If we know the rank of a matrix r, can we assume that will have precisely r non-zero eigenvalues?












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$begingroup$


I have looked at many answers on the internet regarding the relationship between rank and eigenvalues, and all of them contain complex calculations and descriptions too advanced for me, a beginner student of linear algebra, and do not really answer my question at hand, which is : is it possible to precisely determine the amount of non-zero eigenvalues of a matrix A simply by its rank r? There is a question in my textbook and vaguely alludes to this being the case, but it never actually goes into detail about the relationship between rank and eigenvalues.






linear-algebra eigenvalues-eigenvectors matrix-rank







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  • $begingroup$
    Perhaps you may want to look at this. math.stackexchange.com/questions/1349907/…
    $endgroup$
    – Anurag A
    Dec 17 '18 at 0:16
















0












$begingroup$


I have looked at many answers on the internet regarding the relationship between rank and eigenvalues, and all of them contain complex calculations and descriptions too advanced for me, a beginner student of linear algebra, and do not really answer my question at hand, which is : is it possible to precisely determine the amount of non-zero eigenvalues of a matrix A simply by its rank r? There is a question in my textbook and vaguely alludes to this being the case, but it never actually goes into detail about the relationship between rank and eigenvalues.






linear-algebra eigenvalues-eigenvectors matrix-rank







share|cite|improve this question









$endgroup$












  • $begingroup$
    Perhaps you may want to look at this. math.stackexchange.com/questions/1349907/…
    $endgroup$
    – Anurag A
    Dec 17 '18 at 0:16














0












0








0





$begingroup$


I have looked at many answers on the internet regarding the relationship between rank and eigenvalues, and all of them contain complex calculations and descriptions too advanced for me, a beginner student of linear algebra, and do not really answer my question at hand, which is : is it possible to precisely determine the amount of non-zero eigenvalues of a matrix A simply by its rank r? There is a question in my textbook and vaguely alludes to this being the case, but it never actually goes into detail about the relationship between rank and eigenvalues.






linear-algebra eigenvalues-eigenvectors matrix-rank







share|cite|improve this question









$endgroup$




I have looked at many answers on the internet regarding the relationship between rank and eigenvalues, and all of them contain complex calculations and descriptions too advanced for me, a beginner student of linear algebra, and do not really answer my question at hand, which is : is it possible to precisely determine the amount of non-zero eigenvalues of a matrix A simply by its rank r? There is a question in my textbook and vaguely alludes to this being the case, but it never actually goes into detail about the relationship between rank and eigenvalues.






linear-algebra eigenvalues-eigenvectors matrix-rank




linear-algebra eigenvalues-eigenvectors matrix-rank






share|cite|improve this question













share|cite|improve this question











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asked Dec 17 '18 at 0:10









Will BurghardWill Burghard

42




42












  • $begingroup$
    Perhaps you may want to look at this. math.stackexchange.com/questions/1349907/…
    $endgroup$
    – Anurag A
    Dec 17 '18 at 0:16


















  • $begingroup$
    Perhaps you may want to look at this. math.stackexchange.com/questions/1349907/…
    $endgroup$
    – Anurag A
    Dec 17 '18 at 0:16
















$begingroup$
Perhaps you may want to look at this. math.stackexchange.com/questions/1349907/…
$endgroup$
– Anurag A
Dec 17 '18 at 0:16




$begingroup$
Perhaps you may want to look at this. math.stackexchange.com/questions/1349907/…
$endgroup$
– Anurag A
Dec 17 '18 at 0:16










3 Answers
3






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2












$begingroup$

Consider the identity matrix in $3$ dimensions. It has rank $3$, but the characteristic polynomial is $(1-lambda)^3$, which only has one root $lambda=1$ of multiplicity $3$.






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    2












    $begingroup$

    The matrix will have exactly $r$ non-zero singular values, but not necessarily $r$ non-zero eigenvalues.



    Consider the matrix $A = begin{bmatrix}0&1\0&0end{bmatrix}$. We have $text{rank}(A) = 1$, but both eigenvalues of $A$ are $0$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$


      I have looked at many answers on the internet regarding the
      relationship between rank and eigenvalues, and all of them contain
      complex calculations and descriptions too advanced for me, a beginner
      student of linear algebra




      Supposing that the matrix is square and admits an eigendecomp then you would have $r$ eigenvalues right but there matrices with rank $r$ that aren't square and have $r$ singular values.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        Consider the identity matrix in $3$ dimensions. It has rank $3$, but the characteristic polynomial is $(1-lambda)^3$, which only has one root $lambda=1$ of multiplicity $3$.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          Consider the identity matrix in $3$ dimensions. It has rank $3$, but the characteristic polynomial is $(1-lambda)^3$, which only has one root $lambda=1$ of multiplicity $3$.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            Consider the identity matrix in $3$ dimensions. It has rank $3$, but the characteristic polynomial is $(1-lambda)^3$, which only has one root $lambda=1$ of multiplicity $3$.






            share|cite|improve this answer









            $endgroup$



            Consider the identity matrix in $3$ dimensions. It has rank $3$, but the characteristic polynomial is $(1-lambda)^3$, which only has one root $lambda=1$ of multiplicity $3$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 17 '18 at 0:18









            YiFanYiFan

            3,7511527




            3,7511527























                2












                $begingroup$

                The matrix will have exactly $r$ non-zero singular values, but not necessarily $r$ non-zero eigenvalues.



                Consider the matrix $A = begin{bmatrix}0&1\0&0end{bmatrix}$. We have $text{rank}(A) = 1$, but both eigenvalues of $A$ are $0$.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  The matrix will have exactly $r$ non-zero singular values, but not necessarily $r$ non-zero eigenvalues.



                  Consider the matrix $A = begin{bmatrix}0&1\0&0end{bmatrix}$. We have $text{rank}(A) = 1$, but both eigenvalues of $A$ are $0$.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    The matrix will have exactly $r$ non-zero singular values, but not necessarily $r$ non-zero eigenvalues.



                    Consider the matrix $A = begin{bmatrix}0&1\0&0end{bmatrix}$. We have $text{rank}(A) = 1$, but both eigenvalues of $A$ are $0$.






                    share|cite|improve this answer









                    $endgroup$



                    The matrix will have exactly $r$ non-zero singular values, but not necessarily $r$ non-zero eigenvalues.



                    Consider the matrix $A = begin{bmatrix}0&1\0&0end{bmatrix}$. We have $text{rank}(A) = 1$, but both eigenvalues of $A$ are $0$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 17 '18 at 0:39









                    JimmyK4542JimmyK4542

                    41.1k245106




                    41.1k245106























                        1












                        $begingroup$


                        I have looked at many answers on the internet regarding the
                        relationship between rank and eigenvalues, and all of them contain
                        complex calculations and descriptions too advanced for me, a beginner
                        student of linear algebra




                        Supposing that the matrix is square and admits an eigendecomp then you would have $r$ eigenvalues right but there matrices with rank $r$ that aren't square and have $r$ singular values.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$


                          I have looked at many answers on the internet regarding the
                          relationship between rank and eigenvalues, and all of them contain
                          complex calculations and descriptions too advanced for me, a beginner
                          student of linear algebra




                          Supposing that the matrix is square and admits an eigendecomp then you would have $r$ eigenvalues right but there matrices with rank $r$ that aren't square and have $r$ singular values.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$


                            I have looked at many answers on the internet regarding the
                            relationship between rank and eigenvalues, and all of them contain
                            complex calculations and descriptions too advanced for me, a beginner
                            student of linear algebra




                            Supposing that the matrix is square and admits an eigendecomp then you would have $r$ eigenvalues right but there matrices with rank $r$ that aren't square and have $r$ singular values.






                            share|cite|improve this answer









                            $endgroup$




                            I have looked at many answers on the internet regarding the
                            relationship between rank and eigenvalues, and all of them contain
                            complex calculations and descriptions too advanced for me, a beginner
                            student of linear algebra




                            Supposing that the matrix is square and admits an eigendecomp then you would have $r$ eigenvalues right but there matrices with rank $r$ that aren't square and have $r$ singular values.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 17 '18 at 0:18









                            Ryan HoweRyan Howe

                            2,45411323




                            2,45411323






























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