What are the eigenspaces and the minimal polynomial of the “transposing about the anti-diagonal”...












1












$begingroup$


Given a field $K$ of characteristics zero, let the endomorphism $f:M_2(K)to M_2(K)$ be defined by $fleft(begin{bmatrix}a&b\c&dend{bmatrix}right)=begin{bmatrix}d&b\c&aend{bmatrix}$ for every $begin{bmatrix}a&b\c&dend{bmatrix}in M_2(K).$



How can I find the eigenspaces and the minimal polynomial of $f$?



And why is $operatorname{char} K=0$ important?




What I noticed is that $f$ is invertible (hence its eigenvalues are non-zero) and $f^{-1}=f$. But I couldn't go further than this






linear-algebra matrices eigenvalues-eigenvectors linear-transformations transpose







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Given a field $K$ of characteristics zero, let the endomorphism $f:M_2(K)to M_2(K)$ be defined by $fleft(begin{bmatrix}a&b\c&dend{bmatrix}right)=begin{bmatrix}d&b\c&aend{bmatrix}$ for every $begin{bmatrix}a&b\c&dend{bmatrix}in M_2(K).$



    How can I find the eigenspaces and the minimal polynomial of $f$?



    And why is $operatorname{char} K=0$ important?




    What I noticed is that $f$ is invertible (hence its eigenvalues are non-zero) and $f^{-1}=f$. But I couldn't go further than this






    linear-algebra matrices eigenvalues-eigenvectors linear-transformations transpose







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Given a field $K$ of characteristics zero, let the endomorphism $f:M_2(K)to M_2(K)$ be defined by $fleft(begin{bmatrix}a&b\c&dend{bmatrix}right)=begin{bmatrix}d&b\c&aend{bmatrix}$ for every $begin{bmatrix}a&b\c&dend{bmatrix}in M_2(K).$



      How can I find the eigenspaces and the minimal polynomial of $f$?



      And why is $operatorname{char} K=0$ important?




      What I noticed is that $f$ is invertible (hence its eigenvalues are non-zero) and $f^{-1}=f$. But I couldn't go further than this






      linear-algebra matrices eigenvalues-eigenvectors linear-transformations transpose







      share|cite|improve this question









      $endgroup$




      Given a field $K$ of characteristics zero, let the endomorphism $f:M_2(K)to M_2(K)$ be defined by $fleft(begin{bmatrix}a&b\c&dend{bmatrix}right)=begin{bmatrix}d&b\c&aend{bmatrix}$ for every $begin{bmatrix}a&b\c&dend{bmatrix}in M_2(K).$



      How can I find the eigenspaces and the minimal polynomial of $f$?



      And why is $operatorname{char} K=0$ important?




      What I noticed is that $f$ is invertible (hence its eigenvalues are non-zero) and $f^{-1}=f$. But I couldn't go further than this






      linear-algebra matrices eigenvalues-eigenvectors linear-transformations transpose




      linear-algebra matrices eigenvalues-eigenvectors linear-transformations transpose






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 23 at 12:16









      LearnerLearner

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      17510






















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          $begingroup$

          Since as you noted $f^2=1$, and $fneqpm1$, it follows that $X^2-1$ is indeed the minimum polynomial of $f$. The eigenvalues are $+1$ an $-1$. We have $f(A)=1cdot A$ for $Ain$Mat$_2(K)$ if and only if $$begin{pmatrix}d & b \ c & aend{pmatrix}=begin{pmatrix}a & b \ c & dend{pmatrix},$$
          if and only if $a=d$, so the eigenspace for $lambda=+1$ is
          $$E_1=left{begin{pmatrix}a & b \ c & aend{pmatrix} | a,b,cin Kright}.$$



          Similarly, the eigenspace for $lambda=-1$ is found by studying $f(A)=-A$, which gives $d=-a$, $b=-b$, $c=-c$, hence $d=-a$, $b=0=c$ when char$(K)neq2$:
          $$E_{-1}=left{begin{pmatrix}a & 0 \ 0 & -aend{pmatrix} | ain Kright}.$$



          By the way, $f$ is the symplectic involution, and satisfies $f(AB)=f(B)f(A)$ in addition. Moreover $E_1$ is the set of symmetric elements with respect to the involution and $E_{-1}$ is the set of skew-symmetric elements. You can define a symplectic involution on $ntimes n$ matrices whenever $n$ is even.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Very instructive and exhaustive answer, thank you!
            $endgroup$
            – Learner
            Jan 23 at 15:03





















          1












          $begingroup$

          Let$$E_1=begin{bmatrix}1&0\0&0end{bmatrix}, E_2=begin{bmatrix}0&1\0&0end{bmatrix}, E_3=begin{bmatrix}0&0\1&0end{bmatrix},text{ and }E_4=begin{bmatrix}0&0\0&1end{bmatrix}.$$Then ${E_1,E_2,E_3,E_4}$ is a basis of $M_2(mathbb{R})$ and the matrix of $f$ with respect to this basis is$$begin{bmatrix}0&0&0&1\0&1&0&0\0&0&1&0\1&0&0&0end{bmatrix}.$$Can you take it from here?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            The minimal polynomial of your matrix is $(lambda+1)(lambda-1)$ - then it's also the minimal polynomial of $f$, because it's of degree $2$, hence equal to $chi_f$. And, the eigenspaces of your matrix are the ones of $f$. All correct? Also, one thing: $K$ is isomorphic to a subset of $Bbb C$, provided that $|K|$ isn't larger than the continuum; still, I can see cardinality here doesn't make a difference, but is it enough (in an exam) to say "without loss of generality we may take $K=Bbb R$"? Thanks for your help, this cleared my thoughts!
            $endgroup$
            – Learner
            Jan 23 at 15:02








          • 1




            $begingroup$
            Yes, it's all correct.
            $endgroup$
            – José Carlos Santos
            Jan 23 at 15:03











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          2 Answers
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          $begingroup$

          Since as you noted $f^2=1$, and $fneqpm1$, it follows that $X^2-1$ is indeed the minimum polynomial of $f$. The eigenvalues are $+1$ an $-1$. We have $f(A)=1cdot A$ for $Ain$Mat$_2(K)$ if and only if $$begin{pmatrix}d & b \ c & aend{pmatrix}=begin{pmatrix}a & b \ c & dend{pmatrix},$$
          if and only if $a=d$, so the eigenspace for $lambda=+1$ is
          $$E_1=left{begin{pmatrix}a & b \ c & aend{pmatrix} | a,b,cin Kright}.$$



          Similarly, the eigenspace for $lambda=-1$ is found by studying $f(A)=-A$, which gives $d=-a$, $b=-b$, $c=-c$, hence $d=-a$, $b=0=c$ when char$(K)neq2$:
          $$E_{-1}=left{begin{pmatrix}a & 0 \ 0 & -aend{pmatrix} | ain Kright}.$$



          By the way, $f$ is the symplectic involution, and satisfies $f(AB)=f(B)f(A)$ in addition. Moreover $E_1$ is the set of symmetric elements with respect to the involution and $E_{-1}$ is the set of skew-symmetric elements. You can define a symplectic involution on $ntimes n$ matrices whenever $n$ is even.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Very instructive and exhaustive answer, thank you!
            $endgroup$
            – Learner
            Jan 23 at 15:03


















          1












          $begingroup$

          Since as you noted $f^2=1$, and $fneqpm1$, it follows that $X^2-1$ is indeed the minimum polynomial of $f$. The eigenvalues are $+1$ an $-1$. We have $f(A)=1cdot A$ for $Ain$Mat$_2(K)$ if and only if $$begin{pmatrix}d & b \ c & aend{pmatrix}=begin{pmatrix}a & b \ c & dend{pmatrix},$$
          if and only if $a=d$, so the eigenspace for $lambda=+1$ is
          $$E_1=left{begin{pmatrix}a & b \ c & aend{pmatrix} | a,b,cin Kright}.$$



          Similarly, the eigenspace for $lambda=-1$ is found by studying $f(A)=-A$, which gives $d=-a$, $b=-b$, $c=-c$, hence $d=-a$, $b=0=c$ when char$(K)neq2$:
          $$E_{-1}=left{begin{pmatrix}a & 0 \ 0 & -aend{pmatrix} | ain Kright}.$$



          By the way, $f$ is the symplectic involution, and satisfies $f(AB)=f(B)f(A)$ in addition. Moreover $E_1$ is the set of symmetric elements with respect to the involution and $E_{-1}$ is the set of skew-symmetric elements. You can define a symplectic involution on $ntimes n$ matrices whenever $n$ is even.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Very instructive and exhaustive answer, thank you!
            $endgroup$
            – Learner
            Jan 23 at 15:03
















          1












          1








          1





          $begingroup$

          Since as you noted $f^2=1$, and $fneqpm1$, it follows that $X^2-1$ is indeed the minimum polynomial of $f$. The eigenvalues are $+1$ an $-1$. We have $f(A)=1cdot A$ for $Ain$Mat$_2(K)$ if and only if $$begin{pmatrix}d & b \ c & aend{pmatrix}=begin{pmatrix}a & b \ c & dend{pmatrix},$$
          if and only if $a=d$, so the eigenspace for $lambda=+1$ is
          $$E_1=left{begin{pmatrix}a & b \ c & aend{pmatrix} | a,b,cin Kright}.$$



          Similarly, the eigenspace for $lambda=-1$ is found by studying $f(A)=-A$, which gives $d=-a$, $b=-b$, $c=-c$, hence $d=-a$, $b=0=c$ when char$(K)neq2$:
          $$E_{-1}=left{begin{pmatrix}a & 0 \ 0 & -aend{pmatrix} | ain Kright}.$$



          By the way, $f$ is the symplectic involution, and satisfies $f(AB)=f(B)f(A)$ in addition. Moreover $E_1$ is the set of symmetric elements with respect to the involution and $E_{-1}$ is the set of skew-symmetric elements. You can define a symplectic involution on $ntimes n$ matrices whenever $n$ is even.






          share|cite|improve this answer









          $endgroup$



          Since as you noted $f^2=1$, and $fneqpm1$, it follows that $X^2-1$ is indeed the minimum polynomial of $f$. The eigenvalues are $+1$ an $-1$. We have $f(A)=1cdot A$ for $Ain$Mat$_2(K)$ if and only if $$begin{pmatrix}d & b \ c & aend{pmatrix}=begin{pmatrix}a & b \ c & dend{pmatrix},$$
          if and only if $a=d$, so the eigenspace for $lambda=+1$ is
          $$E_1=left{begin{pmatrix}a & b \ c & aend{pmatrix} | a,b,cin Kright}.$$



          Similarly, the eigenspace for $lambda=-1$ is found by studying $f(A)=-A$, which gives $d=-a$, $b=-b$, $c=-c$, hence $d=-a$, $b=0=c$ when char$(K)neq2$:
          $$E_{-1}=left{begin{pmatrix}a & 0 \ 0 & -aend{pmatrix} | ain Kright}.$$



          By the way, $f$ is the symplectic involution, and satisfies $f(AB)=f(B)f(A)$ in addition. Moreover $E_1$ is the set of symmetric elements with respect to the involution and $E_{-1}$ is the set of skew-symmetric elements. You can define a symplectic involution on $ntimes n$ matrices whenever $n$ is even.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 12:56









          Jose BroxJose Brox

          3,15711128




          3,15711128








          • 1




            $begingroup$
            Very instructive and exhaustive answer, thank you!
            $endgroup$
            – Learner
            Jan 23 at 15:03
















          • 1




            $begingroup$
            Very instructive and exhaustive answer, thank you!
            $endgroup$
            – Learner
            Jan 23 at 15:03










          1




          1




          $begingroup$
          Very instructive and exhaustive answer, thank you!
          $endgroup$
          – Learner
          Jan 23 at 15:03






          $begingroup$
          Very instructive and exhaustive answer, thank you!
          $endgroup$
          – Learner
          Jan 23 at 15:03













          1












          $begingroup$

          Let$$E_1=begin{bmatrix}1&0\0&0end{bmatrix}, E_2=begin{bmatrix}0&1\0&0end{bmatrix}, E_3=begin{bmatrix}0&0\1&0end{bmatrix},text{ and }E_4=begin{bmatrix}0&0\0&1end{bmatrix}.$$Then ${E_1,E_2,E_3,E_4}$ is a basis of $M_2(mathbb{R})$ and the matrix of $f$ with respect to this basis is$$begin{bmatrix}0&0&0&1\0&1&0&0\0&0&1&0\1&0&0&0end{bmatrix}.$$Can you take it from here?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            The minimal polynomial of your matrix is $(lambda+1)(lambda-1)$ - then it's also the minimal polynomial of $f$, because it's of degree $2$, hence equal to $chi_f$. And, the eigenspaces of your matrix are the ones of $f$. All correct? Also, one thing: $K$ is isomorphic to a subset of $Bbb C$, provided that $|K|$ isn't larger than the continuum; still, I can see cardinality here doesn't make a difference, but is it enough (in an exam) to say "without loss of generality we may take $K=Bbb R$"? Thanks for your help, this cleared my thoughts!
            $endgroup$
            – Learner
            Jan 23 at 15:02








          • 1




            $begingroup$
            Yes, it's all correct.
            $endgroup$
            – José Carlos Santos
            Jan 23 at 15:03
















          1












          $begingroup$

          Let$$E_1=begin{bmatrix}1&0\0&0end{bmatrix}, E_2=begin{bmatrix}0&1\0&0end{bmatrix}, E_3=begin{bmatrix}0&0\1&0end{bmatrix},text{ and }E_4=begin{bmatrix}0&0\0&1end{bmatrix}.$$Then ${E_1,E_2,E_3,E_4}$ is a basis of $M_2(mathbb{R})$ and the matrix of $f$ with respect to this basis is$$begin{bmatrix}0&0&0&1\0&1&0&0\0&0&1&0\1&0&0&0end{bmatrix}.$$Can you take it from here?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            The minimal polynomial of your matrix is $(lambda+1)(lambda-1)$ - then it's also the minimal polynomial of $f$, because it's of degree $2$, hence equal to $chi_f$. And, the eigenspaces of your matrix are the ones of $f$. All correct? Also, one thing: $K$ is isomorphic to a subset of $Bbb C$, provided that $|K|$ isn't larger than the continuum; still, I can see cardinality here doesn't make a difference, but is it enough (in an exam) to say "without loss of generality we may take $K=Bbb R$"? Thanks for your help, this cleared my thoughts!
            $endgroup$
            – Learner
            Jan 23 at 15:02








          • 1




            $begingroup$
            Yes, it's all correct.
            $endgroup$
            – José Carlos Santos
            Jan 23 at 15:03














          1












          1








          1





          $begingroup$

          Let$$E_1=begin{bmatrix}1&0\0&0end{bmatrix}, E_2=begin{bmatrix}0&1\0&0end{bmatrix}, E_3=begin{bmatrix}0&0\1&0end{bmatrix},text{ and }E_4=begin{bmatrix}0&0\0&1end{bmatrix}.$$Then ${E_1,E_2,E_3,E_4}$ is a basis of $M_2(mathbb{R})$ and the matrix of $f$ with respect to this basis is$$begin{bmatrix}0&0&0&1\0&1&0&0\0&0&1&0\1&0&0&0end{bmatrix}.$$Can you take it from here?






          share|cite|improve this answer









          $endgroup$



          Let$$E_1=begin{bmatrix}1&0\0&0end{bmatrix}, E_2=begin{bmatrix}0&1\0&0end{bmatrix}, E_3=begin{bmatrix}0&0\1&0end{bmatrix},text{ and }E_4=begin{bmatrix}0&0\0&1end{bmatrix}.$$Then ${E_1,E_2,E_3,E_4}$ is a basis of $M_2(mathbb{R})$ and the matrix of $f$ with respect to this basis is$$begin{bmatrix}0&0&0&1\0&1&0&0\0&0&1&0\1&0&0&0end{bmatrix}.$$Can you take it from here?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 12:20









          José Carlos SantosJosé Carlos Santos

          165k22132235




          165k22132235












          • $begingroup$
            The minimal polynomial of your matrix is $(lambda+1)(lambda-1)$ - then it's also the minimal polynomial of $f$, because it's of degree $2$, hence equal to $chi_f$. And, the eigenspaces of your matrix are the ones of $f$. All correct? Also, one thing: $K$ is isomorphic to a subset of $Bbb C$, provided that $|K|$ isn't larger than the continuum; still, I can see cardinality here doesn't make a difference, but is it enough (in an exam) to say "without loss of generality we may take $K=Bbb R$"? Thanks for your help, this cleared my thoughts!
            $endgroup$
            – Learner
            Jan 23 at 15:02








          • 1




            $begingroup$
            Yes, it's all correct.
            $endgroup$
            – José Carlos Santos
            Jan 23 at 15:03


















          • $begingroup$
            The minimal polynomial of your matrix is $(lambda+1)(lambda-1)$ - then it's also the minimal polynomial of $f$, because it's of degree $2$, hence equal to $chi_f$. And, the eigenspaces of your matrix are the ones of $f$. All correct? Also, one thing: $K$ is isomorphic to a subset of $Bbb C$, provided that $|K|$ isn't larger than the continuum; still, I can see cardinality here doesn't make a difference, but is it enough (in an exam) to say "without loss of generality we may take $K=Bbb R$"? Thanks for your help, this cleared my thoughts!
            $endgroup$
            – Learner
            Jan 23 at 15:02








          • 1




            $begingroup$
            Yes, it's all correct.
            $endgroup$
            – José Carlos Santos
            Jan 23 at 15:03
















          $begingroup$
          The minimal polynomial of your matrix is $(lambda+1)(lambda-1)$ - then it's also the minimal polynomial of $f$, because it's of degree $2$, hence equal to $chi_f$. And, the eigenspaces of your matrix are the ones of $f$. All correct? Also, one thing: $K$ is isomorphic to a subset of $Bbb C$, provided that $|K|$ isn't larger than the continuum; still, I can see cardinality here doesn't make a difference, but is it enough (in an exam) to say "without loss of generality we may take $K=Bbb R$"? Thanks for your help, this cleared my thoughts!
          $endgroup$
          – Learner
          Jan 23 at 15:02






          $begingroup$
          The minimal polynomial of your matrix is $(lambda+1)(lambda-1)$ - then it's also the minimal polynomial of $f$, because it's of degree $2$, hence equal to $chi_f$. And, the eigenspaces of your matrix are the ones of $f$. All correct? Also, one thing: $K$ is isomorphic to a subset of $Bbb C$, provided that $|K|$ isn't larger than the continuum; still, I can see cardinality here doesn't make a difference, but is it enough (in an exam) to say "without loss of generality we may take $K=Bbb R$"? Thanks for your help, this cleared my thoughts!
          $endgroup$
          – Learner
          Jan 23 at 15:02






          1




          1




          $begingroup$
          Yes, it's all correct.
          $endgroup$
          – José Carlos Santos
          Jan 23 at 15:03




          $begingroup$
          Yes, it's all correct.
          $endgroup$
          – José Carlos Santos
          Jan 23 at 15:03


















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