Do $ AB $ and $ BA $ have same minimal and characteristic polynomials?












33















Let $ A, B $ be two square matrices of order $n$. Do $ AB $ and $ BA $ have same minimal and characteristic polynomials?




I have a proof only if $ A$ or $ B $ is invertible. Is it true for all cases?










share|cite|improve this question




















  • 2




    The coefficients of the characteristic polynomial are continuous functions in the entries of a matrix, so if the characteristic polynomials of $AB$ and $BA$ coincide for a dense set of $A$ (or a dense set of $B$) then they always coincide. The coefficients of the minimal polynomial, on the other hand...
    – Qiaochu Yuan
    Feb 22 '13 at 21:12










  • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@QiaochuYuan
    – cmi
    Nov 19 '18 at 4:56
















33















Let $ A, B $ be two square matrices of order $n$. Do $ AB $ and $ BA $ have same minimal and characteristic polynomials?




I have a proof only if $ A$ or $ B $ is invertible. Is it true for all cases?










share|cite|improve this question




















  • 2




    The coefficients of the characteristic polynomial are continuous functions in the entries of a matrix, so if the characteristic polynomials of $AB$ and $BA$ coincide for a dense set of $A$ (or a dense set of $B$) then they always coincide. The coefficients of the minimal polynomial, on the other hand...
    – Qiaochu Yuan
    Feb 22 '13 at 21:12










  • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@QiaochuYuan
    – cmi
    Nov 19 '18 at 4:56














33












33








33


31






Let $ A, B $ be two square matrices of order $n$. Do $ AB $ and $ BA $ have same minimal and characteristic polynomials?




I have a proof only if $ A$ or $ B $ is invertible. Is it true for all cases?










share|cite|improve this question
















Let $ A, B $ be two square matrices of order $n$. Do $ AB $ and $ BA $ have same minimal and characteristic polynomials?




I have a proof only if $ A$ or $ B $ is invertible. Is it true for all cases?







linear-algebra matrices






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '15 at 23:49









user26857

39.3k124083




39.3k124083










asked Feb 22 '13 at 17:16









AndyAndy

9901018




9901018








  • 2




    The coefficients of the characteristic polynomial are continuous functions in the entries of a matrix, so if the characteristic polynomials of $AB$ and $BA$ coincide for a dense set of $A$ (or a dense set of $B$) then they always coincide. The coefficients of the minimal polynomial, on the other hand...
    – Qiaochu Yuan
    Feb 22 '13 at 21:12










  • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@QiaochuYuan
    – cmi
    Nov 19 '18 at 4:56














  • 2




    The coefficients of the characteristic polynomial are continuous functions in the entries of a matrix, so if the characteristic polynomials of $AB$ and $BA$ coincide for a dense set of $A$ (or a dense set of $B$) then they always coincide. The coefficients of the minimal polynomial, on the other hand...
    – Qiaochu Yuan
    Feb 22 '13 at 21:12










  • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@QiaochuYuan
    – cmi
    Nov 19 '18 at 4:56








2




2




The coefficients of the characteristic polynomial are continuous functions in the entries of a matrix, so if the characteristic polynomials of $AB$ and $BA$ coincide for a dense set of $A$ (or a dense set of $B$) then they always coincide. The coefficients of the minimal polynomial, on the other hand...
– Qiaochu Yuan
Feb 22 '13 at 21:12




The coefficients of the characteristic polynomial are continuous functions in the entries of a matrix, so if the characteristic polynomials of $AB$ and $BA$ coincide for a dense set of $A$ (or a dense set of $B$) then they always coincide. The coefficients of the minimal polynomial, on the other hand...
– Qiaochu Yuan
Feb 22 '13 at 21:12












Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@QiaochuYuan
– cmi
Nov 19 '18 at 4:56




Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@QiaochuYuan
– cmi
Nov 19 '18 at 4:56










5 Answers
5






active

oldest

votes


















28














If $A$ is invertible then $A^{-1}(AB)A= BA$, so $AB$ and $BA$ are similar, which implies (but is stronger than) $AB$ and $BA$ have the same minimal polynomial and the same characteristic polynomial.
The same goes if $B$ is invertible.



In general, from the above observation, it is not too difficult to show that $AB$, and $BA$ have the same characteristic polynomial, the type of proof could depends on the field considered for the coefficient of your matrices though.
If the matrices are in $mathcal{M}_n(mathbb C)$, you use the fact that $operatorname{GL}_n(mathbb C)$ is dense in $mathcal{M}_n(mathbb C)$ and the continuity of the function which maps a matrix to its characteristic polynomial. There are at least 5 other ways to proceed (especially for other field than $mathbb C$).



In general $AB$ and $BA$ do not have the same minimal polynomial. I'll let you search a bit for a counter example.






share|cite|improve this answer



















  • 9




    5 other ways? I'm quite curious what those ways are. I only know of the continuity argument and an argument involving determinant identities on block matrices. Would it be possible to provide a reference to some other methods?
    – EuYu
    Feb 22 '13 at 17:37










  • @EuYu, I have no reference, sorry.
    – Nathan Portland
    Feb 22 '13 at 22:11










  • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same? @NathanPortland
    – cmi
    Nov 19 '18 at 4:55



















24














Before proving $AB$ and $BA$ have the same characteristic polynomials show that if $A_{mtimes n}$ and $B_{ntimes m} $ then characteristic polynomials of $AB$ and $BA$ satisfy following statement: $$x^n|xI_m-AB|=x^m|xI_n-BA|$$ therefore easily conclude if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.



Define $$C = begin{bmatrix} xI_m & A \B & I_n end{bmatrix}, D = begin{bmatrix} I_m & 0 \-B & xI_n end{bmatrix}.$$ We have
$$
begin{align*}
det CD &= x^n|xI_m-AB|,\
det DC &= x^m|xI_n-BA|.
end{align*}
$$
and we know $det CD=det DC$ if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.






share|cite|improve this answer























  • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@M.H
    – cmi
    Nov 19 '18 at 4:56



















15














Hint: Consider $A = begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}$ and $B = begin{bmatrix} 0 & 0 \ 0 & 1 end{bmatrix}$. What do you get in that case?






share|cite|improve this answer

















  • 2




    This shows that $AB$ and $BA$ have different minimal polynomial. But characteristic polynomials are the same right?
    – Unknown x
    Nov 25 '17 at 1:39










  • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@Jim
    – cmi
    Nov 19 '18 at 4:56



















4














It's not true that their characteristic polynomials will be the same in the general case. The best result in this general vein is the following.



Let $Ainmathbb{F}^{m times n}$ and let $Binmathbb{F}^{n times m}$, and $AB$, $BA$ with minimal polynomials (over $mathbb{F}$) $m_{AB}(x)$ and $m_{BA}(x)$ respectively. Then one of the following holds:



$m_{AB}(x) = m_{BA}(x)$, or $m_{AB}(x) = x cdot m_{BA}(x)$, or $xcdot m_{AB}(x) = m_{BA}(x)$.



It's easy, just use the fact that $(BA)^k=B(AB)^{k-1}A$.






share|cite|improve this answer























  • in general means for m not equal to n??
    – user152715
    Sep 16 '14 at 21:16










  • The question says (and always has said) square matrices $A,B$. If you are answering a more general question, then you should announce this. Also, answering a more general question is only useful if this is no more difficult than the actual question, or if the more general solution sheds more light on the solution.
    – Marc van Leeuwen
    Dec 23 '15 at 21:10












  • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@MarcvanLeeuwen
    – cmi
    Nov 19 '18 at 4:58



















0














For squre matrix, the charateristic polynomials are same, but for $A$ is matrix of size $m times n$ and B be matrix of size $n times m$. Then $x^{m}C_{BA}(x)=x^{n}C_{AB}(x)$ this implies that the nonzero eigenvalue of $AB$, counted with multiplicites, are same as nonzero eigenvalue of $BA$.



That is if A is of size 7×4 and B is of size 4×7 and assume that the 4×4 matrix BA has nonzero eigenvalue 1,1,3 so fourth eigenvalue of BA is 0. Then the 7×7 matrix AB will also have nonzero eigenvalue 1,1,3 and remaining four eigenvalue of AB are zero.






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    5 Answers
    5






    active

    oldest

    votes








    5 Answers
    5






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    28














    If $A$ is invertible then $A^{-1}(AB)A= BA$, so $AB$ and $BA$ are similar, which implies (but is stronger than) $AB$ and $BA$ have the same minimal polynomial and the same characteristic polynomial.
    The same goes if $B$ is invertible.



    In general, from the above observation, it is not too difficult to show that $AB$, and $BA$ have the same characteristic polynomial, the type of proof could depends on the field considered for the coefficient of your matrices though.
    If the matrices are in $mathcal{M}_n(mathbb C)$, you use the fact that $operatorname{GL}_n(mathbb C)$ is dense in $mathcal{M}_n(mathbb C)$ and the continuity of the function which maps a matrix to its characteristic polynomial. There are at least 5 other ways to proceed (especially for other field than $mathbb C$).



    In general $AB$ and $BA$ do not have the same minimal polynomial. I'll let you search a bit for a counter example.






    share|cite|improve this answer



















    • 9




      5 other ways? I'm quite curious what those ways are. I only know of the continuity argument and an argument involving determinant identities on block matrices. Would it be possible to provide a reference to some other methods?
      – EuYu
      Feb 22 '13 at 17:37










    • @EuYu, I have no reference, sorry.
      – Nathan Portland
      Feb 22 '13 at 22:11










    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same? @NathanPortland
      – cmi
      Nov 19 '18 at 4:55
















    28














    If $A$ is invertible then $A^{-1}(AB)A= BA$, so $AB$ and $BA$ are similar, which implies (but is stronger than) $AB$ and $BA$ have the same minimal polynomial and the same characteristic polynomial.
    The same goes if $B$ is invertible.



    In general, from the above observation, it is not too difficult to show that $AB$, and $BA$ have the same characteristic polynomial, the type of proof could depends on the field considered for the coefficient of your matrices though.
    If the matrices are in $mathcal{M}_n(mathbb C)$, you use the fact that $operatorname{GL}_n(mathbb C)$ is dense in $mathcal{M}_n(mathbb C)$ and the continuity of the function which maps a matrix to its characteristic polynomial. There are at least 5 other ways to proceed (especially for other field than $mathbb C$).



    In general $AB$ and $BA$ do not have the same minimal polynomial. I'll let you search a bit for a counter example.






    share|cite|improve this answer



















    • 9




      5 other ways? I'm quite curious what those ways are. I only know of the continuity argument and an argument involving determinant identities on block matrices. Would it be possible to provide a reference to some other methods?
      – EuYu
      Feb 22 '13 at 17:37










    • @EuYu, I have no reference, sorry.
      – Nathan Portland
      Feb 22 '13 at 22:11










    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same? @NathanPortland
      – cmi
      Nov 19 '18 at 4:55














    28












    28








    28






    If $A$ is invertible then $A^{-1}(AB)A= BA$, so $AB$ and $BA$ are similar, which implies (but is stronger than) $AB$ and $BA$ have the same minimal polynomial and the same characteristic polynomial.
    The same goes if $B$ is invertible.



    In general, from the above observation, it is not too difficult to show that $AB$, and $BA$ have the same characteristic polynomial, the type of proof could depends on the field considered for the coefficient of your matrices though.
    If the matrices are in $mathcal{M}_n(mathbb C)$, you use the fact that $operatorname{GL}_n(mathbb C)$ is dense in $mathcal{M}_n(mathbb C)$ and the continuity of the function which maps a matrix to its characteristic polynomial. There are at least 5 other ways to proceed (especially for other field than $mathbb C$).



    In general $AB$ and $BA$ do not have the same minimal polynomial. I'll let you search a bit for a counter example.






    share|cite|improve this answer














    If $A$ is invertible then $A^{-1}(AB)A= BA$, so $AB$ and $BA$ are similar, which implies (but is stronger than) $AB$ and $BA$ have the same minimal polynomial and the same characteristic polynomial.
    The same goes if $B$ is invertible.



    In general, from the above observation, it is not too difficult to show that $AB$, and $BA$ have the same characteristic polynomial, the type of proof could depends on the field considered for the coefficient of your matrices though.
    If the matrices are in $mathcal{M}_n(mathbb C)$, you use the fact that $operatorname{GL}_n(mathbb C)$ is dense in $mathcal{M}_n(mathbb C)$ and the continuity of the function which maps a matrix to its characteristic polynomial. There are at least 5 other ways to proceed (especially for other field than $mathbb C$).



    In general $AB$ and $BA$ do not have the same minimal polynomial. I'll let you search a bit for a counter example.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 17 '15 at 23:50









    user26857

    39.3k124083




    39.3k124083










    answered Feb 22 '13 at 17:30









    Nathan PortlandNathan Portland

    1,015720




    1,015720








    • 9




      5 other ways? I'm quite curious what those ways are. I only know of the continuity argument and an argument involving determinant identities on block matrices. Would it be possible to provide a reference to some other methods?
      – EuYu
      Feb 22 '13 at 17:37










    • @EuYu, I have no reference, sorry.
      – Nathan Portland
      Feb 22 '13 at 22:11










    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same? @NathanPortland
      – cmi
      Nov 19 '18 at 4:55














    • 9




      5 other ways? I'm quite curious what those ways are. I only know of the continuity argument and an argument involving determinant identities on block matrices. Would it be possible to provide a reference to some other methods?
      – EuYu
      Feb 22 '13 at 17:37










    • @EuYu, I have no reference, sorry.
      – Nathan Portland
      Feb 22 '13 at 22:11










    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same? @NathanPortland
      – cmi
      Nov 19 '18 at 4:55








    9




    9




    5 other ways? I'm quite curious what those ways are. I only know of the continuity argument and an argument involving determinant identities on block matrices. Would it be possible to provide a reference to some other methods?
    – EuYu
    Feb 22 '13 at 17:37




    5 other ways? I'm quite curious what those ways are. I only know of the continuity argument and an argument involving determinant identities on block matrices. Would it be possible to provide a reference to some other methods?
    – EuYu
    Feb 22 '13 at 17:37












    @EuYu, I have no reference, sorry.
    – Nathan Portland
    Feb 22 '13 at 22:11




    @EuYu, I have no reference, sorry.
    – Nathan Portland
    Feb 22 '13 at 22:11












    Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same? @NathanPortland
    – cmi
    Nov 19 '18 at 4:55




    Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same? @NathanPortland
    – cmi
    Nov 19 '18 at 4:55











    24














    Before proving $AB$ and $BA$ have the same characteristic polynomials show that if $A_{mtimes n}$ and $B_{ntimes m} $ then characteristic polynomials of $AB$ and $BA$ satisfy following statement: $$x^n|xI_m-AB|=x^m|xI_n-BA|$$ therefore easily conclude if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.



    Define $$C = begin{bmatrix} xI_m & A \B & I_n end{bmatrix}, D = begin{bmatrix} I_m & 0 \-B & xI_n end{bmatrix}.$$ We have
    $$
    begin{align*}
    det CD &= x^n|xI_m-AB|,\
    det DC &= x^m|xI_n-BA|.
    end{align*}
    $$
    and we know $det CD=det DC$ if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.






    share|cite|improve this answer























    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@M.H
      – cmi
      Nov 19 '18 at 4:56
















    24














    Before proving $AB$ and $BA$ have the same characteristic polynomials show that if $A_{mtimes n}$ and $B_{ntimes m} $ then characteristic polynomials of $AB$ and $BA$ satisfy following statement: $$x^n|xI_m-AB|=x^m|xI_n-BA|$$ therefore easily conclude if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.



    Define $$C = begin{bmatrix} xI_m & A \B & I_n end{bmatrix}, D = begin{bmatrix} I_m & 0 \-B & xI_n end{bmatrix}.$$ We have
    $$
    begin{align*}
    det CD &= x^n|xI_m-AB|,\
    det DC &= x^m|xI_n-BA|.
    end{align*}
    $$
    and we know $det CD=det DC$ if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.






    share|cite|improve this answer























    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@M.H
      – cmi
      Nov 19 '18 at 4:56














    24












    24








    24






    Before proving $AB$ and $BA$ have the same characteristic polynomials show that if $A_{mtimes n}$ and $B_{ntimes m} $ then characteristic polynomials of $AB$ and $BA$ satisfy following statement: $$x^n|xI_m-AB|=x^m|xI_n-BA|$$ therefore easily conclude if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.



    Define $$C = begin{bmatrix} xI_m & A \B & I_n end{bmatrix}, D = begin{bmatrix} I_m & 0 \-B & xI_n end{bmatrix}.$$ We have
    $$
    begin{align*}
    det CD &= x^n|xI_m-AB|,\
    det DC &= x^m|xI_n-BA|.
    end{align*}
    $$
    and we know $det CD=det DC$ if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.






    share|cite|improve this answer














    Before proving $AB$ and $BA$ have the same characteristic polynomials show that if $A_{mtimes n}$ and $B_{ntimes m} $ then characteristic polynomials of $AB$ and $BA$ satisfy following statement: $$x^n|xI_m-AB|=x^m|xI_n-BA|$$ therefore easily conclude if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.



    Define $$C = begin{bmatrix} xI_m & A \B & I_n end{bmatrix}, D = begin{bmatrix} I_m & 0 \-B & xI_n end{bmatrix}.$$ We have
    $$
    begin{align*}
    det CD &= x^n|xI_m-AB|,\
    det DC &= x^m|xI_n-BA|.
    end{align*}
    $$
    and we know $det CD=det DC$ if $m=n$ then $AB$ and $BA$ have the same characteristic polynomials.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited May 7 '15 at 6:46









    user26857

    39.3k124083




    39.3k124083










    answered Feb 22 '13 at 17:36









    M.HM.H

    7,2371553




    7,2371553












    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@M.H
      – cmi
      Nov 19 '18 at 4:56


















    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@M.H
      – cmi
      Nov 19 '18 at 4:56
















    Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@M.H
    – cmi
    Nov 19 '18 at 4:56




    Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@M.H
    – cmi
    Nov 19 '18 at 4:56











    15














    Hint: Consider $A = begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}$ and $B = begin{bmatrix} 0 & 0 \ 0 & 1 end{bmatrix}$. What do you get in that case?






    share|cite|improve this answer

















    • 2




      This shows that $AB$ and $BA$ have different minimal polynomial. But characteristic polynomials are the same right?
      – Unknown x
      Nov 25 '17 at 1:39










    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@Jim
      – cmi
      Nov 19 '18 at 4:56
















    15














    Hint: Consider $A = begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}$ and $B = begin{bmatrix} 0 & 0 \ 0 & 1 end{bmatrix}$. What do you get in that case?






    share|cite|improve this answer

















    • 2




      This shows that $AB$ and $BA$ have different minimal polynomial. But characteristic polynomials are the same right?
      – Unknown x
      Nov 25 '17 at 1:39










    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@Jim
      – cmi
      Nov 19 '18 at 4:56














    15












    15








    15






    Hint: Consider $A = begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}$ and $B = begin{bmatrix} 0 & 0 \ 0 & 1 end{bmatrix}$. What do you get in that case?






    share|cite|improve this answer












    Hint: Consider $A = begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}$ and $B = begin{bmatrix} 0 & 0 \ 0 & 1 end{bmatrix}$. What do you get in that case?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Feb 22 '13 at 17:20









    JimJim

    24.3k23270




    24.3k23270








    • 2




      This shows that $AB$ and $BA$ have different minimal polynomial. But characteristic polynomials are the same right?
      – Unknown x
      Nov 25 '17 at 1:39










    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@Jim
      – cmi
      Nov 19 '18 at 4:56














    • 2




      This shows that $AB$ and $BA$ have different minimal polynomial. But characteristic polynomials are the same right?
      – Unknown x
      Nov 25 '17 at 1:39










    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@Jim
      – cmi
      Nov 19 '18 at 4:56








    2




    2




    This shows that $AB$ and $BA$ have different minimal polynomial. But characteristic polynomials are the same right?
    – Unknown x
    Nov 25 '17 at 1:39




    This shows that $AB$ and $BA$ have different minimal polynomial. But characteristic polynomials are the same right?
    – Unknown x
    Nov 25 '17 at 1:39












    Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@Jim
    – cmi
    Nov 19 '18 at 4:56




    Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@Jim
    – cmi
    Nov 19 '18 at 4:56











    4














    It's not true that their characteristic polynomials will be the same in the general case. The best result in this general vein is the following.



    Let $Ainmathbb{F}^{m times n}$ and let $Binmathbb{F}^{n times m}$, and $AB$, $BA$ with minimal polynomials (over $mathbb{F}$) $m_{AB}(x)$ and $m_{BA}(x)$ respectively. Then one of the following holds:



    $m_{AB}(x) = m_{BA}(x)$, or $m_{AB}(x) = x cdot m_{BA}(x)$, or $xcdot m_{AB}(x) = m_{BA}(x)$.



    It's easy, just use the fact that $(BA)^k=B(AB)^{k-1}A$.






    share|cite|improve this answer























    • in general means for m not equal to n??
      – user152715
      Sep 16 '14 at 21:16










    • The question says (and always has said) square matrices $A,B$. If you are answering a more general question, then you should announce this. Also, answering a more general question is only useful if this is no more difficult than the actual question, or if the more general solution sheds more light on the solution.
      – Marc van Leeuwen
      Dec 23 '15 at 21:10












    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@MarcvanLeeuwen
      – cmi
      Nov 19 '18 at 4:58
















    4














    It's not true that their characteristic polynomials will be the same in the general case. The best result in this general vein is the following.



    Let $Ainmathbb{F}^{m times n}$ and let $Binmathbb{F}^{n times m}$, and $AB$, $BA$ with minimal polynomials (over $mathbb{F}$) $m_{AB}(x)$ and $m_{BA}(x)$ respectively. Then one of the following holds:



    $m_{AB}(x) = m_{BA}(x)$, or $m_{AB}(x) = x cdot m_{BA}(x)$, or $xcdot m_{AB}(x) = m_{BA}(x)$.



    It's easy, just use the fact that $(BA)^k=B(AB)^{k-1}A$.






    share|cite|improve this answer























    • in general means for m not equal to n??
      – user152715
      Sep 16 '14 at 21:16










    • The question says (and always has said) square matrices $A,B$. If you are answering a more general question, then you should announce this. Also, answering a more general question is only useful if this is no more difficult than the actual question, or if the more general solution sheds more light on the solution.
      – Marc van Leeuwen
      Dec 23 '15 at 21:10












    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@MarcvanLeeuwen
      – cmi
      Nov 19 '18 at 4:58














    4












    4








    4






    It's not true that their characteristic polynomials will be the same in the general case. The best result in this general vein is the following.



    Let $Ainmathbb{F}^{m times n}$ and let $Binmathbb{F}^{n times m}$, and $AB$, $BA$ with minimal polynomials (over $mathbb{F}$) $m_{AB}(x)$ and $m_{BA}(x)$ respectively. Then one of the following holds:



    $m_{AB}(x) = m_{BA}(x)$, or $m_{AB}(x) = x cdot m_{BA}(x)$, or $xcdot m_{AB}(x) = m_{BA}(x)$.



    It's easy, just use the fact that $(BA)^k=B(AB)^{k-1}A$.






    share|cite|improve this answer














    It's not true that their characteristic polynomials will be the same in the general case. The best result in this general vein is the following.



    Let $Ainmathbb{F}^{m times n}$ and let $Binmathbb{F}^{n times m}$, and $AB$, $BA$ with minimal polynomials (over $mathbb{F}$) $m_{AB}(x)$ and $m_{BA}(x)$ respectively. Then one of the following holds:



    $m_{AB}(x) = m_{BA}(x)$, or $m_{AB}(x) = x cdot m_{BA}(x)$, or $xcdot m_{AB}(x) = m_{BA}(x)$.



    It's easy, just use the fact that $(BA)^k=B(AB)^{k-1}A$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited May 7 '15 at 6:42









    user26857

    39.3k124083




    39.3k124083










    answered Jul 16 '14 at 6:02









    user164626user164626

    411




    411












    • in general means for m not equal to n??
      – user152715
      Sep 16 '14 at 21:16










    • The question says (and always has said) square matrices $A,B$. If you are answering a more general question, then you should announce this. Also, answering a more general question is only useful if this is no more difficult than the actual question, or if the more general solution sheds more light on the solution.
      – Marc van Leeuwen
      Dec 23 '15 at 21:10












    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@MarcvanLeeuwen
      – cmi
      Nov 19 '18 at 4:58


















    • in general means for m not equal to n??
      – user152715
      Sep 16 '14 at 21:16










    • The question says (and always has said) square matrices $A,B$. If you are answering a more general question, then you should announce this. Also, answering a more general question is only useful if this is no more difficult than the actual question, or if the more general solution sheds more light on the solution.
      – Marc van Leeuwen
      Dec 23 '15 at 21:10












    • Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@MarcvanLeeuwen
      – cmi
      Nov 19 '18 at 4:58
















    in general means for m not equal to n??
    – user152715
    Sep 16 '14 at 21:16




    in general means for m not equal to n??
    – user152715
    Sep 16 '14 at 21:16












    The question says (and always has said) square matrices $A,B$. If you are answering a more general question, then you should announce this. Also, answering a more general question is only useful if this is no more difficult than the actual question, or if the more general solution sheds more light on the solution.
    – Marc van Leeuwen
    Dec 23 '15 at 21:10






    The question says (and always has said) square matrices $A,B$. If you are answering a more general question, then you should announce this. Also, answering a more general question is only useful if this is no more difficult than the actual question, or if the more general solution sheds more light on the solution.
    – Marc van Leeuwen
    Dec 23 '15 at 21:10














    Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@MarcvanLeeuwen
    – cmi
    Nov 19 '18 at 4:58




    Is the eigen value argument another one way? Can I say that the eigen values of $AB$ and $BA$ are same so the characteristics equations are same?@MarcvanLeeuwen
    – cmi
    Nov 19 '18 at 4:58











    0














    For squre matrix, the charateristic polynomials are same, but for $A$ is matrix of size $m times n$ and B be matrix of size $n times m$. Then $x^{m}C_{BA}(x)=x^{n}C_{AB}(x)$ this implies that the nonzero eigenvalue of $AB$, counted with multiplicites, are same as nonzero eigenvalue of $BA$.



    That is if A is of size 7×4 and B is of size 4×7 and assume that the 4×4 matrix BA has nonzero eigenvalue 1,1,3 so fourth eigenvalue of BA is 0. Then the 7×7 matrix AB will also have nonzero eigenvalue 1,1,3 and remaining four eigenvalue of AB are zero.






    share|cite|improve this answer




























      0














      For squre matrix, the charateristic polynomials are same, but for $A$ is matrix of size $m times n$ and B be matrix of size $n times m$. Then $x^{m}C_{BA}(x)=x^{n}C_{AB}(x)$ this implies that the nonzero eigenvalue of $AB$, counted with multiplicites, are same as nonzero eigenvalue of $BA$.



      That is if A is of size 7×4 and B is of size 4×7 and assume that the 4×4 matrix BA has nonzero eigenvalue 1,1,3 so fourth eigenvalue of BA is 0. Then the 7×7 matrix AB will also have nonzero eigenvalue 1,1,3 and remaining four eigenvalue of AB are zero.






      share|cite|improve this answer


























        0












        0








        0






        For squre matrix, the charateristic polynomials are same, but for $A$ is matrix of size $m times n$ and B be matrix of size $n times m$. Then $x^{m}C_{BA}(x)=x^{n}C_{AB}(x)$ this implies that the nonzero eigenvalue of $AB$, counted with multiplicites, are same as nonzero eigenvalue of $BA$.



        That is if A is of size 7×4 and B is of size 4×7 and assume that the 4×4 matrix BA has nonzero eigenvalue 1,1,3 so fourth eigenvalue of BA is 0. Then the 7×7 matrix AB will also have nonzero eigenvalue 1,1,3 and remaining four eigenvalue of AB are zero.






        share|cite|improve this answer














        For squre matrix, the charateristic polynomials are same, but for $A$ is matrix of size $m times n$ and B be matrix of size $n times m$. Then $x^{m}C_{BA}(x)=x^{n}C_{AB}(x)$ this implies that the nonzero eigenvalue of $AB$, counted with multiplicites, are same as nonzero eigenvalue of $BA$.



        That is if A is of size 7×4 and B is of size 4×7 and assume that the 4×4 matrix BA has nonzero eigenvalue 1,1,3 so fourth eigenvalue of BA is 0. Then the 7×7 matrix AB will also have nonzero eigenvalue 1,1,3 and remaining four eigenvalue of AB are zero.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Sep 1 '18 at 16:29

























        answered Sep 1 '18 at 10:36









        user499117user499117

        409




        409






























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