How to prove that the query oracle is unitary?












5












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The query oracle: $O_{x}|irangle|brangle = |irangle|b oplus x_{i}rangle$ used in algorithms like Deutsch Jozsa is unitary. How do I prove it is unitary?






algorithm quantum-gate unitarity deutsch-jozsa-algorithm







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  • $begingroup$
    Related: Why are oracles Hermitian by construction?
    $endgroup$
    – Blue♦
    15 hours ago
















5












$begingroup$


The query oracle: $O_{x}|irangle|brangle = |irangle|b oplus x_{i}rangle$ used in algorithms like Deutsch Jozsa is unitary. How do I prove it is unitary?






algorithm quantum-gate unitarity deutsch-jozsa-algorithm







share|improve this question









New contributor




Divyat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Related: Why are oracles Hermitian by construction?
    $endgroup$
    – Blue♦
    15 hours ago














5












5








5





$begingroup$


The query oracle: $O_{x}|irangle|brangle = |irangle|b oplus x_{i}rangle$ used in algorithms like Deutsch Jozsa is unitary. How do I prove it is unitary?






algorithm quantum-gate unitarity deutsch-jozsa-algorithm







share|improve this question









New contributor




Divyat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




The query oracle: $O_{x}|irangle|brangle = |irangle|b oplus x_{i}rangle$ used in algorithms like Deutsch Jozsa is unitary. How do I prove it is unitary?






algorithm quantum-gate unitarity deutsch-jozsa-algorithm




algorithm quantum-gate unitarity deutsch-jozsa-algorithm






share|improve this question









New contributor




Divyat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Divyat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 15 hours ago









Blue♦

6,57541555




6,57541555






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asked 15 hours ago









DivyatDivyat

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283




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New contributor





Divyat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Divyat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Related: Why are oracles Hermitian by construction?
    $endgroup$
    – Blue♦
    15 hours ago


















  • $begingroup$
    Related: Why are oracles Hermitian by construction?
    $endgroup$
    – Blue♦
    15 hours ago
















$begingroup$
Related: Why are oracles Hermitian by construction?
$endgroup$
– Blue♦
15 hours ago




$begingroup$
Related: Why are oracles Hermitian by construction?
$endgroup$
– Blue♦
15 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Notice that $mathcal O_x$ is a permutation matrix.



The matrix elements are
$$langle j, crvertmathcal O_xlvert i,brangle
=delta_{ij}langle crvert boplus x_irangle
=delta_{ij}delta_{c,boplus x_i}.$$
In other words, $mathcal O_x$ is diagonal with respect to the first register, and, for each block corresponding to a given $i$, connects all and only the indices $b,c$ such that $boplus c=x_i$ (remember that here $b,c,x_iinmathbb Z_2^{otimes n}$ are length-$n$ bit strings).
Also, notice that for a given $b$ there is no more than one $c$ such that $boplus c=x_i$ (more precisely, there isn't any such $c$ if $x_i=0$, and there is exactly one if $x_ineq 0$).



It follows that $mathcal O_x$ is a (real) permutation matrix, and such matrices are always unitarily diagonalizable with unit eigenvalues (and therefore unitary).
In this case, we have that the eigenvalues of $mathcal O_x$ are $pm1$, and the eigenvectors are, in the case $x_ineq 0$,
$$lvert irangleotimes(lvert branglepmlvert boplus x_irangle)$$
for all $i$ and $b$. If $x_i=0$ then $mathcal O_x$ is the identity, and therefore its spectrum is trivial${}^dagger$.



This shows explicitly that $mathcal O_x$ is unitarily diagonalizable with unit eigenvalues, and therefore is unitary.




${}^dagger$ I'm actually being a bit sloppy for the sake of simplicity here. This analysis holds for each different block of $mathcal O_x$ corresponding to a given $i$. More precisely, I should say that $mathcal O_x$ is block-diagonal as it doesn't connect spaces with $ineq j$ on the first register, and each block is either the identity in the subspace in which it acts if $x_i=0$, or a permutation matrix that connects different pairs of basis states if $x_ineq 0$.






share|improve this answer











$endgroup$





















    6












    $begingroup$

    Apply it twice:
    $$
    O_xO_x|irangle|brangle=O_x|irangle|boplus x_irangle=|irangle|boplus x_ioplus x_irangle=|irangle|brangle
    $$
    Hence, $O_x$ is its own inverse.



    Since $O_x^2=mathbb{I}$, the eigenvalues of $O_x^2$ are all 1, meaning that the eigenvalues of $O_x$ are all $pm 1$. Hence $O_x^star=O_x$ (where $^star$ represents the Hermitian conjugate), and therefore
    $$
    O_xO_x^star=mathbb{I},
    $$
    as required for a unitary.






    share|improve this answer











    $endgroup$













    • $begingroup$
      Thanks, I now understand $O_{x}$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^{*}=I$. I think we further need to show $O_{x}$ is Hermitian, then it will be done.
      $endgroup$
      – Divyat
      13 hours ago












    • $begingroup$
      One more doubt, I think the answer still assumes that $O_{x}$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
      $endgroup$
      – Divyat
      12 hours ago












    • $begingroup$
      This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
      $endgroup$
      – Danylo Y
      11 hours ago












    • $begingroup$
      I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $begin{pmatrix}costheta & 2sintheta \ sintheta/2 & -costhetaend{pmatrix}$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
      $endgroup$
      – glS
      9 hours ago













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    2 Answers
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    active

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

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    2












    $begingroup$

    Notice that $mathcal O_x$ is a permutation matrix.



    The matrix elements are
    $$langle j, crvertmathcal O_xlvert i,brangle
    =delta_{ij}langle crvert boplus x_irangle
    =delta_{ij}delta_{c,boplus x_i}.$$
    In other words, $mathcal O_x$ is diagonal with respect to the first register, and, for each block corresponding to a given $i$, connects all and only the indices $b,c$ such that $boplus c=x_i$ (remember that here $b,c,x_iinmathbb Z_2^{otimes n}$ are length-$n$ bit strings).
    Also, notice that for a given $b$ there is no more than one $c$ such that $boplus c=x_i$ (more precisely, there isn't any such $c$ if $x_i=0$, and there is exactly one if $x_ineq 0$).



    It follows that $mathcal O_x$ is a (real) permutation matrix, and such matrices are always unitarily diagonalizable with unit eigenvalues (and therefore unitary).
    In this case, we have that the eigenvalues of $mathcal O_x$ are $pm1$, and the eigenvectors are, in the case $x_ineq 0$,
    $$lvert irangleotimes(lvert branglepmlvert boplus x_irangle)$$
    for all $i$ and $b$. If $x_i=0$ then $mathcal O_x$ is the identity, and therefore its spectrum is trivial${}^dagger$.



    This shows explicitly that $mathcal O_x$ is unitarily diagonalizable with unit eigenvalues, and therefore is unitary.




    ${}^dagger$ I'm actually being a bit sloppy for the sake of simplicity here. This analysis holds for each different block of $mathcal O_x$ corresponding to a given $i$. More precisely, I should say that $mathcal O_x$ is block-diagonal as it doesn't connect spaces with $ineq j$ on the first register, and each block is either the identity in the subspace in which it acts if $x_i=0$, or a permutation matrix that connects different pairs of basis states if $x_ineq 0$.






    share|improve this answer











    $endgroup$


















      2












      $begingroup$

      Notice that $mathcal O_x$ is a permutation matrix.



      The matrix elements are
      $$langle j, crvertmathcal O_xlvert i,brangle
      =delta_{ij}langle crvert boplus x_irangle
      =delta_{ij}delta_{c,boplus x_i}.$$
      In other words, $mathcal O_x$ is diagonal with respect to the first register, and, for each block corresponding to a given $i$, connects all and only the indices $b,c$ such that $boplus c=x_i$ (remember that here $b,c,x_iinmathbb Z_2^{otimes n}$ are length-$n$ bit strings).
      Also, notice that for a given $b$ there is no more than one $c$ such that $boplus c=x_i$ (more precisely, there isn't any such $c$ if $x_i=0$, and there is exactly one if $x_ineq 0$).



      It follows that $mathcal O_x$ is a (real) permutation matrix, and such matrices are always unitarily diagonalizable with unit eigenvalues (and therefore unitary).
      In this case, we have that the eigenvalues of $mathcal O_x$ are $pm1$, and the eigenvectors are, in the case $x_ineq 0$,
      $$lvert irangleotimes(lvert branglepmlvert boplus x_irangle)$$
      for all $i$ and $b$. If $x_i=0$ then $mathcal O_x$ is the identity, and therefore its spectrum is trivial${}^dagger$.



      This shows explicitly that $mathcal O_x$ is unitarily diagonalizable with unit eigenvalues, and therefore is unitary.




      ${}^dagger$ I'm actually being a bit sloppy for the sake of simplicity here. This analysis holds for each different block of $mathcal O_x$ corresponding to a given $i$. More precisely, I should say that $mathcal O_x$ is block-diagonal as it doesn't connect spaces with $ineq j$ on the first register, and each block is either the identity in the subspace in which it acts if $x_i=0$, or a permutation matrix that connects different pairs of basis states if $x_ineq 0$.






      share|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Notice that $mathcal O_x$ is a permutation matrix.



        The matrix elements are
        $$langle j, crvertmathcal O_xlvert i,brangle
        =delta_{ij}langle crvert boplus x_irangle
        =delta_{ij}delta_{c,boplus x_i}.$$
        In other words, $mathcal O_x$ is diagonal with respect to the first register, and, for each block corresponding to a given $i$, connects all and only the indices $b,c$ such that $boplus c=x_i$ (remember that here $b,c,x_iinmathbb Z_2^{otimes n}$ are length-$n$ bit strings).
        Also, notice that for a given $b$ there is no more than one $c$ such that $boplus c=x_i$ (more precisely, there isn't any such $c$ if $x_i=0$, and there is exactly one if $x_ineq 0$).



        It follows that $mathcal O_x$ is a (real) permutation matrix, and such matrices are always unitarily diagonalizable with unit eigenvalues (and therefore unitary).
        In this case, we have that the eigenvalues of $mathcal O_x$ are $pm1$, and the eigenvectors are, in the case $x_ineq 0$,
        $$lvert irangleotimes(lvert branglepmlvert boplus x_irangle)$$
        for all $i$ and $b$. If $x_i=0$ then $mathcal O_x$ is the identity, and therefore its spectrum is trivial${}^dagger$.



        This shows explicitly that $mathcal O_x$ is unitarily diagonalizable with unit eigenvalues, and therefore is unitary.




        ${}^dagger$ I'm actually being a bit sloppy for the sake of simplicity here. This analysis holds for each different block of $mathcal O_x$ corresponding to a given $i$. More precisely, I should say that $mathcal O_x$ is block-diagonal as it doesn't connect spaces with $ineq j$ on the first register, and each block is either the identity in the subspace in which it acts if $x_i=0$, or a permutation matrix that connects different pairs of basis states if $x_ineq 0$.






        share|improve this answer











        $endgroup$



        Notice that $mathcal O_x$ is a permutation matrix.



        The matrix elements are
        $$langle j, crvertmathcal O_xlvert i,brangle
        =delta_{ij}langle crvert boplus x_irangle
        =delta_{ij}delta_{c,boplus x_i}.$$
        In other words, $mathcal O_x$ is diagonal with respect to the first register, and, for each block corresponding to a given $i$, connects all and only the indices $b,c$ such that $boplus c=x_i$ (remember that here $b,c,x_iinmathbb Z_2^{otimes n}$ are length-$n$ bit strings).
        Also, notice that for a given $b$ there is no more than one $c$ such that $boplus c=x_i$ (more precisely, there isn't any such $c$ if $x_i=0$, and there is exactly one if $x_ineq 0$).



        It follows that $mathcal O_x$ is a (real) permutation matrix, and such matrices are always unitarily diagonalizable with unit eigenvalues (and therefore unitary).
        In this case, we have that the eigenvalues of $mathcal O_x$ are $pm1$, and the eigenvectors are, in the case $x_ineq 0$,
        $$lvert irangleotimes(lvert branglepmlvert boplus x_irangle)$$
        for all $i$ and $b$. If $x_i=0$ then $mathcal O_x$ is the identity, and therefore its spectrum is trivial${}^dagger$.



        This shows explicitly that $mathcal O_x$ is unitarily diagonalizable with unit eigenvalues, and therefore is unitary.




        ${}^dagger$ I'm actually being a bit sloppy for the sake of simplicity here. This analysis holds for each different block of $mathcal O_x$ corresponding to a given $i$. More precisely, I should say that $mathcal O_x$ is block-diagonal as it doesn't connect spaces with $ineq j$ on the first register, and each block is either the identity in the subspace in which it acts if $x_i=0$, or a permutation matrix that connects different pairs of basis states if $x_ineq 0$.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 9 hours ago

























        answered 10 hours ago









        glSglS

        4,303739




        4,303739

























            6












            $begingroup$

            Apply it twice:
            $$
            O_xO_x|irangle|brangle=O_x|irangle|boplus x_irangle=|irangle|boplus x_ioplus x_irangle=|irangle|brangle
            $$
            Hence, $O_x$ is its own inverse.



            Since $O_x^2=mathbb{I}$, the eigenvalues of $O_x^2$ are all 1, meaning that the eigenvalues of $O_x$ are all $pm 1$. Hence $O_x^star=O_x$ (where $^star$ represents the Hermitian conjugate), and therefore
            $$
            O_xO_x^star=mathbb{I},
            $$
            as required for a unitary.






            share|improve this answer











            $endgroup$













            • $begingroup$
              Thanks, I now understand $O_{x}$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^{*}=I$. I think we further need to show $O_{x}$ is Hermitian, then it will be done.
              $endgroup$
              – Divyat
              13 hours ago












            • $begingroup$
              One more doubt, I think the answer still assumes that $O_{x}$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
              $endgroup$
              – Divyat
              12 hours ago












            • $begingroup$
              This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
              $endgroup$
              – Danylo Y
              11 hours ago












            • $begingroup$
              I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $begin{pmatrix}costheta & 2sintheta \ sintheta/2 & -costhetaend{pmatrix}$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
              $endgroup$
              – glS
              9 hours ago


















            6












            $begingroup$

            Apply it twice:
            $$
            O_xO_x|irangle|brangle=O_x|irangle|boplus x_irangle=|irangle|boplus x_ioplus x_irangle=|irangle|brangle
            $$
            Hence, $O_x$ is its own inverse.



            Since $O_x^2=mathbb{I}$, the eigenvalues of $O_x^2$ are all 1, meaning that the eigenvalues of $O_x$ are all $pm 1$. Hence $O_x^star=O_x$ (where $^star$ represents the Hermitian conjugate), and therefore
            $$
            O_xO_x^star=mathbb{I},
            $$
            as required for a unitary.






            share|improve this answer











            $endgroup$













            • $begingroup$
              Thanks, I now understand $O_{x}$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^{*}=I$. I think we further need to show $O_{x}$ is Hermitian, then it will be done.
              $endgroup$
              – Divyat
              13 hours ago












            • $begingroup$
              One more doubt, I think the answer still assumes that $O_{x}$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
              $endgroup$
              – Divyat
              12 hours ago












            • $begingroup$
              This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
              $endgroup$
              – Danylo Y
              11 hours ago












            • $begingroup$
              I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $begin{pmatrix}costheta & 2sintheta \ sintheta/2 & -costhetaend{pmatrix}$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
              $endgroup$
              – glS
              9 hours ago
















            6












            6








            6





            $begingroup$

            Apply it twice:
            $$
            O_xO_x|irangle|brangle=O_x|irangle|boplus x_irangle=|irangle|boplus x_ioplus x_irangle=|irangle|brangle
            $$
            Hence, $O_x$ is its own inverse.



            Since $O_x^2=mathbb{I}$, the eigenvalues of $O_x^2$ are all 1, meaning that the eigenvalues of $O_x$ are all $pm 1$. Hence $O_x^star=O_x$ (where $^star$ represents the Hermitian conjugate), and therefore
            $$
            O_xO_x^star=mathbb{I},
            $$
            as required for a unitary.






            share|improve this answer











            $endgroup$



            Apply it twice:
            $$
            O_xO_x|irangle|brangle=O_x|irangle|boplus x_irangle=|irangle|boplus x_ioplus x_irangle=|irangle|brangle
            $$
            Hence, $O_x$ is its own inverse.



            Since $O_x^2=mathbb{I}$, the eigenvalues of $O_x^2$ are all 1, meaning that the eigenvalues of $O_x$ are all $pm 1$. Hence $O_x^star=O_x$ (where $^star$ represents the Hermitian conjugate), and therefore
            $$
            O_xO_x^star=mathbb{I},
            $$
            as required for a unitary.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 13 hours ago

























            answered 15 hours ago









            DaftWullieDaftWullie

            15.2k1541




            15.2k1541












            • $begingroup$
              Thanks, I now understand $O_{x}$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^{*}=I$. I think we further need to show $O_{x}$ is Hermitian, then it will be done.
              $endgroup$
              – Divyat
              13 hours ago












            • $begingroup$
              One more doubt, I think the answer still assumes that $O_{x}$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
              $endgroup$
              – Divyat
              12 hours ago












            • $begingroup$
              This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
              $endgroup$
              – Danylo Y
              11 hours ago












            • $begingroup$
              I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $begin{pmatrix}costheta & 2sintheta \ sintheta/2 & -costhetaend{pmatrix}$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
              $endgroup$
              – glS
              9 hours ago




















            • $begingroup$
              Thanks, I now understand $O_{x}$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^{*}=I$. I think we further need to show $O_{x}$ is Hermitian, then it will be done.
              $endgroup$
              – Divyat
              13 hours ago












            • $begingroup$
              One more doubt, I think the answer still assumes that $O_{x}$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
              $endgroup$
              – Divyat
              12 hours ago












            • $begingroup$
              This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
              $endgroup$
              – Danylo Y
              11 hours ago












            • $begingroup$
              I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $begin{pmatrix}costheta & 2sintheta \ sintheta/2 & -costhetaend{pmatrix}$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
              $endgroup$
              – glS
              9 hours ago


















            $begingroup$
            Thanks, I now understand $O_{x}$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^{*}=I$. I think we further need to show $O_{x}$ is Hermitian, then it will be done.
            $endgroup$
            – Divyat
            13 hours ago






            $begingroup$
            Thanks, I now understand $O_{x}$ is its own inverse but how does that make it unitary? For a unitary matrix, one would need to show that its inverse is equal to its conjugate transpose, $MM^{*}=I$. I think we further need to show $O_{x}$ is Hermitian, then it will be done.
            $endgroup$
            – Divyat
            13 hours ago














            $begingroup$
            One more doubt, I think the answer still assumes that $O_{x}$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
            $endgroup$
            – Divyat
            12 hours ago






            $begingroup$
            One more doubt, I think the answer still assumes that $O_{x}$ is normal matrix, as then with real eigenvalues we can claim it to be Hermitian. Please tell whether it is okay to assume oracle as normal or some justifications for it?
            $endgroup$
            – Divyat
            12 hours ago














            $begingroup$
            This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
            $endgroup$
            – Danylo Y
            11 hours ago






            $begingroup$
            This explanation is not quite correct, because you assumed that $O_x$ is diagonalizable, but it is not follow from $O_x^2=I$. The thing is $O_x$ maps basis vectors to basis vectors. Since $O_x^2=I$ then $Q_x$ is a permutation of basis vectors.
            $endgroup$
            – Danylo Y
            11 hours ago














            $begingroup$
            I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $begin{pmatrix}costheta & 2sintheta \ sintheta/2 & -costhetaend{pmatrix}$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
            $endgroup$
            – glS
            9 hours ago






            $begingroup$
            I actually asked some time ago on math.SE whether one can deduce that $sqrt A$ is diagonalizable from the fact that $A$ is, see here. As far as I understood the answers, $A^2=I$ is not enough to imply that $A$ is unitarily diagonalizable, a counterexample being $begin{pmatrix}costheta & 2sintheta \ sintheta/2 & -costhetaend{pmatrix}$, which is a square root of the identity, but not (unitarily) diagonalizable, and not unitary, even though its eigenvalues are $pm 1$.
            $endgroup$
            – glS
            9 hours ago












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