Eigenvalues of diagonal matrix












0












$begingroup$


Problem:




Let $A$ ∈ $Bbb C^{n×n}$ and let $A$ be a diagonal Matrix with entries $lambda_1, ldots, lambda_n$ ∈ $mathbb{C}$ Determine spec($A$*$A$)




I think it is clear that the spec ($A$ *$A$) = {$lambda_1, ldots, lambda_n$}, as the eigenvalues of a diagonal matix are just the elements on the diagonal. Could someone appove my thoughts?






linear-algebra eigenvalues-eigenvectors diagonalization







share|cite|improve this question









$endgroup$












  • $begingroup$
    have you tried verifying with particular examples?
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 16:18










  • $begingroup$
    well if we have a diagonal matrix 3x3 with 1,2,3 on the diagonal then the eigenvalues are just 1,2,3 so I suppose that for every diagonal matrix the eigenvalues are all the elements on the diagonal or am I wrong?
    $endgroup$
    – Kai
    Jan 6 at 16:49










  • $begingroup$
    your above statement is correct, but have you computed the corresponding $A^*A$ and see what do you get?
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 16:54










  • $begingroup$
    I think that A*A is the same as A trransponse times A and that gives us also a diagonal matrix or???
    $endgroup$
    – Kai
    Jan 6 at 17:51












  • $begingroup$
    yes, it's a diagonal matrix, refer to Fred's solution.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 17:54
















0












$begingroup$


Problem:




Let $A$ ∈ $Bbb C^{n×n}$ and let $A$ be a diagonal Matrix with entries $lambda_1, ldots, lambda_n$ ∈ $mathbb{C}$ Determine spec($A$*$A$)




I think it is clear that the spec ($A$ *$A$) = {$lambda_1, ldots, lambda_n$}, as the eigenvalues of a diagonal matix are just the elements on the diagonal. Could someone appove my thoughts?






linear-algebra eigenvalues-eigenvectors diagonalization







share|cite|improve this question









$endgroup$












  • $begingroup$
    have you tried verifying with particular examples?
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 16:18










  • $begingroup$
    well if we have a diagonal matrix 3x3 with 1,2,3 on the diagonal then the eigenvalues are just 1,2,3 so I suppose that for every diagonal matrix the eigenvalues are all the elements on the diagonal or am I wrong?
    $endgroup$
    – Kai
    Jan 6 at 16:49










  • $begingroup$
    your above statement is correct, but have you computed the corresponding $A^*A$ and see what do you get?
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 16:54










  • $begingroup$
    I think that A*A is the same as A trransponse times A and that gives us also a diagonal matrix or???
    $endgroup$
    – Kai
    Jan 6 at 17:51












  • $begingroup$
    yes, it's a diagonal matrix, refer to Fred's solution.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 17:54














0












0








0





$begingroup$


Problem:




Let $A$ ∈ $Bbb C^{n×n}$ and let $A$ be a diagonal Matrix with entries $lambda_1, ldots, lambda_n$ ∈ $mathbb{C}$ Determine spec($A$*$A$)




I think it is clear that the spec ($A$ *$A$) = {$lambda_1, ldots, lambda_n$}, as the eigenvalues of a diagonal matix are just the elements on the diagonal. Could someone appove my thoughts?






linear-algebra eigenvalues-eigenvectors diagonalization







share|cite|improve this question









$endgroup$




Problem:




Let $A$ ∈ $Bbb C^{n×n}$ and let $A$ be a diagonal Matrix with entries $lambda_1, ldots, lambda_n$ ∈ $mathbb{C}$ Determine spec($A$*$A$)




I think it is clear that the spec ($A$ *$A$) = {$lambda_1, ldots, lambda_n$}, as the eigenvalues of a diagonal matix are just the elements on the diagonal. Could someone appove my thoughts?






linear-algebra eigenvalues-eigenvectors diagonalization




linear-algebra eigenvalues-eigenvectors diagonalization






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 16:17









KaiKai

686




686












  • $begingroup$
    have you tried verifying with particular examples?
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 16:18










  • $begingroup$
    well if we have a diagonal matrix 3x3 with 1,2,3 on the diagonal then the eigenvalues are just 1,2,3 so I suppose that for every diagonal matrix the eigenvalues are all the elements on the diagonal or am I wrong?
    $endgroup$
    – Kai
    Jan 6 at 16:49










  • $begingroup$
    your above statement is correct, but have you computed the corresponding $A^*A$ and see what do you get?
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 16:54










  • $begingroup$
    I think that A*A is the same as A trransponse times A and that gives us also a diagonal matrix or???
    $endgroup$
    – Kai
    Jan 6 at 17:51












  • $begingroup$
    yes, it's a diagonal matrix, refer to Fred's solution.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 17:54


















  • $begingroup$
    have you tried verifying with particular examples?
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 16:18










  • $begingroup$
    well if we have a diagonal matrix 3x3 with 1,2,3 on the diagonal then the eigenvalues are just 1,2,3 so I suppose that for every diagonal matrix the eigenvalues are all the elements on the diagonal or am I wrong?
    $endgroup$
    – Kai
    Jan 6 at 16:49










  • $begingroup$
    your above statement is correct, but have you computed the corresponding $A^*A$ and see what do you get?
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 16:54










  • $begingroup$
    I think that A*A is the same as A trransponse times A and that gives us also a diagonal matrix or???
    $endgroup$
    – Kai
    Jan 6 at 17:51












  • $begingroup$
    yes, it's a diagonal matrix, refer to Fred's solution.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 17:54
















$begingroup$
have you tried verifying with particular examples?
$endgroup$
– Siong Thye Goh
Jan 6 at 16:18




$begingroup$
have you tried verifying with particular examples?
$endgroup$
– Siong Thye Goh
Jan 6 at 16:18












$begingroup$
well if we have a diagonal matrix 3x3 with 1,2,3 on the diagonal then the eigenvalues are just 1,2,3 so I suppose that for every diagonal matrix the eigenvalues are all the elements on the diagonal or am I wrong?
$endgroup$
– Kai
Jan 6 at 16:49




$begingroup$
well if we have a diagonal matrix 3x3 with 1,2,3 on the diagonal then the eigenvalues are just 1,2,3 so I suppose that for every diagonal matrix the eigenvalues are all the elements on the diagonal or am I wrong?
$endgroup$
– Kai
Jan 6 at 16:49












$begingroup$
your above statement is correct, but have you computed the corresponding $A^*A$ and see what do you get?
$endgroup$
– Siong Thye Goh
Jan 6 at 16:54




$begingroup$
your above statement is correct, but have you computed the corresponding $A^*A$ and see what do you get?
$endgroup$
– Siong Thye Goh
Jan 6 at 16:54












$begingroup$
I think that A*A is the same as A trransponse times A and that gives us also a diagonal matrix or???
$endgroup$
– Kai
Jan 6 at 17:51






$begingroup$
I think that A*A is the same as A trransponse times A and that gives us also a diagonal matrix or???
$endgroup$
– Kai
Jan 6 at 17:51














$begingroup$
yes, it's a diagonal matrix, refer to Fred's solution.
$endgroup$
– Siong Thye Goh
Jan 6 at 17:54




$begingroup$
yes, it's a diagonal matrix, refer to Fred's solution.
$endgroup$
– Siong Thye Goh
Jan 6 at 17:54










2 Answers
2






active

oldest

votes


















2












$begingroup$

$A^*A= diag(|lambda_1|^2,....,|lambda_n|^2)$



Thus the eigenvalues of $A^*A$ are ?






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    In fact when you ll multiply twice the matrix, if you apply an eigenvector, you ll get the eigen value squared. It s quite obvious when you do the computation.
    So you ll get



    spec ($A$ *$A$) = {$lambda_1^2 , ldots, lambda_n^2 $}



    Moreover, I think this is true for any matrix, not only diagonals one.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Failing absolute values? The e-values of the conjugate matrix are conjugate to those of the initial matrix.
      $endgroup$
      – user376343
      Jan 6 at 16:39










    • $begingroup$
      Conjugate? I was thinkikg that the star was the symbol of product of matrices
      $endgroup$
      – Marine Galantin
      Jan 6 at 16:44










    • $begingroup$
      Written in the way of Fred there is no ambiguity but according to the question idk
      $endgroup$
      – Marine Galantin
      Jan 6 at 16:45










    • $begingroup$
      it suffices to add a note in your solution explaining how you perceive A* A. Usually, A* denotes the complex conjugate of A (especially, with a space in A* A)
      $endgroup$
      – user376343
      Jan 6 at 17:05













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    $A^*A= diag(|lambda_1|^2,....,|lambda_n|^2)$



    Thus the eigenvalues of $A^*A$ are ?






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      $A^*A= diag(|lambda_1|^2,....,|lambda_n|^2)$



      Thus the eigenvalues of $A^*A$ are ?






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        $A^*A= diag(|lambda_1|^2,....,|lambda_n|^2)$



        Thus the eigenvalues of $A^*A$ are ?






        share|cite|improve this answer









        $endgroup$



        $A^*A= diag(|lambda_1|^2,....,|lambda_n|^2)$



        Thus the eigenvalues of $A^*A$ are ?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 16:21









        FredFred

        48.7k11849




        48.7k11849























            0












            $begingroup$

            In fact when you ll multiply twice the matrix, if you apply an eigenvector, you ll get the eigen value squared. It s quite obvious when you do the computation.
            So you ll get



            spec ($A$ *$A$) = {$lambda_1^2 , ldots, lambda_n^2 $}



            Moreover, I think this is true for any matrix, not only diagonals one.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Failing absolute values? The e-values of the conjugate matrix are conjugate to those of the initial matrix.
              $endgroup$
              – user376343
              Jan 6 at 16:39










            • $begingroup$
              Conjugate? I was thinkikg that the star was the symbol of product of matrices
              $endgroup$
              – Marine Galantin
              Jan 6 at 16:44










            • $begingroup$
              Written in the way of Fred there is no ambiguity but according to the question idk
              $endgroup$
              – Marine Galantin
              Jan 6 at 16:45










            • $begingroup$
              it suffices to add a note in your solution explaining how you perceive A* A. Usually, A* denotes the complex conjugate of A (especially, with a space in A* A)
              $endgroup$
              – user376343
              Jan 6 at 17:05


















            0












            $begingroup$

            In fact when you ll multiply twice the matrix, if you apply an eigenvector, you ll get the eigen value squared. It s quite obvious when you do the computation.
            So you ll get



            spec ($A$ *$A$) = {$lambda_1^2 , ldots, lambda_n^2 $}



            Moreover, I think this is true for any matrix, not only diagonals one.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Failing absolute values? The e-values of the conjugate matrix are conjugate to those of the initial matrix.
              $endgroup$
              – user376343
              Jan 6 at 16:39










            • $begingroup$
              Conjugate? I was thinkikg that the star was the symbol of product of matrices
              $endgroup$
              – Marine Galantin
              Jan 6 at 16:44










            • $begingroup$
              Written in the way of Fred there is no ambiguity but according to the question idk
              $endgroup$
              – Marine Galantin
              Jan 6 at 16:45










            • $begingroup$
              it suffices to add a note in your solution explaining how you perceive A* A. Usually, A* denotes the complex conjugate of A (especially, with a space in A* A)
              $endgroup$
              – user376343
              Jan 6 at 17:05
















            0












            0








            0





            $begingroup$

            In fact when you ll multiply twice the matrix, if you apply an eigenvector, you ll get the eigen value squared. It s quite obvious when you do the computation.
            So you ll get



            spec ($A$ *$A$) = {$lambda_1^2 , ldots, lambda_n^2 $}



            Moreover, I think this is true for any matrix, not only diagonals one.






            share|cite|improve this answer









            $endgroup$



            In fact when you ll multiply twice the matrix, if you apply an eigenvector, you ll get the eigen value squared. It s quite obvious when you do the computation.
            So you ll get



            spec ($A$ *$A$) = {$lambda_1^2 , ldots, lambda_n^2 $}



            Moreover, I think this is true for any matrix, not only diagonals one.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 at 16:20









            Marine GalantinMarine Galantin

            940419




            940419












            • $begingroup$
              Failing absolute values? The e-values of the conjugate matrix are conjugate to those of the initial matrix.
              $endgroup$
              – user376343
              Jan 6 at 16:39










            • $begingroup$
              Conjugate? I was thinkikg that the star was the symbol of product of matrices
              $endgroup$
              – Marine Galantin
              Jan 6 at 16:44










            • $begingroup$
              Written in the way of Fred there is no ambiguity but according to the question idk
              $endgroup$
              – Marine Galantin
              Jan 6 at 16:45










            • $begingroup$
              it suffices to add a note in your solution explaining how you perceive A* A. Usually, A* denotes the complex conjugate of A (especially, with a space in A* A)
              $endgroup$
              – user376343
              Jan 6 at 17:05




















            • $begingroup$
              Failing absolute values? The e-values of the conjugate matrix are conjugate to those of the initial matrix.
              $endgroup$
              – user376343
              Jan 6 at 16:39










            • $begingroup$
              Conjugate? I was thinkikg that the star was the symbol of product of matrices
              $endgroup$
              – Marine Galantin
              Jan 6 at 16:44










            • $begingroup$
              Written in the way of Fred there is no ambiguity but according to the question idk
              $endgroup$
              – Marine Galantin
              Jan 6 at 16:45










            • $begingroup$
              it suffices to add a note in your solution explaining how you perceive A* A. Usually, A* denotes the complex conjugate of A (especially, with a space in A* A)
              $endgroup$
              – user376343
              Jan 6 at 17:05


















            $begingroup$
            Failing absolute values? The e-values of the conjugate matrix are conjugate to those of the initial matrix.
            $endgroup$
            – user376343
            Jan 6 at 16:39




            $begingroup$
            Failing absolute values? The e-values of the conjugate matrix are conjugate to those of the initial matrix.
            $endgroup$
            – user376343
            Jan 6 at 16:39












            $begingroup$
            Conjugate? I was thinkikg that the star was the symbol of product of matrices
            $endgroup$
            – Marine Galantin
            Jan 6 at 16:44




            $begingroup$
            Conjugate? I was thinkikg that the star was the symbol of product of matrices
            $endgroup$
            – Marine Galantin
            Jan 6 at 16:44












            $begingroup$
            Written in the way of Fred there is no ambiguity but according to the question idk
            $endgroup$
            – Marine Galantin
            Jan 6 at 16:45




            $begingroup$
            Written in the way of Fred there is no ambiguity but according to the question idk
            $endgroup$
            – Marine Galantin
            Jan 6 at 16:45












            $begingroup$
            it suffices to add a note in your solution explaining how you perceive A* A. Usually, A* denotes the complex conjugate of A (especially, with a space in A* A)
            $endgroup$
            – user376343
            Jan 6 at 17:05






            $begingroup$
            it suffices to add a note in your solution explaining how you perceive A* A. Usually, A* denotes the complex conjugate of A (especially, with a space in A* A)
            $endgroup$
            – user376343
            Jan 6 at 17:05




















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