Why don't similar matrices have same eigenvectors and eigenvalues?












2














What is wrong with this proof:
Suppose R and T are similar operators and R has eigenvalue 2 for some eigenvector $v$.
By property of similar matrices:



$R=STS^{-1}$



Therefore:



$Rv = STS^{-1}v$



$2v = STS^{-1}v$



$2S^{-1}v = TS^{-1}v$



$2S^{-1}Sv = Tv$



$2Iv = Tv$



$Tv = 2v$



Thus $v$ is also an eigenvector of $T$ with eigenvalue 2.
Clearly this proof is incorrect, but where does it go wrong?






linear-algebra matrices eigenvalues-eigenvectors inverse







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  • 1




    Protip: if you're wondering why a calculation fails to be valid, and you know of a counterexample, try evaluating each step using your counterexample, and pick the first line where the two sides fail to be equal. Then you know that this step is the problem, and often the counterexample will show you why the step was problematic. You can use this technique to defeat every $-1 = 1$ fake proof out there!
    – Theo Bendit
    Dec 17 at 23:46
















2














What is wrong with this proof:
Suppose R and T are similar operators and R has eigenvalue 2 for some eigenvector $v$.
By property of similar matrices:



$R=STS^{-1}$



Therefore:



$Rv = STS^{-1}v$



$2v = STS^{-1}v$



$2S^{-1}v = TS^{-1}v$



$2S^{-1}Sv = Tv$



$2Iv = Tv$



$Tv = 2v$



Thus $v$ is also an eigenvector of $T$ with eigenvalue 2.
Clearly this proof is incorrect, but where does it go wrong?






linear-algebra matrices eigenvalues-eigenvectors inverse







share|cite|improve this question







New contributor




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
















  • 1




    Protip: if you're wondering why a calculation fails to be valid, and you know of a counterexample, try evaluating each step using your counterexample, and pick the first line where the two sides fail to be equal. Then you know that this step is the problem, and often the counterexample will show you why the step was problematic. You can use this technique to defeat every $-1 = 1$ fake proof out there!
    – Theo Bendit
    Dec 17 at 23:46














2












2








2


1





What is wrong with this proof:
Suppose R and T are similar operators and R has eigenvalue 2 for some eigenvector $v$.
By property of similar matrices:



$R=STS^{-1}$



Therefore:



$Rv = STS^{-1}v$



$2v = STS^{-1}v$



$2S^{-1}v = TS^{-1}v$



$2S^{-1}Sv = Tv$



$2Iv = Tv$



$Tv = 2v$



Thus $v$ is also an eigenvector of $T$ with eigenvalue 2.
Clearly this proof is incorrect, but where does it go wrong?






linear-algebra matrices eigenvalues-eigenvectors inverse







share|cite|improve this question







New contributor




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











What is wrong with this proof:
Suppose R and T are similar operators and R has eigenvalue 2 for some eigenvector $v$.
By property of similar matrices:



$R=STS^{-1}$



Therefore:



$Rv = STS^{-1}v$



$2v = STS^{-1}v$



$2S^{-1}v = TS^{-1}v$



$2S^{-1}Sv = Tv$



$2Iv = Tv$



$Tv = 2v$



Thus $v$ is also an eigenvector of $T$ with eigenvalue 2.
Clearly this proof is incorrect, but where does it go wrong?






linear-algebra matrices eigenvalues-eigenvectors inverse




linear-algebra matrices eigenvalues-eigenvectors inverse






share|cite|improve this question







New contributor




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











share|cite|improve this question







New contributor




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









share|cite|improve this question




share|cite|improve this question






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Check out our Code of Conduct.









asked Dec 17 at 23:25









Justin Sanders

111




111




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






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








  • 1




    Protip: if you're wondering why a calculation fails to be valid, and you know of a counterexample, try evaluating each step using your counterexample, and pick the first line where the two sides fail to be equal. Then you know that this step is the problem, and often the counterexample will show you why the step was problematic. You can use this technique to defeat every $-1 = 1$ fake proof out there!
    – Theo Bendit
    Dec 17 at 23:46














  • 1




    Protip: if you're wondering why a calculation fails to be valid, and you know of a counterexample, try evaluating each step using your counterexample, and pick the first line where the two sides fail to be equal. Then you know that this step is the problem, and often the counterexample will show you why the step was problematic. You can use this technique to defeat every $-1 = 1$ fake proof out there!
    – Theo Bendit
    Dec 17 at 23:46








1




1




Protip: if you're wondering why a calculation fails to be valid, and you know of a counterexample, try evaluating each step using your counterexample, and pick the first line where the two sides fail to be equal. Then you know that this step is the problem, and often the counterexample will show you why the step was problematic. You can use this technique to defeat every $-1 = 1$ fake proof out there!
– Theo Bendit
Dec 17 at 23:46




Protip: if you're wondering why a calculation fails to be valid, and you know of a counterexample, try evaluating each step using your counterexample, and pick the first line where the two sides fail to be equal. Then you know that this step is the problem, and often the counterexample will show you why the step was problematic. You can use this technique to defeat every $-1 = 1$ fake proof out there!
– Theo Bendit
Dec 17 at 23:46










3 Answers
3






active

oldest

votes


















4














When you went from
$$
2S^{-1}v=TS^{-1}v
$$
to
$$
2S^{-1}Sv=Tv
$$
you completely ignored the rules of matrix multiplcation; you cannot arbitrarily commute matrices in a product! It should have read as follows:
$$
2S^{-1}vS=TS^{-1}vS
$$
Also not that it is not even clear that this is well defined as written, since $v$ is probably a column vector.






share|cite|improve this answer








New contributor




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


























    3














    Since the error in our OP Justin Sanders' argument has been thoroughly vetted in other answers, here I will directly address the title question.



    Similar matrices do have the same eigenvalues, to wit:



    if



    $B = PAP^{-1}, tag 1$



    then



    $B - lambda I = PAP^{-1} - lambda I = PAP^{-1} - lambda PIP^{-1} = P(A - lambda I)P^{-1}, tag 2$



    whence



    $det(B - lambda I) = det(P(A - lambda I)P^{-1}) = det(P) det(A - lambda I) det (P^{-1}) = det(A - lambda I), tag 3$



    since $det(P^{-1}) = (det(P))^{-1}; tag 4$



    thus $A$ and $B$, having the same characteristic polynomials, also share the roots of these polynomials, i.e., their eigenvalues.



    However, similar matrices do not in general share eigenvectors; if



    $B vec v = lambda vec v, tag 5$



    then



    $PAP^{-1} vec v = lambda vec v, tag 6$



    or



    $AP^{-1} vec v = lambda P^{-1} vec v, tag 7$



    that is, $P^{-1} vec v$ is an eigenvector of $A$ corresponding to $lambda$; indeed, if $lambda$ is a root of (3) of multiplicity one, then $P^{-1} vec v$ is, up to a scale factor, the eigenvector of $A$ associated with $lambda$. Since we can in general choose $P$ so that $P^{-1} vec v ne vec v$, the eigenvectors will not be shared 'twixt' $A$ and $B$.



    So while the eigenvalues are similarity invariant, the eigenvectors transform according to $vec v mapsto P^{-1} vec v$.






    share|cite|improve this answer





























      2














      It's wrong when you jump from $2S^{-1}v=TS^{-1}v$ to $2S^{-1}Sv=Tv$.






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

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        active

        oldest

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        active

        oldest

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        4














        When you went from
        $$
        2S^{-1}v=TS^{-1}v
        $$
        to
        $$
        2S^{-1}Sv=Tv
        $$
        you completely ignored the rules of matrix multiplcation; you cannot arbitrarily commute matrices in a product! It should have read as follows:
        $$
        2S^{-1}vS=TS^{-1}vS
        $$
        Also not that it is not even clear that this is well defined as written, since $v$ is probably a column vector.






        share|cite|improve this answer








        New contributor




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























          4














          When you went from
          $$
          2S^{-1}v=TS^{-1}v
          $$
          to
          $$
          2S^{-1}Sv=Tv
          $$
          you completely ignored the rules of matrix multiplcation; you cannot arbitrarily commute matrices in a product! It should have read as follows:
          $$
          2S^{-1}vS=TS^{-1}vS
          $$
          Also not that it is not even clear that this is well defined as written, since $v$ is probably a column vector.






          share|cite|improve this answer








          New contributor




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





















            4












            4








            4






            When you went from
            $$
            2S^{-1}v=TS^{-1}v
            $$
            to
            $$
            2S^{-1}Sv=Tv
            $$
            you completely ignored the rules of matrix multiplcation; you cannot arbitrarily commute matrices in a product! It should have read as follows:
            $$
            2S^{-1}vS=TS^{-1}vS
            $$
            Also not that it is not even clear that this is well defined as written, since $v$ is probably a column vector.






            share|cite|improve this answer








            New contributor




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









            When you went from
            $$
            2S^{-1}v=TS^{-1}v
            $$
            to
            $$
            2S^{-1}Sv=Tv
            $$
            you completely ignored the rules of matrix multiplcation; you cannot arbitrarily commute matrices in a product! It should have read as follows:
            $$
            2S^{-1}vS=TS^{-1}vS
            $$
            Also not that it is not even clear that this is well defined as written, since $v$ is probably a column vector.







            share|cite|improve this answer








            New contributor




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









            share|cite|improve this answer



            share|cite|improve this answer






            New contributor




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









            answered Dec 17 at 23:28









            GenericMathMan

            511




            511




            New contributor




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            New contributor





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























                3














                Since the error in our OP Justin Sanders' argument has been thoroughly vetted in other answers, here I will directly address the title question.



                Similar matrices do have the same eigenvalues, to wit:



                if



                $B = PAP^{-1}, tag 1$



                then



                $B - lambda I = PAP^{-1} - lambda I = PAP^{-1} - lambda PIP^{-1} = P(A - lambda I)P^{-1}, tag 2$



                whence



                $det(B - lambda I) = det(P(A - lambda I)P^{-1}) = det(P) det(A - lambda I) det (P^{-1}) = det(A - lambda I), tag 3$



                since $det(P^{-1}) = (det(P))^{-1}; tag 4$



                thus $A$ and $B$, having the same characteristic polynomials, also share the roots of these polynomials, i.e., their eigenvalues.



                However, similar matrices do not in general share eigenvectors; if



                $B vec v = lambda vec v, tag 5$



                then



                $PAP^{-1} vec v = lambda vec v, tag 6$



                or



                $AP^{-1} vec v = lambda P^{-1} vec v, tag 7$



                that is, $P^{-1} vec v$ is an eigenvector of $A$ corresponding to $lambda$; indeed, if $lambda$ is a root of (3) of multiplicity one, then $P^{-1} vec v$ is, up to a scale factor, the eigenvector of $A$ associated with $lambda$. Since we can in general choose $P$ so that $P^{-1} vec v ne vec v$, the eigenvectors will not be shared 'twixt' $A$ and $B$.



                So while the eigenvalues are similarity invariant, the eigenvectors transform according to $vec v mapsto P^{-1} vec v$.






                share|cite|improve this answer


























                  3














                  Since the error in our OP Justin Sanders' argument has been thoroughly vetted in other answers, here I will directly address the title question.



                  Similar matrices do have the same eigenvalues, to wit:



                  if



                  $B = PAP^{-1}, tag 1$



                  then



                  $B - lambda I = PAP^{-1} - lambda I = PAP^{-1} - lambda PIP^{-1} = P(A - lambda I)P^{-1}, tag 2$



                  whence



                  $det(B - lambda I) = det(P(A - lambda I)P^{-1}) = det(P) det(A - lambda I) det (P^{-1}) = det(A - lambda I), tag 3$



                  since $det(P^{-1}) = (det(P))^{-1}; tag 4$



                  thus $A$ and $B$, having the same characteristic polynomials, also share the roots of these polynomials, i.e., their eigenvalues.



                  However, similar matrices do not in general share eigenvectors; if



                  $B vec v = lambda vec v, tag 5$



                  then



                  $PAP^{-1} vec v = lambda vec v, tag 6$



                  or



                  $AP^{-1} vec v = lambda P^{-1} vec v, tag 7$



                  that is, $P^{-1} vec v$ is an eigenvector of $A$ corresponding to $lambda$; indeed, if $lambda$ is a root of (3) of multiplicity one, then $P^{-1} vec v$ is, up to a scale factor, the eigenvector of $A$ associated with $lambda$. Since we can in general choose $P$ so that $P^{-1} vec v ne vec v$, the eigenvectors will not be shared 'twixt' $A$ and $B$.



                  So while the eigenvalues are similarity invariant, the eigenvectors transform according to $vec v mapsto P^{-1} vec v$.






                  share|cite|improve this answer
























                    3












                    3








                    3






                    Since the error in our OP Justin Sanders' argument has been thoroughly vetted in other answers, here I will directly address the title question.



                    Similar matrices do have the same eigenvalues, to wit:



                    if



                    $B = PAP^{-1}, tag 1$



                    then



                    $B - lambda I = PAP^{-1} - lambda I = PAP^{-1} - lambda PIP^{-1} = P(A - lambda I)P^{-1}, tag 2$



                    whence



                    $det(B - lambda I) = det(P(A - lambda I)P^{-1}) = det(P) det(A - lambda I) det (P^{-1}) = det(A - lambda I), tag 3$



                    since $det(P^{-1}) = (det(P))^{-1}; tag 4$



                    thus $A$ and $B$, having the same characteristic polynomials, also share the roots of these polynomials, i.e., their eigenvalues.



                    However, similar matrices do not in general share eigenvectors; if



                    $B vec v = lambda vec v, tag 5$



                    then



                    $PAP^{-1} vec v = lambda vec v, tag 6$



                    or



                    $AP^{-1} vec v = lambda P^{-1} vec v, tag 7$



                    that is, $P^{-1} vec v$ is an eigenvector of $A$ corresponding to $lambda$; indeed, if $lambda$ is a root of (3) of multiplicity one, then $P^{-1} vec v$ is, up to a scale factor, the eigenvector of $A$ associated with $lambda$. Since we can in general choose $P$ so that $P^{-1} vec v ne vec v$, the eigenvectors will not be shared 'twixt' $A$ and $B$.



                    So while the eigenvalues are similarity invariant, the eigenvectors transform according to $vec v mapsto P^{-1} vec v$.






                    share|cite|improve this answer












                    Since the error in our OP Justin Sanders' argument has been thoroughly vetted in other answers, here I will directly address the title question.



                    Similar matrices do have the same eigenvalues, to wit:



                    if



                    $B = PAP^{-1}, tag 1$



                    then



                    $B - lambda I = PAP^{-1} - lambda I = PAP^{-1} - lambda PIP^{-1} = P(A - lambda I)P^{-1}, tag 2$



                    whence



                    $det(B - lambda I) = det(P(A - lambda I)P^{-1}) = det(P) det(A - lambda I) det (P^{-1}) = det(A - lambda I), tag 3$



                    since $det(P^{-1}) = (det(P))^{-1}; tag 4$



                    thus $A$ and $B$, having the same characteristic polynomials, also share the roots of these polynomials, i.e., their eigenvalues.



                    However, similar matrices do not in general share eigenvectors; if



                    $B vec v = lambda vec v, tag 5$



                    then



                    $PAP^{-1} vec v = lambda vec v, tag 6$



                    or



                    $AP^{-1} vec v = lambda P^{-1} vec v, tag 7$



                    that is, $P^{-1} vec v$ is an eigenvector of $A$ corresponding to $lambda$; indeed, if $lambda$ is a root of (3) of multiplicity one, then $P^{-1} vec v$ is, up to a scale factor, the eigenvector of $A$ associated with $lambda$. Since we can in general choose $P$ so that $P^{-1} vec v ne vec v$, the eigenvectors will not be shared 'twixt' $A$ and $B$.



                    So while the eigenvalues are similarity invariant, the eigenvectors transform according to $vec v mapsto P^{-1} vec v$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 18 at 0:09









                    Robert Lewis

                    43.3k22863




                    43.3k22863























                        2














                        It's wrong when you jump from $2S^{-1}v=TS^{-1}v$ to $2S^{-1}Sv=Tv$.






                        share|cite|improve this answer


























                          2














                          It's wrong when you jump from $2S^{-1}v=TS^{-1}v$ to $2S^{-1}Sv=Tv$.






                          share|cite|improve this answer
























                            2












                            2








                            2






                            It's wrong when you jump from $2S^{-1}v=TS^{-1}v$ to $2S^{-1}Sv=Tv$.






                            share|cite|improve this answer












                            It's wrong when you jump from $2S^{-1}v=TS^{-1}v$ to $2S^{-1}Sv=Tv$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 17 at 23:29









                            José Carlos Santos

                            148k22117218




                            148k22117218






















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                                When can things happen in Etherscan, such as the picture below?

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                                2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  ...
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                                格利澤436b

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                                格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP...
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