Eigenvalues of a real orthogonal matrix.












4












$begingroup$


Let $A$ be a real orthogonal matrix. Then $A^{text T} A = I.$ Let $lambda in Bbb C$ be an eigenvalue of $A$ corresponding to the eigenvector $X in Bbb C^n.$ Then we have





$$begin{align*} X^{text T} A^{text T} A X = X^{text T} X. \ implies (AX)^{text T} AX & = X^{text T} X. \ implies (lambda X)^{text T} lambda X & = X^{text T} X. \ implies {lambda}^2 X^{text T} X & = X^{text T} X. \ implies ({lambda}^2 - 1) X^{text T} X & = 0. end{align*}$$





Since $X$ is an eigenvector $X neq 0.$ Therefore ${|X|_2}^2 = X^{text T} X neq 0.$ Hence we must have ${lambda}^2 - 1 = 0$ i.e. ${lambda}^2 = 1.$ So $lambda = pm 1.$



So according to my argument above it follows that eigenvalues of a real orthogonal matrix are $pm 1.$ But I think that I am wrong as I know that the eigenvalues of an orthogonal matrix are unit modulus i.e. they lie on the unit circle.



What's going wrong in my argument above. Please help me in this regard.



Thank you very much for your valuable time.






linear-algebra eigenvalues-eigenvectors orthogonal-matrices







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    Let $A$ be a real orthogonal matrix. Then $A^{text T} A = I.$ Let $lambda in Bbb C$ be an eigenvalue of $A$ corresponding to the eigenvector $X in Bbb C^n.$ Then we have





    $$begin{align*} X^{text T} A^{text T} A X = X^{text T} X. \ implies (AX)^{text T} AX & = X^{text T} X. \ implies (lambda X)^{text T} lambda X & = X^{text T} X. \ implies {lambda}^2 X^{text T} X & = X^{text T} X. \ implies ({lambda}^2 - 1) X^{text T} X & = 0. end{align*}$$





    Since $X$ is an eigenvector $X neq 0.$ Therefore ${|X|_2}^2 = X^{text T} X neq 0.$ Hence we must have ${lambda}^2 - 1 = 0$ i.e. ${lambda}^2 = 1.$ So $lambda = pm 1.$



    So according to my argument above it follows that eigenvalues of a real orthogonal matrix are $pm 1.$ But I think that I am wrong as I know that the eigenvalues of an orthogonal matrix are unit modulus i.e. they lie on the unit circle.



    What's going wrong in my argument above. Please help me in this regard.



    Thank you very much for your valuable time.






    linear-algebra eigenvalues-eigenvectors orthogonal-matrices







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      0



      $begingroup$


      Let $A$ be a real orthogonal matrix. Then $A^{text T} A = I.$ Let $lambda in Bbb C$ be an eigenvalue of $A$ corresponding to the eigenvector $X in Bbb C^n.$ Then we have





      $$begin{align*} X^{text T} A^{text T} A X = X^{text T} X. \ implies (AX)^{text T} AX & = X^{text T} X. \ implies (lambda X)^{text T} lambda X & = X^{text T} X. \ implies {lambda}^2 X^{text T} X & = X^{text T} X. \ implies ({lambda}^2 - 1) X^{text T} X & = 0. end{align*}$$





      Since $X$ is an eigenvector $X neq 0.$ Therefore ${|X|_2}^2 = X^{text T} X neq 0.$ Hence we must have ${lambda}^2 - 1 = 0$ i.e. ${lambda}^2 = 1.$ So $lambda = pm 1.$



      So according to my argument above it follows that eigenvalues of a real orthogonal matrix are $pm 1.$ But I think that I am wrong as I know that the eigenvalues of an orthogonal matrix are unit modulus i.e. they lie on the unit circle.



      What's going wrong in my argument above. Please help me in this regard.



      Thank you very much for your valuable time.






      linear-algebra eigenvalues-eigenvectors orthogonal-matrices







      share|cite|improve this question











      $endgroup$




      Let $A$ be a real orthogonal matrix. Then $A^{text T} A = I.$ Let $lambda in Bbb C$ be an eigenvalue of $A$ corresponding to the eigenvector $X in Bbb C^n.$ Then we have





      $$begin{align*} X^{text T} A^{text T} A X = X^{text T} X. \ implies (AX)^{text T} AX & = X^{text T} X. \ implies (lambda X)^{text T} lambda X & = X^{text T} X. \ implies {lambda}^2 X^{text T} X & = X^{text T} X. \ implies ({lambda}^2 - 1) X^{text T} X & = 0. end{align*}$$





      Since $X$ is an eigenvector $X neq 0.$ Therefore ${|X|_2}^2 = X^{text T} X neq 0.$ Hence we must have ${lambda}^2 - 1 = 0$ i.e. ${lambda}^2 = 1.$ So $lambda = pm 1.$



      So according to my argument above it follows that eigenvalues of a real orthogonal matrix are $pm 1.$ But I think that I am wrong as I know that the eigenvalues of an orthogonal matrix are unit modulus i.e. they lie on the unit circle.



      What's going wrong in my argument above. Please help me in this regard.



      Thank you very much for your valuable time.






      linear-algebra eigenvalues-eigenvectors orthogonal-matrices




      linear-algebra eigenvalues-eigenvectors orthogonal-matrices






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 31 at 11:14









      Yanko

      8,4692830




      8,4692830










      asked Mar 31 at 5:00









      math maniac.math maniac.

      1647




      1647






















          1 Answer
          1






          active

          oldest

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          4












          $begingroup$

          The mistake is your assumption that $X^TXne0$. Consider a simple example:
          $$A=pmatrix{0&1\-1&0}.$$
          It is orthogonal, and its eigenvalues are $pm i$. One eigenvector is
          $$X=pmatrix{1\i}.$$
          It satisfies $X^TX=0$.



          However, replacing $X^T$ in your argument by $X^H$ (complex conjugate
          of transpose) will give you the correct conclusion that $|lambda|^2=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            how can Euclidean norm of non zero vector be zero?
            $endgroup$
            – math maniac.
            Mar 31 at 5:08






          • 2




            $begingroup$
            @mathmaniac. How can $1^2+i^2$ equal zero?
            $endgroup$
            – Lord Shark the Unknown
            Mar 31 at 5:12










          • $begingroup$
            I think the Euclidean norm of $X in Bbb C^n$ is $sqrt {X^{text T} overline X} text {or} sqrt {{overline X}^{text T} X},$ not $sqrt {X^{text T} X}.$ Am I right?
            $endgroup$
            – math maniac.
            Mar 31 at 5:13








          • 1




            $begingroup$
            Which is same as $sqrt {X^{text H}X},$ as you have rightly pointed out.
            $endgroup$
            – math maniac.
            Mar 31 at 5:19














          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          4












          $begingroup$

          The mistake is your assumption that $X^TXne0$. Consider a simple example:
          $$A=pmatrix{0&1\-1&0}.$$
          It is orthogonal, and its eigenvalues are $pm i$. One eigenvector is
          $$X=pmatrix{1\i}.$$
          It satisfies $X^TX=0$.



          However, replacing $X^T$ in your argument by $X^H$ (complex conjugate
          of transpose) will give you the correct conclusion that $|lambda|^2=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            how can Euclidean norm of non zero vector be zero?
            $endgroup$
            – math maniac.
            Mar 31 at 5:08






          • 2




            $begingroup$
            @mathmaniac. How can $1^2+i^2$ equal zero?
            $endgroup$
            – Lord Shark the Unknown
            Mar 31 at 5:12










          • $begingroup$
            I think the Euclidean norm of $X in Bbb C^n$ is $sqrt {X^{text T} overline X} text {or} sqrt {{overline X}^{text T} X},$ not $sqrt {X^{text T} X}.$ Am I right?
            $endgroup$
            – math maniac.
            Mar 31 at 5:13








          • 1




            $begingroup$
            Which is same as $sqrt {X^{text H}X},$ as you have rightly pointed out.
            $endgroup$
            – math maniac.
            Mar 31 at 5:19


















          4












          $begingroup$

          The mistake is your assumption that $X^TXne0$. Consider a simple example:
          $$A=pmatrix{0&1\-1&0}.$$
          It is orthogonal, and its eigenvalues are $pm i$. One eigenvector is
          $$X=pmatrix{1\i}.$$
          It satisfies $X^TX=0$.



          However, replacing $X^T$ in your argument by $X^H$ (complex conjugate
          of transpose) will give you the correct conclusion that $|lambda|^2=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            how can Euclidean norm of non zero vector be zero?
            $endgroup$
            – math maniac.
            Mar 31 at 5:08






          • 2




            $begingroup$
            @mathmaniac. How can $1^2+i^2$ equal zero?
            $endgroup$
            – Lord Shark the Unknown
            Mar 31 at 5:12










          • $begingroup$
            I think the Euclidean norm of $X in Bbb C^n$ is $sqrt {X^{text T} overline X} text {or} sqrt {{overline X}^{text T} X},$ not $sqrt {X^{text T} X}.$ Am I right?
            $endgroup$
            – math maniac.
            Mar 31 at 5:13








          • 1




            $begingroup$
            Which is same as $sqrt {X^{text H}X},$ as you have rightly pointed out.
            $endgroup$
            – math maniac.
            Mar 31 at 5:19
















          4












          4








          4





          $begingroup$

          The mistake is your assumption that $X^TXne0$. Consider a simple example:
          $$A=pmatrix{0&1\-1&0}.$$
          It is orthogonal, and its eigenvalues are $pm i$. One eigenvector is
          $$X=pmatrix{1\i}.$$
          It satisfies $X^TX=0$.



          However, replacing $X^T$ in your argument by $X^H$ (complex conjugate
          of transpose) will give you the correct conclusion that $|lambda|^2=1$.






          share|cite|improve this answer









          $endgroup$



          The mistake is your assumption that $X^TXne0$. Consider a simple example:
          $$A=pmatrix{0&1\-1&0}.$$
          It is orthogonal, and its eigenvalues are $pm i$. One eigenvector is
          $$X=pmatrix{1\i}.$$
          It satisfies $X^TX=0$.



          However, replacing $X^T$ in your argument by $X^H$ (complex conjugate
          of transpose) will give you the correct conclusion that $|lambda|^2=1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 31 at 5:04









          Lord Shark the UnknownLord Shark the Unknown

          108k1162136




          108k1162136












          • $begingroup$
            how can Euclidean norm of non zero vector be zero?
            $endgroup$
            – math maniac.
            Mar 31 at 5:08






          • 2




            $begingroup$
            @mathmaniac. How can $1^2+i^2$ equal zero?
            $endgroup$
            – Lord Shark the Unknown
            Mar 31 at 5:12










          • $begingroup$
            I think the Euclidean norm of $X in Bbb C^n$ is $sqrt {X^{text T} overline X} text {or} sqrt {{overline X}^{text T} X},$ not $sqrt {X^{text T} X}.$ Am I right?
            $endgroup$
            – math maniac.
            Mar 31 at 5:13








          • 1




            $begingroup$
            Which is same as $sqrt {X^{text H}X},$ as you have rightly pointed out.
            $endgroup$
            – math maniac.
            Mar 31 at 5:19




















          • $begingroup$
            how can Euclidean norm of non zero vector be zero?
            $endgroup$
            – math maniac.
            Mar 31 at 5:08






          • 2




            $begingroup$
            @mathmaniac. How can $1^2+i^2$ equal zero?
            $endgroup$
            – Lord Shark the Unknown
            Mar 31 at 5:12










          • $begingroup$
            I think the Euclidean norm of $X in Bbb C^n$ is $sqrt {X^{text T} overline X} text {or} sqrt {{overline X}^{text T} X},$ not $sqrt {X^{text T} X}.$ Am I right?
            $endgroup$
            – math maniac.
            Mar 31 at 5:13








          • 1




            $begingroup$
            Which is same as $sqrt {X^{text H}X},$ as you have rightly pointed out.
            $endgroup$
            – math maniac.
            Mar 31 at 5:19


















          $begingroup$
          how can Euclidean norm of non zero vector be zero?
          $endgroup$
          – math maniac.
          Mar 31 at 5:08




          $begingroup$
          how can Euclidean norm of non zero vector be zero?
          $endgroup$
          – math maniac.
          Mar 31 at 5:08




          2




          2




          $begingroup$
          @mathmaniac. How can $1^2+i^2$ equal zero?
          $endgroup$
          – Lord Shark the Unknown
          Mar 31 at 5:12




          $begingroup$
          @mathmaniac. How can $1^2+i^2$ equal zero?
          $endgroup$
          – Lord Shark the Unknown
          Mar 31 at 5:12












          $begingroup$
          I think the Euclidean norm of $X in Bbb C^n$ is $sqrt {X^{text T} overline X} text {or} sqrt {{overline X}^{text T} X},$ not $sqrt {X^{text T} X}.$ Am I right?
          $endgroup$
          – math maniac.
          Mar 31 at 5:13






          $begingroup$
          I think the Euclidean norm of $X in Bbb C^n$ is $sqrt {X^{text T} overline X} text {or} sqrt {{overline X}^{text T} X},$ not $sqrt {X^{text T} X}.$ Am I right?
          $endgroup$
          – math maniac.
          Mar 31 at 5:13






          1




          1




          $begingroup$
          Which is same as $sqrt {X^{text H}X},$ as you have rightly pointed out.
          $endgroup$
          – math maniac.
          Mar 31 at 5:19






          $begingroup$
          Which is same as $sqrt {X^{text H}X},$ as you have rightly pointed out.
          $endgroup$
          – math maniac.
          Mar 31 at 5:19




















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