tell me where can I find the proof of the formula of determinant you have used here?Moreover, I din't really get what you meant by saying "A is a matrix of $M_3(mathbb{R})$ so it has real eigen values," because a real matrix may not have real eigen values.
            – Dastan
            yesterday












          • Is there any way to solve the problem without having resorting to the theory of eigen values?
            – Dastan
            yesterday










          • Anyway, thanks a lot!!
            – Dastan
            yesterday










          • You can find more information on the coefficients of characteristic polynomials with some references here: math.stackexchange.com/questions/28650/…. In your case, you can always directly expand $Det(tI-A)$ without using this formula. Otherwise, i meant that any matrix of $M_3(mathbb{R})$ have at least one real eigenvalue. That's because the eigenvalues of an $ntimes n$ matrix are the roots of the characteristic polynomial with degree $n$ and when $n$ is odd then this polynomial has at least one real eigenvalue.
            – rezoons
            23 hours ago


















          • sorry, my bad! I missed the integer 2
            – Dastan
            yesterday










          • Can you tell me where can I find the proof of the formula of determinant you have used here?Moreover, I din't really get what you meant by saying "A is a matrix of $M_3(mathbb{R})$ so it has real eigen values," because a real matrix may not have real eigen values.
            – Dastan
            yesterday












          • Is there any way to solve the problem without having resorting to the theory of eigen values?
            – Dastan
            yesterday










          • Anyway, thanks a lot!!
            – Dastan
            yesterday










          • You can find more information on the coefficients of characteristic polynomials with some references here: math.stackexchange.com/questions/28650/…. In your case, you can always directly expand $Det(tI-A)$ without using this formula. Otherwise, i meant that any matrix of $M_3(mathbb{R})$ have at least one real eigenvalue. That's because the eigenvalues of an $ntimes n$ matrix are the roots of the characteristic polynomial with degree $n$ and when $n$ is odd then this polynomial has at least one real eigenvalue.
            – rezoons
            23 hours ago
















          sorry, my bad! I missed the integer 2
          – Dastan
          yesterday




          sorry, my bad! I missed the integer 2
          – Dastan
          yesterday












          Can you tell me where can I find the proof of the formula of determinant you have used here?Moreover, I din't really get what you meant by saying "A is a matrix of $M_3(mathbb{R})$ so it has real eigen values," because a real matrix may not have real eigen values.
          – Dastan
          yesterday






          Can you tell me where can I find the proof of the formula of determinant you have used here?Moreover, I din't really get what you meant by saying "A is a matrix of $M_3(mathbb{R})$ so it has real eigen values," because a real matrix may not have real eigen values.
          – Dastan
          yesterday














          Is there any way to solve the problem without having resorting to the theory of eigen values?
          – Dastan
          yesterday




          Is there any way to solve the problem without having resorting to the theory of eigen values?
          – Dastan
          yesterday












          Anyway, thanks a lot!!
          – Dastan
          yesterday




          Anyway, thanks a lot!!
          – Dastan
          yesterday












          You can find more information on the coefficients of characteristic polynomials with some references here: math.stackexchange.com/questions/28650/…. In your case, you can always directly expand $Det(tI-A)$ without using this formula. Otherwise, i meant that any matrix of $M_3(mathbb{R})$ have at least one real eigenvalue. That's because the eigenvalues of an $ntimes n$ matrix are the roots of the characteristic polynomial with degree $n$ and when $n$ is odd then this polynomial has at least one real eigenvalue.
          – rezoons
          23 hours ago




          You can find more information on the coefficients of characteristic polynomials with some references here: math.stackexchange.com/questions/28650/…. In your case, you can always directly expand $Det(tI-A)$ without using this formula. Otherwise, i meant that any matrix of $M_3(mathbb{R})$ have at least one real eigenvalue. That's because the eigenvalues of an $ntimes n$ matrix are the roots of the characteristic polynomial with degree $n$ and when $n$ is odd then this polynomial has at least one real eigenvalue.
          – rezoons
          23 hours ago


















           

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