For an orthogonal matrix $Q$, prove $operatornamecond(Q)=1$Orthogonal matrix normWhy is the matrix product of 2 orthogonal matrices also an orthogonal matrix?Finding an orthogonal Matrix with distinct eigenvaluesOrthogonal Matrix Eigenvaluehow to prove this matrix is orthogonalShow that for $n times n$ orthogonal matrix $A$ that $operatornameCond(A)leq n$Can we prove that the eigenvalues of an $ntimes n$ orthogonal matrix are $pm 1$ by only using the definition of orthogonal matrix?Show non-singularity of orthogonal matrixProve that the multiples of two orthogonal eigenvectors with a matrix are also orthogonalProve that $-1$ is an eigenvalue of an orthogonal matrix $A in M_4 times 4 (Bbb R)$ with $det(A)=-1$Band matrix conjugation relation between orthogonal matrices

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For an orthogonal matrix $Q$, prove $operatornamecond(Q)=1$


Orthogonal matrix normWhy is the matrix product of 2 orthogonal matrices also an orthogonal matrix?Finding an orthogonal Matrix with distinct eigenvaluesOrthogonal Matrix Eigenvaluehow to prove this matrix is orthogonalShow that for $n times n$ orthogonal matrix $A$ that $operatornameCond(A)leq n$Can we prove that the eigenvalues of an $ntimes n$ orthogonal matrix are $pm 1$ by only using the definition of orthogonal matrix?Show non-singularity of orthogonal matrixProve that the multiples of two orthogonal eigenvectors with a matrix are also orthogonalProve that $-1$ is an eigenvalue of an orthogonal matrix $A in M_4 times 4 (Bbb R)$ with $det(A)=-1$Band matrix conjugation relation between orthogonal matrices













0












$begingroup$



Given an orthogonal matrix $Q$, prove $$|Q|_2cdot |Q^-1|_2=1$$




I succeed to solve it with eigenvalues but I'm looking for an easier way.






matrices orthogonal-matrices condition-number spectral-norm







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    An orthogonal matrix actually has operator norm equal to $1$. And if $Q$ is orthogonal, then so is $Q^-1$.
    $endgroup$
    – Minus One-Twelfth
    yesterday
















0












$begingroup$



Given an orthogonal matrix $Q$, prove $$|Q|_2cdot |Q^-1|_2=1$$




I succeed to solve it with eigenvalues but I'm looking for an easier way.






matrices orthogonal-matrices condition-number spectral-norm







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    An orthogonal matrix actually has operator norm equal to $1$. And if $Q$ is orthogonal, then so is $Q^-1$.
    $endgroup$
    – Minus One-Twelfth
    yesterday














0












0








0





$begingroup$



Given an orthogonal matrix $Q$, prove $$|Q|_2cdot |Q^-1|_2=1$$




I succeed to solve it with eigenvalues but I'm looking for an easier way.






matrices orthogonal-matrices condition-number spectral-norm







share|cite|improve this question











$endgroup$





Given an orthogonal matrix $Q$, prove $$|Q|_2cdot |Q^-1|_2=1$$




I succeed to solve it with eigenvalues but I'm looking for an easier way.






matrices orthogonal-matrices condition-number spectral-norm




matrices orthogonal-matrices condition-number spectral-norm






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









Rodrigo de Azevedo

13.1k41960




13.1k41960










asked yesterday









J. DoeJ. Doe

16811




16811







  • 1




    $begingroup$
    An orthogonal matrix actually has operator norm equal to $1$. And if $Q$ is orthogonal, then so is $Q^-1$.
    $endgroup$
    – Minus One-Twelfth
    yesterday













  • 1




    $begingroup$
    An orthogonal matrix actually has operator norm equal to $1$. And if $Q$ is orthogonal, then so is $Q^-1$.
    $endgroup$
    – Minus One-Twelfth
    yesterday








1




1




$begingroup$
An orthogonal matrix actually has operator norm equal to $1$. And if $Q$ is orthogonal, then so is $Q^-1$.
$endgroup$
– Minus One-Twelfth
yesterday





$begingroup$
An orthogonal matrix actually has operator norm equal to $1$. And if $Q$ is orthogonal, then so is $Q^-1$.
$endgroup$
– Minus One-Twelfth
yesterday











1 Answer
1






active

oldest

votes


















0












$begingroup$

An orthogonal matrix preserves distance (it's an isometry). It means that for any $mathbf x$ we have $|Qmathbf x|_2=|mathbf x|_2$.



So:
$$|Q|_2 oversettextdef= sup_mathbf xne mathbf 0frac_2 = sup_mathbf xne mathbf 0frac_2_2=1$$
Since $Q^-1$ is orthogonal as well, the result follows.






share|cite|improve this answer









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    0












    $begingroup$

    An orthogonal matrix preserves distance (it's an isometry). It means that for any $mathbf x$ we have $|Qmathbf x|_2=|mathbf x|_2$.



    So:
    $$|Q|_2 oversettextdef= sup_mathbf xne mathbf 0frac_2 = sup_mathbf xne mathbf 0frac_2_2=1$$
    Since $Q^-1$ is orthogonal as well, the result follows.






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      An orthogonal matrix preserves distance (it's an isometry). It means that for any $mathbf x$ we have $|Qmathbf x|_2=|mathbf x|_2$.



      So:
      $$|Q|_2 oversettextdef= sup_mathbf xne mathbf 0frac_2 = sup_mathbf xne mathbf 0frac_2_2=1$$
      Since $Q^-1$ is orthogonal as well, the result follows.






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        An orthogonal matrix preserves distance (it's an isometry). It means that for any $mathbf x$ we have $|Qmathbf x|_2=|mathbf x|_2$.



        So:
        $$|Q|_2 oversettextdef= sup_mathbf xne mathbf 0frac_2 = sup_mathbf xne mathbf 0frac_2_2=1$$
        Since $Q^-1$ is orthogonal as well, the result follows.






        share|cite|improve this answer









        $endgroup$



        An orthogonal matrix preserves distance (it's an isometry). It means that for any $mathbf x$ we have $|Qmathbf x|_2=|mathbf x|_2$.



        So:
        $$|Q|_2 oversettextdef= sup_mathbf xne mathbf 0frac_2 = sup_mathbf xne mathbf 0frac_2_2=1$$
        Since $Q^-1$ is orthogonal as well, the result follows.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        I like SerenaI like Serena

        4,2721722




        4,2721722



























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